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PASSED | deb4ed1ed77be1b1c17bc8765614f93a | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class ModerateModularMode {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long X = io.nextLong();
final long Y = io.nextLong();
if (X < Y) {
io.println(Y - (Y % X) / 2);
} else if (X > Y) {
io.println(Y * (X + 1));
} else {
io.println(X);
}
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
// TODO: read this byte-by-byte like the other read functions.
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 884bed54d8161a23a58f531e8a07ba73 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class ModerateModularMode {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long X = io.nextLong();
final long Y = io.nextLong();
if (X < Y) {
io.println(Y - ((Y % X) >> 1));
} else if (X > Y) {
io.println(Y * (X + 1));
} else {
io.println(X);
}
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
// TODO: read this byte-by-byte like the other read functions.
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 4644f66fad063ab2dcfb5bcf8e8f2697 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class ModerateModularMode {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long X = io.nextLong();
final long Y = io.nextLong();
if (X < Y) {
io.println(Y - ((Y % X) >> 1));
} else if (X > Y) {
io.println(X * Y + Y);
} else {
io.println(X);
}
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
// TODO: read this byte-by-byte like the other read functions.
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 1119c57dac20ee0ddfc261f676ad1be8 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class ModerateModularMode {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long X = io.nextLong();
final long Y = io.nextLong();
if (X < Y) {
io.println(Y - (Y % X) / 2);
} else if (X > Y) {
io.println(X * Y + Y);
} else {
io.println(X);
}
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
// TODO: read this byte-by-byte like the other read functions.
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 6ef706eb572655ca2a26ddb440c90415 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.util.concurrent.LinkedBlockingDeque;
import javax.print.attribute.IntegerSyntax;
import javax.sql.rowset.spi.SyncResolver;
import java.io.*;
import java.nio.channels.NonReadableChannelException;
import java.text.DateFormatSymbols;
import static java.lang.System.*;
public class CpTemp{
static FastScanner fs = null;
static ArrayList<Integer> al[];
static HashMap<Long,Long> hm;
static long fy;
static long fx;
static long ix;
static long iy;
static long prex[];
static long prey[];
public static void main(String[] args) {
fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t= fs.nextInt();
outer:
while (t-- > 0) {
int n = fs.nextInt();
int m = fs.nextInt();
if (n>m) out.println(n+m);
else
{
if (n == m){
out.println(n);
}else{
n = m - (m%n);
out.println((m+n)/2);
}
}
}
out.close();
}
public static void solve(long n, ArrayList<Long>al){
for(long d = 2; d*d<=(n);d++){
if(n%d==0){
al.add(d);
if(d!=(n/d)){
al.add(-1l);
return;
}
}
}
}
// 0 6 1 3 2 m = 7
public static boolean check(long t,int a[],int m) {
int prev = -1;
for(int i=0;i<a.length;i++){
if(i==0){
if(a[i]+t >= m){
prev = 0;
}else{
prev = a[i];
}
}else{
if(a[i]==prev){
continue;
}else if(a[i]<prev){
if(prev - a[i] <= t){
continue;
}else{
return false;
}
}else{
if(a[i]+t >= m && (a[i]+t)%m < prev){
prev = a[i];
}else if(a[i]+t >=m && (a[i]+t)%m >= prev ){
continue;
}else{
prev = a[i];
}
}
}
}
return true;
}
public static int CeilIndex(int A[], int l, int r, int key)
{
while (r - l > 1) {
int m = l + (r - l) / 2;
if (A[m] >= key)
r = m;
else
l = m;
}
return r;
}
public static int LongestIncreasingSubsequenceLength(int A[], int size)
{
// Add boundary case, when array size is one
int[] tailTable = new int[size];
int len; // always points empty slot
tailTable[0] = A[0];
len = 1;
for (int i = 1; i < size; i++) {
if (A[i] < tailTable[0])
// new smallest value
tailTable[0] = A[i];
else if (A[i] > tailTable[len - 1])
// A[i] wants to extend largest subsequence
tailTable[len++] = A[i];
else
// A[i] wants to be current end candidate of an existing
// subsequence. It will replace ceil value in tailTable
tailTable[CeilIndex(tailTable, -1, len - 1, A[i])] = A[i];
}
return len;
}
public static int ask(int st,int end){
if(st>=end) {
return -1;
}
out.println("? "+(st+1)+" "+(end+1));
int g = fs.nextInt();
out.flush();
return g-1;
}
static long bit[]; // '1' index based array
public static void update(long bit[],int i,int x){
for(;i<bit.length;i+=(i&(-i))){
bit[i] += x;
}
}
public static long sum(int i){
long sum=0;
for(;i>0 ;i -= (i&(-i))){
sum += bit[i];
}
return sum;
}
static class Pair implements Comparable<Pair> {
int x;
int y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
public int compareTo(Pair o){
return this.y-o.y;
}
}
static class Pair1 implements Comparable<Pair1> {
int x;
int y;
int z;
Pair1(int x, int y,int z) {
this.x = x;
this.y = y;
this.z = z;
}
public int compareTo(Pair1 o){
return this.z-o.z;
}
}
static int power(int x, int y, int p) {
if (y == 0)
return 1;
if (x == 0)
return 0;
int res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static long power(long x, long y) {
if (y == 0)
return 1;
if (x == 0)
return 0;
long res = 1;
while (y > 0) {
if (y % 2 == 1)
res = (res * x);
y = y >> 1;
x = (x * x);
}
return res;
}
static void sort(long[] a) {
ArrayList<Long> l = new ArrayList<>();
for (long i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readlongArray(int n){
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
static boolean prime[];
static void sieveOfEratosthenes(int n) {
prime = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
public static int log2(int x) {
return (int) (Math.log(x) / Math.log(2));
}
public static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
static long nCk(int n, int k) {
long res = 1;
for (int i = n - k + 1; i <= n; ++i)
res *= i;
for (int i = 2; i <= k; ++i)
res /= i;
return res;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | c1216e8253bc2480db6d333b3860784e | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static int i, j, k, n, m, t, y, x, sum = 0;
static long mod = 1000000007;
static FastScanner fs = new FastScanner();
static PrintWriter out = new PrintWriter(System.out);
static String str;
static long ans;
public static void main(String[] args) {
t =fs.nextInt();
while(t-- >0){
long x = fs.nextLong();
long y = fs.nextLong();
if(x==y)
out.println(x);
else if(x>y){
out.println(x+y);
}
else{
long c = y%x;
out.println(y-(c/2));
}
}
out.close();
}
/**
* Returns the gretaest index of closest element less than x in arr
* returns -1 if all elemets are greater
*
*/
public static int lowerBound(int x, List<Integer> arr, int l, int r){
if(arr.get(l)>=x)
return -1;
if(arr.get(r)<x)
return r;
int mid = (l+r)/2;
if(arr.get(mid)<x && arr.get(mid+1)>x)
return mid;
if(arr.get(mid)>=x)
return lowerBound(x,arr,l,mid-1);
return lowerBound(x,arr,mid+1,r);
}
/**
* Returns the lowest index of closest element greater than or equal to x in arr
* returns -1 if all elements are lesser
*
*/
public static int upperBound(int x, List<Integer> arr, int l, int r){
if(arr.get(r) <=x)
return -1;
if(arr.get(l)>x)
return l;
int mid = (l+r)/2;
if(arr.get(mid)>x && arr.get(mid-1)<=x)
return mid;
if(arr.get(mid)<=x)
return upperBound(x,arr,mid+1,r);
return upperBound(x,arr,l,mid-1);
}
/**
* Returns the index of element if present else -1.
*
*/
public static int binSearch(int x, List<Integer> arr){
int y = Collections.binarySearch(arr, x);
if(y<0)
return -1;
return y;
}
/**
* Gets prime factorisation of a number in list of pairs.
* x is the factor and y is the occurrence.
*
*/
public static List<Pair> primeFactorization(int num){
List<Pair> ans = new ArrayList<>();
for(int i=2;i*i<=num;i++){
if(num % i ==0){
int count = 0;
while(num%i==0){
count++;
num=num/i;
}
ans.add(new Pair(i,count));
}
}
if(num!=1)
ans.add(new Pair(num, 1));
return ans;
}
/*static long nck(int n , int k){
long a = fact[n];
long b = modInv(fact[k]);
b*= modInv(fact[n-k]);
b%=mod;
return (a*b)%mod;
}
static void populateFact(){
fact[0]=1;
fact[1] = 1;
for(i=2;i<300005;i++){
fact[i]=i*fact[i-1];
fact[i]%=mod;
}
}
*/
static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long exp(long base, long pow) {
if (pow == 0) return 1;
long half = exp(base, pow / 2);
if (pow % 2 == 0) return mul(half, half);
return mul(half, mul(half, base));
}
static long mul(long a, long b) {
return ((a % mod) * (b % mod)) % mod;
}
static long add(long a, long b) {
return ((a % mod) + (b % mod)) % mod;
}
static long modInv(long x) {
return exp(x, mod - 2);
}
static void ruffleSort(int[] a) {
//ruffle
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n), temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
//then sort
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
//ruffle
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n);
long temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
Arrays.sort(a);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
static class Pair implements Comparable<Pair> {
public int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
if (x == o.x)
return Integer.compare(y, o.y);
return Integer.compare(x, o.x);
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 9721d570fbfae785ad09a5ec5bdc05c7 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class CF1603_D1_B {
public static void main(String[] args) {
FastScanner scanner = new FastScanner();
int t = scanner.nextInt();
for (int i = 0; i < t; i++) {
solve(scanner);
}
}
private static void solve(FastScanner scanner) {
int x = scanner.nextInt();
int y = scanner.nextInt();
if (x < y) {
long div = y / x;
long n = div * x + (y % x) / 2;
System.out.println(n);
// for (int i = 0; i < x; i++) {
// if (i == y % x) {
// System.out.println(x + i);
// }
// }
} else if (x > y) {
System.out.println(x + y);
} else {
System.out.println(x);
}
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] nextArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
Integer[] nextArray(int n, boolean object) {
Integer[] arr = new Integer[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | e78ac9326c0d270abdd58d0988717b7b | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
private static final MyWriter writer = new MyWriter();
private static final MyReader scan = new MyReader();
public static void main(String[] args) throws Exception {
Main main = new Main();
int q = scan.nextInt(); while (q-- > 0)
main.solve(); writer.close();
}
void solve() throws Exception {
int x = scan.nextInt();
int y = scan.nextInt();
if (x < y) {
writer.println(y - (y % x / 2));
} else if(x > y) {
writer.println(x + y);
} else {
writer.println(x);
}
}
}
class MyWriter implements Closeable {
private BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
public void print(Object object) throws IOException { writer.write(object.toString()); }
public void println(Object object) throws IOException { writer.write(object.toString());writer.write(System.lineSeparator()); }
public void close() throws IOException { writer.close(); }
}
class MyReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() { try { return reader.readLine(); } catch (IOException ex) { return null;}}
public boolean hasNext() { while (!tokenizer.hasMoreTokens()) { String nextLine = innerNextLine();if (nextLine == null) return false;tokenizer = new StringTokenizer(nextLine); }return true; }
public String nextLine() { tokenizer = new StringTokenizer("");return innerNextLine(); }
public String next() { hasNext();return tokenizer.nextToken(); }
public int nextInt() { return Integer.parseInt(next()); }
public long nextLong() { return Long.parseLong(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 3d8196e03a545b0deb9738c57d45cd5c | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main{
public static void main(String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine());
while(T-- > 0) {
StringTokenizer st = new StringTokenizer(br.readLine());
int x = Integer.parseInt(st.nextToken());
int y = Integer.parseInt(st.nextToken());
if(x > y) System.out.println(x+y);
else if(x==y) System.out.println(x);
else {
int t;
if(y % x == 0) {
t = (y/x-1)*x;
}
else t=y/x*x;
System.out.println((y+t) / 2);
}
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 41dfa5be05946b7713e8eef44a3eda97 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class modmode{
public static void main(String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine());
while(T-- > 0){
StringTokenizer st = new StringTokenizer(br.readLine());
int x = Integer.parseInt(st.nextToken());
int y = Integer.parseInt(st.nextToken());
if(x == y){
System.out.println(x);
continue;
}
else if(x > y){
long ans = x + y;
System.out.println(ans);
continue;
}
else{
x = y / x * x;
long med = (x + y) / 2;
System.out.println(med);
}
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | f3f10ba1974300d572d4fa7276a8d41f | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /*
Challenge 1: Newbie to CM in 1year (Dec 2021 - Nov 2022) 🔥 5* Codechef
Challenge 2: CM to IM in 1 year (Dec 2022 - Nov 2023) 🔥🔥 6* Codechef
Challenge 3: IM to GM in 1 year (Dec 2023 - Nov 2024) 🔥🔥🔥 7* Codechef
Goal: Become better in CP!
Key: Consistency!
*/
import java.util.*;
import java.io.*;
import java.math.*;
public class Coder {
static StringBuffer str=new StringBuffer();
static long x,y;
static void solve(){
long n=0;
if(x==y) n=x;
else if(x<y){
if(y%x==0) n=x;
else n=y-y%x/2;
}
else n=x+y;
str.append(n).append("\n");
}
public static void main(String[] args) throws java.lang.Exception {
BufferedReader bf;
PrintWriter pw;
boolean lenv=false;
if(lenv){
bf = new BufferedReader(
new FileReader("input.txt"));
pw=new PrintWriter(new
BufferedWriter(new FileWriter("output.txt")));
}else{
bf = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new OutputStreamWriter(System.out));
}
int q = Integer.parseInt(bf.readLine().trim());
while (q-- > 0) {
String s[]=bf.readLine().trim().split("\\s+");
x=Long.parseLong(s[0]);
y=Long.parseLong(s[1]);
solve();
}
pw.println(str);
pw.flush();
// System.outin.print(str);
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 0417c3fb5e71ac1d658facf1829a2199 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter writer = new PrintWriter(System.out);
int t = Integer.parseInt(reader.readLine());
while (t-- > 0) {
String[] line = reader.readLine().split(" ");
int x = Integer.parseInt(line[0]);
int y = Integer.parseInt(line[1]);
if (x == y) {
writer.println(x);
} else if (x > y) {
writer.println(x + y);
} else {
writer.println(y - (y % x) / 2);
}
}
reader.close();
writer.close();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | ee1dab84bf66571d73237bee125c3025 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.io.*;
import java.util.*;
public class solution {
public static void main(String[] args) {
FScanner sc = new FScanner();
//Arrays.fill(prime, true);
//sieve();
int t=sc.nextInt();
while(t-->0) {
long x=sc.nextInt();
long y=sc.nextInt();
long ans=0;
if(x==2)
ans=2;
else if(x<y)
{
if(y%x==0)
ans=x;
else
ans=y- (y%x)/2 ;
}
else if(x==y)
ans=x;
else
ans=x+y;
System.out.println(ans);
}
}
public static boolean check(int k,int n,ArrayList<pair> arr)
{
int c=0;
for(int i=0;i<n;i++)
{
if( k-1-arr.get(i).a<=c && arr.get(i).b>=c )
c++;
// System.out.println(c);
}
if(c>=k)
return true;
else
return false;
}
/* public static void sieve()
{ prime[0]=prime[1]=false;
int n=1000000;
for(int p = 2; p*p <=n; p++)
{
if(prime[p] == true)
{
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}
*/
}
class pair {
int a;int b;
public pair(int f,int ind)
{
a=f;
b=ind;
}
}
class FScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer sb = new StringTokenizer("");
String next(){
while(!sb.hasMoreTokens()){
try{
sb = new StringTokenizer(br.readLine());
} catch(IOException e){ }
}
return sb.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for(int i=0;i<n;i++) a[i] = nextInt();
return a;
}
float nextFloat(){
return Float.parseFloat(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | a47b2f5ee8831bd36d7a94842b6bc151 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
/**
* #
*
* @author pttrung
*/
public class B_Round_752_Div1 {
public static int MOD = 1000000007;
public static void main(String[] args) throws FileNotFoundException {
// PrintWriter out = new PrintWriter(new FileOutputStream(new File(
// "output.txt")));
PrintWriter out = new PrintWriter(System.out);
Scanner in = new Scanner();
int T = in.nextInt();
for (int t = 0; t < T; t++) {
int x = in.nextInt();
int y = in.nextInt();
if (y % x == 0) {
out.println(x);
continue;
}
if (x > y) {
out.println((x + y));
continue;
}
int a = (y / x);
int b = (y % x);
for (int i = 1; (long) i * (long) i <= a; i++) {
if (a % i != 0) {
continue;
}
int o = a / i;
if (b % (i + 1) == 0) {
out.println((x * (a / i) + (b / (i + 1))));
break;
}
if (b % (o + 1) == 0) {
out.println((x * (a / o) + (b / (o + 1))));
break;
}
}
}
out.close();
}
static int abs(int a) {
return a < 0 ? -a : a;
}
public static int[] KMP(String val) {
int i = 0;
int j = -1;
int[] result = new int[val.length() + 1];
result[0] = -1;
while (i < val.length()) {
while (j >= 0 && val.charAt(j) != val.charAt(i)) {
j = result[j];
}
j++;
i++;
result[i] = j;
}
return result;
}
public static boolean nextPer(int[] data) {
int i = data.length - 1;
while (i > 0 && data[i] < data[i - 1]) {
i--;
}
if (i == 0) {
return false;
}
int j = data.length - 1;
while (data[j] < data[i - 1]) {
j--;
}
int temp = data[i - 1];
data[i - 1] = data[j];
data[j] = temp;
Arrays.sort(data, i, data.length);
return true;
}
public static int digit(long n) {
int result = 0;
while (n > 0) {
n /= 10;
result++;
}
return result;
}
public static double dist(long a, long b, long x, long y) {
double val = (b - a) * (b - a) + (x - y) * (x - y);
val = Math.sqrt(val);
double other = x * x + a * a;
other = Math.sqrt(other);
return val + other;
}
public static class Point {
int x;
int y;
public Point(int start, int end) {
this.x = start;
this.y = end;
}
public String toString() {
return x + " " + y;
}
}
public static class FT {
long[] data;
FT(int n) {
data = new long[n];
}
public void update(int index, long value) {
while (index < data.length) {
data[index] += value;
index += (index & (-index));
}
}
public long get(int index) {
long result = 0;
while (index > 0) {
result += data[index];
index -= (index & (-index));
}
return result;
}
}
public static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
public static long pow(long a, int b) {
if (b == 0) {
return 1;
}
if (b == 1) {
return a;
}
long val = pow(a, b / 2);
if (b % 2 == 0) {
return (val * val);
} else {
return (val * ((val * a)));
}
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() throws FileNotFoundException {
// System.setOut(new PrintStream(new BufferedOutputStream(System.out), true));
br = new BufferedReader(new InputStreamReader(System.in));
//br = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt"))));
}
public String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
throw new RuntimeException();
}
}
return st.nextToken();
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
st = null;
try {
return br.readLine();
} catch (Exception e) {
throw new RuntimeException();
}
}
public boolean endLine() {
try {
String next = br.readLine();
while (next != null && next.trim().isEmpty()) {
next = br.readLine();
}
if (next == null) {
return true;
}
st = new StringTokenizer(next);
return st.hasMoreTokens();
} catch (Exception e) {
throw new RuntimeException();
}
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 7d9eb38f55db584696790c24cb294888 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.text.DecimalFormat;
import java.util.*;
public class Solution {
static BufferedReader bf;
static PrintWriter out;
static Scanner sc;
static StringTokenizer st;
static long mod = (long)(1e9+7);
static long mod2 = 998244353;
static long fact[] = new long[1000001];
static long inverse[] = new long[1000001];
public static void main (String[] args)throws IOException {
bf = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
sc = new Scanner(System.in);
// fact[0] = 1;
// inverse[0] = 1;
// for(int i = 1;i<fact.length;i++){
// fact[i] = (fact[i-1] * i)%mod;
// // inverse[i] = binaryExpo(fact[i], mod-2);a
// }
int t = nextInt();
while(t-->0){
solve();
}
out.flush();
}
public static void solve() throws IOException{
int x = nextInt();
int y = nextInt();
if(x < y){
out.println((y - (y % x)/2));
}
else if(x == y){
out.println(x);
}
else{
out.println((x + y));
}
}
public static int height(List<List<Integer>>list,int cur,int parent){
List<Integer>temp = list.get(cur);
int max = 0;
for(int i = 0;i<temp.size();i++){
int next = temp.get(i);
if(next != parent){
max = Math.max(max,height(list,next,cur));
}
}
return max+1;
}
public static void req(int i,int j){
out.println("? " + i + " " + j);
out.flush();
}
public static int maxy(int node, int l,int r,int tl,int tr,int[][]tree){
if(l>=tl && r <= tr)return tree[node][0];
if(r < tl || l > tr)return Integer.MIN_VALUE;
int mid = (l + r)/2;
return Math.max(maxy(node*2,l,mid,tl,tr,tree),maxy(node*2+1,mid+1,r,tl,tr,tree));
}
public static int mini(int node,int l,int r,int tl,int tr,int[][]tree){
if(l >= tl && r <= tr)return tree[node][1];
if(r < tl || l > tr)return Integer.MAX_VALUE;
int mid = (l + r)/2;
return Math.min(mini(node*2 , l , mid ,tl ,tr,tree),mini(node*2 + 1, mid + 1, r, tl, tr, tree));
}
public static void fillParent(int[][]parent,List<List<Integer>>list,int cur ,int p,int[]depths){
parent[cur][0] = p;
List<Integer>temp = list.get(cur);
for(int i = 0;i<temp.size();i++){
int next = temp.get(i);
if(next != p){
depths[next] = depths[cur] + 1;
fillParent(parent,list,next,cur,depths);
}
}
}
public static int lca(int[][]parent , int u, int v){
if(u == v)return u;
for(int i = 18;i>=0;i--){
if(parent[u][i] != parent[v][i]){
u = parent[u][i];
v = parent[v][i];
}
}
return parent[u][0];
}
public static int bringSame(int[][]parent,int[]depths,int u, int v){
if(depths[u] < depths[v]){
int temp = u;
u = v;
v = temp;
}
int k = depths[u] - depths[v];
for(int i = 0;i<=18;i++){
if((k & (1<<i)) != 0){
u = parent[u][i];
}
}
return u;
}
public static long nck(int n,int k){
return fact[n] * inverse[n-k] %mod * inverse[k]%mod;
}
public static void plus(int node,int low,int high,int tlow,int thigh,int[]tree){
if(low >= tlow && high <= thigh){
tree[node]++;
return;
}
if(high < tlow || low > thigh)return;
int mid = (low + high)/2;
plus(node*2,low,mid,tlow,thigh,tree);
plus(node*2 + 1 , mid + 1, high,tlow, thigh,tree);
}
public static boolean allEqual(int[]arr,int x){
for(int i = 0;i<arr.length;i++){
if(arr[i] != x)return false;
}
return true;
}
public static long helper(StringBuilder sb){
return Long.parseLong(sb.toString());
}
public static int min(int node,int low,int high,int tlow,int thigh,int[][]tree){
if(low >= tlow&& high <= thigh)return tree[node][0];
if(high < tlow || low > thigh)return Integer.MAX_VALUE;
int mid = (low + high)/2;
// println(low+" "+high+" "+tlow+" "+thigh);
return Math.min(min(node*2,low,mid,tlow,thigh,tree) ,min(node*2+1,mid+1,high,tlow,thigh,tree));
}
public static int max(int node,int low,int high,int tlow,int thigh,int[][]tree){
if(low >= tlow && high <= thigh)return tree[node][1];
if(high < tlow || low > thigh)return Integer.MIN_VALUE;
int mid = (low + high)/2;
return Math.max(max(node*2,low,mid,tlow,thigh,tree) ,max(node*2+1,mid+1,high,tlow,thigh,tree));
}
public static long[] help(List<List<Integer>>list,int[][]range,int cur){
List<Integer>temp = list.get(cur);
if(temp.size() == 0)return new long[]{range[cur][1],1};
long sum = 0;
long count = 0;
for(int i = 0;i<temp.size();i++){
long []arr = help(list,range,temp.get(i));
sum += arr[0];
count += arr[1];
}
if(sum < range[cur][0]){
count++;
sum = range[cur][1];
}
return new long[]{sum,count};
}
public static long findSum(int node,int low, int high,int tlow,int thigh,long[]tree,long mod){
if(low >= tlow && high <= thigh)return tree[node]%mod;
if(low > thigh || high < tlow)return 0;
int mid = (low + high)/2;
return((findSum(node*2,low,mid,tlow,thigh,tree,mod) % mod) + (findSum(node*2+1,mid+1,high,tlow,thigh,tree,mod)))%mod;
}
public static boolean allzero(long[]arr){
for(int i =0 ;i<arr.length;i++){
if(arr[i]!=0)return false;
}
return true;
}
public static long count(long[][]dp,int i,int[]arr,int drank,long sum){
if(i == arr.length)return 0;
if(dp[i][drank]!=-1 && arr[i]+sum >=0)return dp[i][drank];
if(sum + arr[i] >= 0){
long count1 = 1 + count(dp,i+1,arr,drank+1,sum+arr[i]);
long count2 = count(dp,i+1,arr,drank,sum);
return dp[i][drank] = Math.max(count1,count2);
}
return dp[i][drank] = count(dp,i+1,arr,drank,sum);
}
public static void help(int[]arr,char[]signs,int l,int r){
if( l < r){
int mid = (l+r)/2;
help(arr,signs,l,mid);
help(arr,signs,mid+1,r);
merge(arr,signs,l,mid,r);
}
}
public static void merge(int[]arr,char[]signs,int l,int mid,int r){
int n1 = mid - l + 1;
int n2 = r - mid;
int[]a = new int[n1];
int []b = new int[n2];
char[]asigns = new char[n1];
char[]bsigns = new char[n2];
for(int i = 0;i<n1;i++){
a[i] = arr[i+l];
asigns[i] = signs[i+l];
}
for(int i = 0;i<n2;i++){
b[i] = arr[i+mid+1];
bsigns[i] = signs[i+mid+1];
}
int i = 0;
int j = 0;
int k = l;
boolean opp = false;
while(i<n1 && j<n2){
if(a[i] <= b[j]){
arr[k] = a[i];
if(opp){
if(asigns[i] == 'R'){
signs[k] = 'L';
}
else{
signs[k] = 'R';
}
}
else{
signs[k] = asigns[i];
}
i++;
k++;
}
else{
arr[k] = b[j];
int times = n1 - i;
if(times%2 == 1){
if(bsigns[j] == 'R'){
signs[k] = 'L';
}
else{
signs[k] = 'R';
}
}
else{
signs[k] = bsigns[j];
}
j++;
opp = !opp;
k++;
}
}
while(i<n1){
arr[k] = a[i];
if(opp){
if(asigns[i] == 'R'){
signs[k] = 'L';
}
else{
signs[k] = 'R';
}
}
else{
signs[k] = asigns[i];
}
i++;
k++;
}
while(j<n2){
arr[k] = b[j];
signs[k] = bsigns[j];
j++;
k++;
}
}
public static long binaryExpo(long base,long expo){
if(expo == 0)return 1;
long val;
if(expo%2 == 1){
val = (binaryExpo(base, expo-1)*base)%mod;
}
else{
val = binaryExpo(base, expo/2);
val = (val*val)%mod;
}
return (val%mod);
}
public static boolean isSorted(int[]arr){
for(int i =1;i<arr.length;i++){
if(arr[i] < arr[i-1]){
return false;
}
}
return true;
}
//function to find the topological sort of the a DAG
public static boolean hasCycle(int[]indegree,List<List<Integer>>list,int n,List<Integer>topo){
Queue<Integer>q = new LinkedList<>();
for(int i =1;i<indegree.length;i++){
if(indegree[i] == 0){
q.add(i);
topo.add(i);
}
}
while(!q.isEmpty()){
int cur = q.poll();
List<Integer>l = list.get(cur);
for(int i = 0;i<l.size();i++){
indegree[l.get(i)]--;
if(indegree[l.get(i)] == 0){
q.add(l.get(i));
topo.add(l.get(i));
}
}
}
if(topo.size() == n)return false;
return true;
}
// function to find the parent of any given node with path compression in DSU
public static int find(int val,int[]parent){
if(val == parent[val])return val;
return parent[val] = find(parent[val],parent);
}
// function to connect two components
public static void union(int[]rank,int[]parent,int u,int v){
int a = find(u,parent);
int b= find(v,parent);
if(a == b)return;
if(rank[a] == rank[b]){
parent[b] = a;
rank[a]++;
}
else{
if(rank[a] > rank[b]){
parent[b] = a;
}
else{
parent[a] = b;
}
}
}
//
public static int findN(int n){
int num = 1;
while(num < n){
num *=2;
}
return num;
}
// code for input
public static void print(String s ){
System.out.print(s);
}
public static void print(int num ){
System.out.print(num);
}
public static void print(long num ){
System.out.print(num);
}
public static void println(String s){
System.out.println(s);
}
public static void println(int num){
System.out.println(num);
}
public static void println(long num){
System.out.println(num);
}
public static void println(){
System.out.println();
}
public static int Int(String s){
return Integer.parseInt(s);
}
public static long Long(String s){
return Long.parseLong(s);
}
public static String[] nextStringArray()throws IOException{
return bf.readLine().split(" ");
}
public static String nextString()throws IOException{
return bf.readLine();
}
public static long[] nextLongArray(int n)throws IOException{
String[]str = bf.readLine().split(" ");
long[]arr = new long[n];
for(int i =0;i<n;i++){
arr[i] = Long.parseLong(str[i]);
}
return arr;
}
public static int[][] newIntMatrix(int r,int c)throws IOException{
int[][]arr = new int[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Integer.parseInt(str[j]);
}
}
return arr;
}
public static long[][] newLongMatrix(int r,int c)throws IOException{
long[][]arr = new long[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Long.parseLong(str[j]);
}
}
return arr;
}
static class pair{
int one;
int two;
pair(int one,int two){
this.one = one ;
this.two =two;
}
}
public static long gcd(long a,long b){
if(b == 0)return a;
return gcd(b,a%b);
}
public static long lcm(long a,long b){
return (a*b)/(gcd(a,b));
}
public static boolean isPalindrome(String s){
int i = 0;
int j = s.length()-1;
while(i<=j){
if(s.charAt(i) != s.charAt(j)){
return false;
}
i++;
j--;
}
return true;
}
// these functions are to calculate the number of smaller elements after self
public static void sort(int[]arr,int l,int r){
if(l < r){
int mid = (l+r)/2;
sort(arr,l,mid);
sort(arr,mid+1,r);
smallerNumberAfterSelf(arr, l, mid, r);
}
}
public static void smallerNumberAfterSelf(int[]arr,int l,int mid,int r){
int n1 = mid - l +1;
int n2 = r - mid;
int []a = new int[n1];
int[]b = new int[n2];
for(int i = 0;i<n1;i++){
a[i] = arr[l+i];
}
for(int i =0;i<n2;i++){
b[i] = arr[mid+i+1];
}
int i = 0;
int j =0;
int k = l;
while(i<n1 && j < n2){
if(a[i] < b[j]){
arr[k++] = a[i++];
}
else{
arr[k++] = b[j++];
}
}
while(i<n1){
arr[k++] = a[i++];
}
while(j<n2){
arr[k++] = b[j++];
}
}
public static String next(){
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(bf.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
static int nextInt() {
return Integer.parseInt(next());
}
static long nextLong() {
return Long.parseLong(next());
}
static double nextDouble() {
return Double.parseDouble(next());
}
static String nextLine(){
String str = "";
try {
str = bf.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
static String interpolate(int n,List<Integer>items,List<Float>prices){
List<Integer>item = new ArrayList<>();
List<Float>price = new ArrayList<>();
for(int i =0 ;i<items.size();i++){
if(items.get(i)> 0){
item.add(items.get(i));
price.add(prices.get(i));
}
}
DecimalFormat df = new DecimalFormat("0.00");
if(item.size() == 1){
return df.format(price.get(0))+"";
}
for(int i = 0;i<item.size();i++){
if(n == item.get(i)){
return df.format(price.get(i))+"";
}
}
for(int i = 0;i<item.size()-1;i++){
if(item.get(i) < n && n < item.get(i+1)){
float x1 = item.get(i);
float x2 = item.get(i+1);
float y1 = price.get(i);
float y2 = price.get(i+1);
float m = (y2 - y1)/(x2 - x1);
float c = y1 - m * x1;
float ans = m * n + c;
return df.format(ans)+"";
}
}
if(item.get(0) > n){
float x1 = item.get(0);
float x2 = item.get(1);
float y1 = price.get(0);
float y2 = price.get(1);
float m = (y2 - y1)/(x2 - x1);
float c = y1 - m * x1;
float ans = m * n + c;
return df.format(ans)+"";
}
int size = item.size();
float x1 = item.get(size-2);
float x2 = item.get(size-1);
float y1 = price.get(size-2);
float y2 = price.get(size-1);
float m = (y2 - y1)/(x2 - x1);
float c = y1 - m * x1;
float ans = m * n + c;
return df.format(ans)+"";
}
}
// use some math tricks it might help
// sometimes just try to think in straightforward plan in A and B problems don't always complecate the questions with thinking too much differently
// always use long number to do 10^9+7 modulo
// if a problem is related to binary string it could also be related to parenthesis
// *****try to use binary search(it is a very beautiful thing it can work in some of the very unexpected problems ) in the question it might work******
// try sorting
// try to think in opposite direction of question it might work in your way
// if a problem is related to maths try to relate some of the continuous subarray with variables like - > a+b+c+d or a,b,c,d in general
// if the question is to much related to left and/or right side of any element in an array then try monotonic stack it could work.
// in range query sums try to do binary search it could work
// analyse the time complexity of program thoroughly
// anylyse the test cases properly
// if we divide any number by 2 till it gets 1 then there will be (number - 1) operation required
// try to do the opposite operation of what is given in the problem
//think about the base cases properly
//If a question is related to numbers try prime factorisation or something related to number theory
// keep in mind unique strings
//you can calculate the number of inversion in O(n log n)
// in a matrix you could sometimes think about row and cols indenpendentaly.
// Try to think in more constructive(means a way to look through various cases of a problem) way.
// observe the problem carefully the answer could be hidden in the given test cases itself. (A, B , C);
// when we have equations like (a+b = N) and we have to find the max of (a*b) then the values near to the N/2 must be chosen as (a and b);
// give emphasis to the number of occurences of elements it might help.
// if a number is even then we can find the pair for each and every number in that range whose bitwise AND is zero.
// if a you find a problem related to the graph make a graph | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 9f5981138ed1aec54f5b5d522a98293a | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that :()
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x1603B
{
public static void main(String omkar[]) throws Exception
{
BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(infile.readLine());
int T = Integer.parseInt(st.nextToken());
StringBuilder sb = new StringBuilder();
while(T-->0)
{
st = new StringTokenizer(infile.readLine());
long X = Integer.parseInt(st.nextToken());
long Y = Integer.parseInt(st.nextToken());
if(X > Y)
{
long res = X+Y;
sb.append(res);
}
else if(X < Y)
{
long res = (Y%X)/2;
res = Y-res;
sb.append(res);
}
else
sb.append(Y);
sb.append("\n");
}
System.out.print(sb);
}
}
/*
3
121210 333
332 145510
168 204
*/ | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | e9612a817ac15ea867a79e7468c75589 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.io.*;
import java.rmi.MarshalException;
import java.util.*;
public class ModerateModular{
static long mod = 1000000007L;
static MyScanner sc = new MyScanner();
static void solve() {
long x = sc.nextLong();
long y = sc.nextLong();
if(x>y){
out.println(x+y);
}else{
out.println(y-(y%x)/2);
}
}
static class Pair implements Comparable<Pair>{
char c;
int fre;
Pair(char c,int f){
this.c= c;
fre = f;
}
public int compareTo(Pair p){
return this.fre - p.fre;
}
}
static void reverse(int arr[]){
int i = 0;int j = arr.length-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
}
static void reverse(long arr[]){
int i = 0;int j = arr.length-1;
while(i<j){
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
}
static long pow(long a, long b) {
if (b == 0) return 1;
long res = pow(a, b / 2);
res = (res * res) % 1_000_000_007;
if (b % 2 == 1) {
res = (res * a) % 1_000_000_007;
}
return res;
}
static int lis(int arr[],int n){
int lis[] = new int[n];
lis[0] = 1;
for(int i = 1;i<n;i++){
lis[i] = 1;
for(int j = 0;j<i;j++){
if(arr[i]>arr[j]){
lis[i] = Math.max(lis[i],lis[j]+1);
}
}
}
int max = Integer.MIN_VALUE;
for(int i = 0;i<n;i++){
max = Math.max(lis[i],max);
}
return max;
}
static boolean isPali(String str){
int i = 0;
int j = str.length()-1;
while(i<j){
if(str.charAt(i)!=str.charAt(j)){
return false;
}
i++;
j--;
}
return true;
}
static long gcd(long a,long b){
if(b==0) return a;
return gcd(b,a%b);
}
static String reverse(String str){
char arr[] = str.toCharArray();
int i = 0;
int j = arr.length-1;
while(i<j){
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
String st = new String(arr);
return st;
}
static boolean isprime(int n){
if(n==1) return false;
if(n==3 || n==2) return true;
if(n%2==0 || n%3==0) return false;
for(int i = 5;i*i<=n;i+= 6){
if(n%i== 0 || n%(i+2)==0){
return false;
}
}
return true;
}
public static void main(String[] args) {
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
// int t= 1;
while(t-- >0){
solve();
// solve2();
// solve3();
}
// Stop writing your solution here. -------------------------------------
out.close();
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int[] readIntArray(int n){
int arr[] = new int[n];
for(int i = 0;i<n;i++){
arr[i] = Integer.parseInt(next());
}
return arr;
}
int[] reverse(int arr[]){
int n= arr.length;
int i = 0;
int j = n-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
j--;i++;
}
return arr;
}
long[] readLongArray(int n){
long arr[] = new long[n];
for(int i = 0;i<n;i++){
arr[i] = Long.parseLong(next());
}
return arr;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
private static void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
private static void sort(long[] arr) {
List<Long> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 8cf7525bc8e43930c4c144fe49f678f2 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.lang.*;
public class Main {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-->0){
int x = scn.nextInt();
int y = scn.nextInt();
if(x>y){
System.out.println(x+y);
}
else{
System.out.println(y-(y%x)/2);
}
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 7ba6d28ab7fb66a082661a16dc205e70 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef {
public static void main(String[] args) throws java.lang.Exception {
FastReader in = new FastReader(System.in);
StringBuilder sb = new StringBuilder();
int t = 1;
t = in.nextInt();
while (t > 0) {
--t;
int x = in.nextInt();
int y = in.nextInt();
if(x == y)
sb.append(x+"\n");
else if(x>y)
sb.append((x+y)+"\n");
else {
int rem = y%x;
int count = (y/x)*x + rem/2;
sb.append(count+"\n");
}
}
System.out.print(sb);
}
static long gcd(long a, long b) {
if (a == 0)
return b;
else
return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
static void sort(long[] a) {
ArrayList<Long> l = new ArrayList<>();
for (long i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
}
class FastReader {
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan())
;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
String nextLine() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan())
;
StringBuilder sb = new StringBuilder();
for (; c != 10 && c != 13; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
char nextChar() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan())
;
return (char) c;
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan())
;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan())
;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 673fce6b512bd7cf30cdadbfa8452b04 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class aa {
//--------------------------INPUT READER--------------------------------//
static class fs {
public BufferedReader br;
StringTokenizer st = new StringTokenizer("");
public fs() { this(System.in); }
public fs(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
String next() {
while (!st.hasMoreTokens()) {
try { st = new StringTokenizer(br.readLine()); }
catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String ns() { return next(); }
int[] na(long nn) {
int n = (int) nn;
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = ni();
return a;
}
long[] nal(long nn) {
int n = (int) nn;
long[] l = new long[n];
for(int i = 0; i < n; i++) l[i] = nl();
return l;
}
}
//-----------------------------------------------------------------------//
//---------------------------PRINTER-------------------------------------//
static class Printer {
static PrintWriter w;
public Printer() {this(System.out);}
public Printer(OutputStream os) {
w = new PrintWriter(os);
}
public void p(int i) {w.println(i);};
public void p(long l) {w.println(l);};
public void p(double d) {w.println(d);};
public void p(String s) { w.println(s);};
public void pr(int i) {w.print(i);};
public void pr(long l) {w.print(l);};
public void pr(double d) {w.print(d);};
public void pr(String s) { w.print(s);};
public void pl() {w.println();};
public void close() {w.close();};
}
//------------------------------------------------------------------------//
//--------------------------VARIABLES------------------------------------//
static fs sc = new fs();
static OutputStream outputStream = System.out;
static Printer w = new Printer(outputStream);
//-----------------------------------------------------------------------//
//--------------------------ADMIN_MODE-----------------------------------//
private static void ADMIN_MODE() throws IOException {
if (System.getProperty("ONLINE_JUDGE") == null) {
w = new Printer(new FileOutputStream("output.txt"));
sc = new fs(new FileInputStream("input.txt"));
}
}
//-----------------------------------------------------------------------//
//----------------------------START--------------------------------------//
public static void main(String[] args)
throws IOException {
ADMIN_MODE();
int t = sc.ni();while(t-->0)
solve();
w.close();
}
static void solve() throws IOException {
long x = sc.ni(), y = sc.ni();
if(x>y) {
w.p(x+y);
return;
} else if(x==y) {
w.p(x);
return;
}
long q = y/x;
long reach = q*x;
long n = (y+ reach)/2;
w.p(n);
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 802eddbfcfff33d7eb9a6e31f9c50073 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static PrintWriter pw;
static Scanner sc;
static long ceildiv(long x, long y) { return (x+y-1) / y; }
static int mod(long x, int m) { return (int) ((x%m + m) % m); }
static void put(Map<Integer, Integer> map, Integer p){if(map.containsKey(p)) map.replace(p, map.get(p)+1); else map.put(p, 1); }
static void rem(Map<Integer, Integer> map, Integer p){ if(map.get(p)==1) map.remove(p); else map.replace(p, map.get(p)-1); }
static int Int(boolean x){ return x ? 1 : 0; }
static final int inf=(int)1e9, mod= inf + 7;
static final long infL=inf*1l*inf;
static final double eps=1e-9;
public static long gcd(long x, long y) { return y==0? x: gcd(y, x%y); }
public static void main(String[] args) throws Exception {
sc = new Scanner(System.in);
pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0)
testCase();
pw.close();
}
static void testCase() throws Exception{
int x = sc.nextInt(), y = sc.nextInt();
if(x > y){
pw.println((x+1) * 1l * y);
return;
}
pw.println(y - (y%x)/2);
}
static void printArr(int[] arr) {
for (int i = 0; i < arr.length; i++)
pw.print(arr[i] + " ");
pw.println();
}
static void printArr(long[] arr) {
for (int i = 0; i < arr.length; i++)
pw.print(arr[i] + " ");
pw.println();
}
static void printArr(double[] arr) {
for (int i = 0; i < arr.length; i++)
pw.print(arr[i] + " ");
pw.println();
}
static void printArr(Integer[] arr) {
for (int i = 0; i < arr.length; i++)
pw.print(arr[i] + " ");
pw.println();
}
static void printIter(Iterable list) {
for (Object o : list)
pw.print(o + " ");
pw.println();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public int[] nextDigits() throws IOException {
String s = nextLine();
int[] arr = new int[s.length()];
for (int i = 0; i < arr.length; i++)
arr[i] = s.charAt(i) - '0';
return arr;
}
public int[] nextArr(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < arr.length; i++)
arr[i] = nextInt();
return arr;
}
public Integer[] nextsort(int n) throws IOException {
Integer[] arr = new Integer[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
public Pair nextPair() throws IOException {
return new Pair(nextInt(), nextInt());
}
public long[] nextLongArr(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
public Pair[] nextPairArr(int n) throws IOException {
Pair[] arr = new Pair[n];
for (int i = 0; i < n; i++)
arr[i] = nextPair();
return arr;
}
public boolean hasNext() throws IOException {
return (st != null && st.hasMoreTokens()) || br.ready();
}
}
static class Pair implements Comparable<Pair> {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
public Pair(Map.Entry<Integer, Integer> a) {
x = a.getKey();
y = a.getValue();
}
public int hashCode() {
return (int) ((this.x*1l*100003 + this.y) % mod);
}
public int compareTo(Pair p) {
if(x != p.x)
return x - p.x;
return y - p.y;
}
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (this.getClass() != obj.getClass()) {
return false;
}
Pair p = (Pair) obj;
return this.x == p.x && this.y == p.y;
}
public Pair clone() {
return new Pair(x, y);
}
public String toString() {
return this.x + " " + this.y;
}
public void subtract(Pair p) {
x -= p.x;
y -= p.y;
}
public void add(Pair p) {
x += p.x;
y += p.y;
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 2dbb9f6893544eb4732a435e8d697c0b | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
public class B1603 {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
long x = sc.nextLong();
long y = sc.nextLong();
// long x = (int) (Math.random() * (10)) + 1;
// long y = (int) (Math.random() * (10)) + 1;
// x *= 2;
// y *= 2;
long n;
if (x > y) {
n = (x + y);
} else if (x == y) {
n = (x);
} else {
// long b = y / x;
long a = (y - y / x * x) / (2);
n = y / x * x + a;
}
// if (n % x != y % n) {
// System.out.println(x + " " + y);
// }
// pw.println(n % x == y % n);
pw.println(n);
}
pw.close();
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader f) {
br = new BufferedReader(f);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public int[] nextIntArr(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(next());
}
return arr;
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 45171cf0de823a1d197641df4c131ed5 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.Scanner;
public class B1603 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
StringBuilder out = new StringBuilder();
int T = in.nextInt();
for (int t=0; t<T; t++) {
int x = in.nextInt();
int y = in.nextInt();
int n;
if (x <= y) {
n = y - (y%x)/2;
} else {
n = x+y;
}
out.append(n).append('\n');
}
System.out.println(out);
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 1d04f8bd5fcb599b21c803ff04352138 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws Throwable {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
for (int tc = 0; tc < t; ++tc) {
st = new StringTokenizer(br.readLine());
int x = Integer.parseInt(st.nextToken());
int y = Integer.parseInt(st.nextToken());
System.out.println(solve(x, y));
}
}
static int solve(int x, int y) {
if (x == y) {
return x;
}
if (x > y) {
return x + y;
}
return (y / x * x + y) / 2;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 3ee2c883e974e6dc18b5eb8080f5b727 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | // package faltu;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.Map.Entry;
public class Main {
public static int upperBound(long[] arr, long m, int l, int r) {
while(l<=r) {
int mid=(l+r)/2;
if(arr[mid]<=m) l=mid+1;
else r=mid-1;
}
return l;
}
public static int lowerBound(long[] a, long m, int l, int r) {
while(l<=r) {
int mid=(l+r)/2;
if(a[mid]<m) l=mid+1;
else r=mid-1;
}
return l;
}
public static long getClosest(long val1, long val2,long target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
static void ruffleSort(long[] a) {
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
long oi=r.nextInt(n), temp=a[i];
a[i]=a[(int)oi];
a[(int)oi]=temp;
}
Arrays.sort(a);
}
static void ruffleSort(int[] a){
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n), temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
Arrays.sort(a);
}
int ceilIndex(int input[], int T[], int end, int s){
int start = 0;
int middle;
int len = end;
while(start <= end){
middle = (start + end)/2;
if(middle < len && input[T[middle]] < s && s <= input[T[middle+1]]){
return middle+1;
}else if(input[T[middle]] < s){
start = middle+1;
}else{
end = middle-1;
}
}
return -1;
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a,long b)
{
return (a / gcd(a, b)) * b;
}
public static int[] reverse(int a[], int left, int right)
{
// Reverse the sub-array
while (left < right) {
int temp = a[left];
a[left++] = a[right];
a[right--] = temp;
}
return a;
}
static int lowerLimitBinarySearch(ArrayList<Long> v,long k) {
int n =v.size();
int first = 0,second = n;
while(first <second) {
int mid = first + (second-first)/2;
if(v.get(mid) > k) {
second = mid;
}else {
first = mid+1;
}
}
if(first < n && v.get(first) < k) {
first++;
}
return first; //1 index
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
// *******----segement tree implement---*****
// -------------START--------------------------
void buildTree (int[] arr,int[] tree,int start,int end,int treeNode)
{
if(start==end)
{
tree[treeNode]=arr[start];
return;
}
buildTree(arr,tree,start,end,2*treeNode);
buildTree(arr,tree,start,end,2*treeNode+1);
tree[treeNode]=tree[treeNode*2]+tree[2*treeNode+1];
}
void updateTree(int[] arr,int[] tree,int start,int end,int treeNode,int idx,int value)
{
if(start==end)
{
arr[idx]=value;
tree[treeNode]=value;
return;
}
int mid=(start+end)/2;
if(idx>mid)
{
updateTree(arr,tree,mid+1,end,2*treeNode+1,idx,value);
}
else
{
updateTree(arr,tree,start,mid,2*treeNode,idx,value);
}
tree[treeNode]=tree[2*treeNode]+tree[2*treeNode+1];
}
// disjoint set implementation --start
static void makeSet(int n)
{
parent=new int[n];
rank=new int[n];
for(int i=0;i<n;i++)
{
parent[i]=i;
rank[i]=0;
}
}
static void union(int u,int v)
{
u=findpar(u);
v=findpar(v);
if(rank[u]<rank[v])parent[u]=v;
else if(rank[v]<rank[u])parent[v]=u;
else
{
parent[v]=u;
rank[u]++;
}
}
private static int findpar(int node)
{
if(node==parent[node])return node;
return parent[node]=findpar(parent[node]);
}
static int parent[];
static int rank[];
// *************end
static void presumbit(int[][]prebitsum) {
for(int i=1;i<=200000;i++) {
int z=i;
int j=0;
while(z>0) {
if((z&1)==1) {
prebitsum[i][j]+=(prebitsum[i-1][j]+1);
}else {
prebitsum[i][j]=prebitsum[i-1][j];
}
z=z>>1;
j++;
}
}
}
static void countOfSetBit(long[]a) {
for(int j=30;j>=0;j--) {
int cnt=0;
for(long i:a) {
if((i&1<<j)==1)cnt++;
}
// printing the current no set bit in all array element
System.out.println(cnt);
}
}
public static int[] sort(int[] arr) {
ArrayList<Integer> al = new ArrayList<>();
for(int i=0;i<arr.length;i++) al.add(arr[i]);
Collections.sort(al);
for(int i=0;i<arr.length;i++) arr[i]=al.get(i);
return arr;
}
static ArrayList<String>powof2s;
static void powof2S() {
long i=1;
while(i<(long)2e18) {
powof2s.add(String.valueOf(i));
i*=2;
}
}
static boolean coprime(int a, long l){
return (gcd(a, l) == 1);
}
static int[][] dirs8 = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
static int[][] dirs4 = {{1,0},{-1,0},{0,1},{0,-1}};
static Long MOD=(long) (1e9+7);
static int prebitsum[][];
static ArrayList<Integer>arr;
static boolean[] vis;
static ArrayList<ArrayList<Integer>>adj;
static long[]a;
static FastReader s = new FastReader();
public static void main(String[] args) throws IOException
{
// sieve();
// prebitsum=new int[2147483648][31];
// presumbit(prebitsum);
// powof2S();
int tt = s.nextInt();
while(tt-->0) {
solver();
}
}
private static void solver() {
long x=s.nextLong();
long y=s.nextLong();
if(x==y) {
System.out.println(x);return;
}
if(x>y) {
System.out.println((x+y));
return;
}
else {
long rem=y%x;
long ans=((y/x)*x)+ rem/2;
System.out.println(ans);
}
}
public static int searchindex(long arr[], long t){
int index = Arrays.binarySearch(arr, t);
return (index < 0) ? -1 : index;
}
static void pc2d(char[][]a) {
int n=a.length;
int m=a[0].length;
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
static void pi2d(long[][] d) {
int n=d.length;
int m=d[0].length;
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
System.out.print(d[i][j]+" ");
}
System.out.println();
}
}
static void DFSUtil(int v, boolean[] vis)
{
vis[v] = true;
Iterator<Integer> it = adj.get(v).iterator();
while (it.hasNext()) {
int n = it.next();
if (!vis[n])
DFSUtil(n, vis);
}
}
static long DFS(int n)
{
vis = new boolean[n+1];
long cnt=0;
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
DFSUtil(i, vis);
cnt++;
}
}
return cnt;
}
public static String revStr(String str){
String input = str;
StringBuilder input1 = new StringBuilder();
input1.append(input);
input1.reverse();
return input1.toString();
}
static long myPow(long n, long i){
if(i==0) return 1;
if(i%2==0) return (myPow(n,i/2)%MOD * myPow(n,i/2)%MOD)%MOD;
return (n%MOD* myPow(n,i-i)%MOD)%MOD;
}
static boolean isPalindrome(String str,int i,int j)
{
while (i < j) {
if (str.charAt(i) != str.charAt(j))
return false;
i++;
j--;
}
return true;
}
static boolean sign(long num) {
return num>0;
}
static boolean isSquare(long x){
if(x==1)return true;
long y=(long) Math.sqrt(x);
return y*y==x;
}
static long Power(long a,long b) {
if(b == 0){
return 1;
}
long ans = Power(a,b/2);
ans *= ans%MOD;
if(b % 2!=0){
ans *= a%MOD;
}
return ans%MOD;
}
static boolean issafe(int i, int j, int r,int c, char ch)
{
if (i < 0 || j < 0 || i >= r || j >= c|| ch!= '1')return false;
else return true;
}
static long power(long a, long b)
{
a %=MOD;
long out = 1;
while (b > 0) {
if((b&1)!=0)out = out * a % MOD;
a = a * a % MOD;
b >>= 1;
a*=a;
}
return out;
}
static long[] sieve;
public static void sieve()
{
int nnn=(int) 1e6+1;
long nn=(int) 1e6;
sieve=new long[(int) nnn];
int[] freq=new int[(int) nnn];
sieve[0]=0;
sieve[1]=1;
for(int i=2;i<=nn;i++)
{
sieve[i]=1;
freq[i]=1;
}
for(int i=2;i*i<=nn;i++)
{
if(sieve[i]==1)
{
for(int j=i*i;j<=nn;j+=i)
{
if(sieve[j]==1)
{
sieve[j]=0;
}
}
}
}
}
}
class decrease implements Comparator<Long> {
// Used for sorting in ascending order of
// roll number
public int compare(long a, long b)
{
return (int) (b - a);
}
@Override
public int compare(Long o1, Long o2) {
// TODO Auto-generated method stub
return (int) (o2-o1);
}
}
class pair{
double x;
double y;
// public pair(long x,long y) {
// this.x=x;
// this.y=y;
// }
public pair(double x,double y) {
this.x=x;
this.y=y;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 0c54c63b774c9976ed720fbcc230dbc5 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
// cd C:\Users\Lenovo\Desktop\New
//ArrayList<Integer> a=new ArrayList<>();
//List<Integer> lis=new ArrayList<>();
//StringBuilder ans = new StringBuilder();
//HashMap<Integer,Integer> map=new HashMap<>();
public class cf {
static FastReader in=new FastReader();
static final Random random=new Random();
static int mod=1000000007;
//static long dp[]=new long[200002];
public static void main(String args[]) throws IOException {
FastReader sc=new FastReader();
//Scanner s=new Scanner(System.in);
int tt=sc.nextInt();
//int tt=1;
while(tt-->0){
//PriorityQueue<Integer> p = new PriorityQueue<Integer>(Collections.reverseOrder());
int a=sc.nextInt();
int b=sc.nextInt();
if(a>b){
System.out.println(a+b);
}
else{
System.out.println(b-(b%a)/2);
}
}
}
static long comb(int n,int k){
return factorial(n) * pow(factorial(k), mod-2) % mod * pow(factorial(n-k), mod-2) % mod;
}
static long pow(long a, long b) {
// long mod=1000000007;
long res = 1;
while (b != 0) {
if ((b & 1) != 0) {
res = (res * a) % mod;
}
a = (a * a) % mod;
b /= 2;
}
return res;
}
static boolean powOfTwo(long n){
while(n%2==0){
n=n/2;
}
if(n!=1){
return false;
}
return true;
}
static int upper_bound(long arr[], long key)
{
int mid, N = arr.length;
int low = 0;
int high = N;
// Till low is less than high
while (low < high && low != N) {
mid = low + (high - low) / 2;
if (key >= arr[mid]) {
low = mid + 1;
}
else {
high = mid;
}
}
if (low == N ) {
return -1;
}
return low;
}
static boolean prime(int n){
for(int i=2;i<=Math.sqrt(n);i++){
if(n%i==0){
return false;
}
}
return true;
}
static long factorial(int n){
long ret = 1;
while(n > 0){
ret = ret * n % mod;
n--;
}
return ret;
}
static long find(ArrayList<Long> arr,long n){
int l=0;
int r=arr.size();
while(l+1<r){
int mid=(l+r)/2;
if(arr.get(mid)<n){
l=mid;
}
else{
r=mid;
}
}
return arr.get(l);
}
static void rotate(int ans[]){
int last=ans[0];
for(int i=0;i<ans.length-1;i++){
ans[i]=ans[i+1];
}
ans[ans.length-1]=last;
}
static int countprimefactors(int n){
int ans=0;
int z=(int)Math.sqrt(n);
for(int i=2;i<=z;i++){
while(n%i==0){
ans++;
n=n/i;
}
}
if(n>1){
ans++;
}
return ans;
}
static String reverse(String s){
String ans="";
for(int i=s.length()-1;i>=0;i--){
ans+=s.charAt(i);
}
return ans;
}
static int msb(int x){
int ans=0;
while(x!=0){
x=x/2;
ans++;
}
return ans;
}
static void ruffleSort(int[] a) {
int n=a.length;
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static int gcd(int a,int b)
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
/* Iterator<Map.Entry<Integer, Integer>> iterator = map.entrySet().iterator();
while(iterator.hasNext()){
Map.Entry<Integer, Integer> entry = iterator.next();
int value = entry.getValue();
if(value==1){
iterator.remove();
}
else{
entry.setValue(value-1);
}
}
*/
static class Pair implements Comparable
{
int a,b;
public String toString()
{
return a+" " + b;
}
public Pair(int x , int y)
{
a=x;b=y;
}
@Override
public int compareTo(Object o) {
Pair p = (Pair)o;
if(a!=p.a){
return a-p.a;
}
else{
return b-p.b;
}
/*if(p.a!=a){
return a-p.a;//in
}
else{
return b-p.b;//
}*/
}
}
/* public static boolean checkAP(List<Integer> lis){
for(int i=1;i<lis.size()-1;i++){
if(2*lis.get(i)!=lis.get(i-1)+lis.get(i+1)){
return false;
}
}
return true;
}
public static int minBS(int[]arr,int val){
int l=-1;
int r=arr.length;
while(r>l+1){
int mid=(l+r)/2;
if(arr[mid]>=val){
r=mid;
}
else{
l=mid;
}
}
return r;
}
public static int maxBS(int[]arr,int val){
int l=-1;
int r=arr.length;
while(r>l+1){
int mid=(l+r)/2;
if(arr[mid]<=val){
l=mid;
}
else{
r=mid;
}
}
return l;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}*/
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | bf6ff0bc30d81d7762f43f0ad39dc80e | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /* DHUOJ solution #554790 @ 2022-03-26 17:12:32.571 */
import java.util.Scanner;
public class text1 {
public static void main(String[] args) {
Scanner reader=new Scanner(System.in);
int N = reader.nextInt();
long x, y;
long []a=new long[N];
for (int i = 0; i < N; i++) {
x=reader.nextLong();
y=reader.nextLong();
if(x>y){
a[i] = x+y;
}
else
a[i] = y-(y%x)/2;
}
for (int i = 0; i < N; i++) {
System.out.println(a[i]);
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 57a8199747f27888cd1e5e4098ae2ca3 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /* DHUOJ solution #554777 @ 2022-03-26 17:12:28.175 */
import java.util.Scanner;
public class text1 {
public static void main(String[] args) {
Scanner reader=new Scanner(System.in);
int N = reader.nextInt();
long x, y;
long []a=new long[N];
for (int i = 0; i < N; i++) {
x=reader.nextLong();
y=reader.nextLong();
if(x>y){
a[i] = x+y;
}
else
a[i] = y-(y%x)/2;
}
for (int i = 0; i < N; i++) {
System.out.println(a[i]);
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | bebb7fbfb129273b183def982a7df81a | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /* DHUOJ solution #554776 @ 2022-03-26 17:12:28.165 */
import java.util.Scanner;
public class text1 {
public static void main(String[] args) {
Scanner reader=new Scanner(System.in);
int N = reader.nextInt();
long x, y;
long []a=new long[N];
for (int i = 0; i < N; i++) {
x=reader.nextLong();
y=reader.nextLong();
if(x>y){
a[i] = x+y;
}
else
a[i] = y-(y%x)/2;
}
for (int i = 0; i < N; i++) {
System.out.println(a[i]);
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | c1f86e5e61133678e8949f442e86add2 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) throws Exception {
int tc = io.nextInt();
for (int i = 0; i < tc; i++) {
solve();
}
io.close();
}
private static void solve() throws Exception {
int x = io.nextInt();
int y = io.nextInt();
int res;
if (x <= y) {
res = (y - (y % x) / 2);
} else {
res = x + y;
}
io.println(res);
}
static void sort(long[] a) {
ArrayList<Long> l = new ArrayList<>(a.length);
for (long i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
//-----------PrintWriter for faster output---------------------------------
public static FastIO io = new FastIO();
//-----------MyScanner class for faster input----------
static class FastIO extends PrintWriter {
private InputStream stream;
private byte[] buf = new byte[1 << 16];
private int curChar, numChars;
// standard input
public FastIO() {
this(System.in, System.out);
}
public FastIO(InputStream i, OutputStream o) {
super(o);
stream = i;
}
// file input
public FastIO(String i, String o) throws IOException {
super(new FileWriter(o));
stream = new FileInputStream(i);
}
// throws InputMismatchException() if previously detected end of file
private int nextByte() {
if (numChars == -1) throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars == -1) return -1; // end of file
}
return buf[curChar++];
}
// to read in entire lines, replace c <= ' '
// with a function that checks whether c is a line break
public String next() {
int c;
do {
c = nextByte();
} while (c <= ' ');
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = nextByte();
} while (c > ' ');
return res.toString();
}
public String nextLine() {
int c;
do {
c = nextByte();
} while (c < '\n');
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = nextByte();
} while (c > '\n');
return res.toString();
}
public int nextInt() {
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public long nextLong() {
int c;
do {
c = nextByte();
} while (c <= ' ');
int sgn = 1;
if (c == '-') {
sgn = -1;
c = nextByte();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10 * res + c - '0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
int[] nextInts(int n) {
int[] data = new int[n];
for (int i = 0; i < n; i++) {
data[i] = io.nextInt();
}
return data;
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
//--------------------------------------------------------
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 317cd32c9282856529258fa1edbcef28 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class r752b {
public static void main(String[] args) {
FastScanner scan=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
//evenness property means they will never cross without intersecting
int t=scan.nextInt();
for(int tt=0;tt<t;tt++) {
int x=scan.nextInt(), y=scan.nextInt();
// for(int i=x+1;i<y;i++) {
// out.println(i+": "+i%x+" "+y%i);
// }
int n;
if(x>y) n=x+y;
else if(y%x==0) {
n=x;
}
else {
int enda=y%x-1;
n=y-1-(enda-1)/2;
}
out.println(n);
}
out.close();
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
try {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer(br.readLine());
} catch (Exception e){e.printStackTrace();}
}
public String next() {
if (st.hasMoreTokens()) return st.nextToken();
try {st = new StringTokenizer(br.readLine());}
catch (Exception e) {e.printStackTrace();}
return st.nextToken();
}
public int nextInt() {return Integer.parseInt(next());}
public long nextLong() {return Long.parseLong(next());}
public double nextDouble() {return Double.parseDouble(next());}
public String nextLine() {
String line = "";
if(st.hasMoreTokens()) line = st.nextToken();
else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
while(st.hasMoreTokens()) line += " "+st.nextToken();
return line;
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 8c016fb72f9c97d9f17710c17a408e06 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
DModerateModularMode solver = new DModerateModularMode();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class DModerateModularMode {
public void solve(int testNumber, InputReader in, OutputWriter out) {
long x = in.nextInt();
long y = in.nextInt();
if (y < x) {
out.println(x + y);
} else if (y % x == 0) {
out.println(x);
} else {
out.println(y - (y % x) / 2);
}
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void println(long i) {
writer.println(i);
}
}
static class InputReader {
final private int BUFFER_SIZE = 1 << 16;
private InputStream stream;
private byte[] buf = new byte[BUFFER_SIZE];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return nextString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 90a5f0f6cc5c1b7fd69cd19114dfce82 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class D {
public static void main(String args[]){
FScanner in = new FScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-->0) {
int x=in.nextInt(),y=in.nextInt();
if(x==y)
out.println(x);
else if(x>y)
out.println((long)x+y);
else
{
int rem=y%x;
int new_num=y-rem;
int avg=(new_num+y)/2;
out.println(avg);
}
}
out.close();
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static boolean checkprime(int n1)
{
if(n1%2==0||n1%3==0)
return false;
else
{
for(int i=5;i*i<=n1;i+=6)
{
if(n1%i==0||n1%(i+2)==0)
return false;
}
return true;
}
}
static class FScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer sb = new StringTokenizer("");
String next(){
while(!sb.hasMoreTokens()){
try{
sb = new StringTokenizer(br.readLine());
} catch(IOException e){ }
}
return sb.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for(int i=0;i<n;i++) a[i] = nextInt();
return a;
}
float nextFloat(){
return Float.parseFloat(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 8d627188ab383453727e6208e68a1ac0 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class code{
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static int GCD(int a, int b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
//@SuppressWarnings("unchecked")
public static void main(String[] arg) throws IOException{
//Reader in=new Reader();
PrintWriter out = new PrintWriter(System.out);
Scanner in = new Scanner(System.in);
int t=in.nextInt();
while(t-- > 0){
long x=in.nextLong();
long y=in.nextLong();
if(y>=x){
long res=y-(y%x)/2;
out.println(res);
}
else{
long res=x+y;
out.println((res));
}
}
out.flush();
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | ce09155f505af00d1d95b33840956db4 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.io.*;
import java.util.*;
public final class Main {
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
static Pair[] moves = new Pair[]{new Pair(-1, 0), new Pair(0, 1), new Pair(1, 0), new Pair(0, -1)};
static int mod = (int) (1e9 + 7);
static int mod2 = 998244353;
public static void main(String[] args) {
int tt = i();
while (tt-- > 0) {
solve();
}
out.flush();
}
public static void solve() {
long x = l();
long y = l();
if (x > y) {
out.println(x + y);
} else if (x == y) {
out.println(x);
} else {
long k = y / x;
long d = y - k * x;
if (k == 1) {
out.println(x + d / 2);
} else {
long h = x / 2;
out.println(y - d / 2 - h);
}
}
}
static long[] pre(int[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static void print(char A[]) {
for (char c : A) {
out.print(c);
}
out.println();
}
static void print(boolean A[]) {
for (boolean c : A) {
out.print(c + " ");
}
out.println();
}
static void print(int A[]) {
for (int c : A) {
out.print(c + " ");
}
out.println();
}
static void print(long A[]) {
for (long i : A) {
out.print(i + " ");
}
out.println();
}
static void print(List<Integer> A) {
for (int a : A) {
out.print(a + " ");
}
}
static int i() {
return in.nextInt();
}
static long l() {
return in.nextLong();
}
static double d() {
return in.nextDouble();
}
static String s() {
return in.nextLine();
}
static int[][] inputWithIdx(int N) {
int A[][] = new int[N][2];
for (int i = 0; i < N; i++) {
A[i] = new int[]{i, in.nextInt()};
}
return A;
}
static int[] input(int N) {
int A[] = new int[N];
for (int i = 0; i < N; i++) {
A[i] = in.nextInt();
}
return A;
}
static long[] inputLong(int N) {
long A[] = new long[N];
for (int i = 0; i < A.length; i++) {
A[i] = in.nextLong();
}
return A;
}
static int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long GCD(long a, long b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long LCM(int a, int b) {
return (long) a / GCD(a, b) * b;
}
static long LCM(long a, long b) {
return a / GCD(a, b) * b;
}
static void shuffleAndSort(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static void shuffleAndSort(int[][] arr, Comparator<? super int[]> comparator) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int[] temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr, comparator);
}
static void shuffleAndSort(long[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
long temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static boolean isPerfectSquare(double number) {
double sqrt = Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static void swap(int A[], int a, int b) {
int t = A[a];
A[a] = A[b];
A[b] = t;
}
static long pow(long a, long b, int mod) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow = (pow * x) % mod;
}
x = (x * x) % mod;
b /= 2;
}
return pow;
}
static long pow(long a, long b) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow *= x;
}
x = x * x;
b /= 2;
}
return pow;
}
static long modInverse(long x, int mod) {
return pow(x, mod - 2, mod);
}
static boolean isPrime(long N) {
if (N <= 1) {
return false;
}
if (N <= 3) {
return true;
}
if (N % 2 == 0 || N % 3 == 0) {
return false;
}
for (int i = 5; i * i <= N; i = i + 6) {
if (N % i == 0 || N % (i + 2) == 0) {
return false;
}
}
return true;
}
public static String reverse(String str) {
if (str == null) {
return null;
}
return new StringBuilder(str).reverse().toString();
}
public static String repeat(char ch, int repeat) {
if (repeat <= 0) {
return "";
}
final char[] buf = new char[repeat];
for (int i = repeat - 1; i >= 0; i--) {
buf[i] = ch;
}
return new String(buf);
}
public static int[] manacher(String s) {
char[] chars = s.toCharArray();
int n = s.length();
int[] d1 = new int[n];
for (int i = 0, l = 0, r = -1; i < n; i++) {
int k = (i > r) ? 1 : Math.min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && chars[i - k] == chars[i + k]) {
k++;
}
d1[i] = k--;
if (i + k > r) {
l = i - k;
r = i + k;
}
}
return d1;
}
public static int[] kmp(String s) {
int n = s.length();
int[] res = new int[n];
for (int i = 1; i < n; ++i) {
int j = res[i - 1];
while (j > 0 && s.charAt(i) != s.charAt(j)) {
j = res[j - 1];
}
if (s.charAt(i) == s.charAt(j)) {
++j;
}
res[i] = j;
}
return res;
}
}
class Pair {
int i, j;
Pair(int i, int j) {
this.i = i;
this.j = j;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
Pair pair = (Pair) o;
return i == pair.i && j == pair.j;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
class Node {
int data, open, close;
Node() {
this(0, 0, 0);
}
Node(int data, int open, int close) {
this.data = data;
this.open = open;
this.close = close;
}
}
class ST {
int n;
Node[] st;
ST(int n) {
this.n = n;
st = new Node[4 * Integer.highestOneBit(n)];
}
void build(Node[] nodes) {
build(0, 0, n - 1, nodes);
}
private void build(int id, int l, int r, Node[] nodes) {
if (l == r) {
st[id] = nodes[l];
return;
}
int mid = (l + r) >> 1;
build((id << 1) + 1, l, mid, nodes);
build((id << 1) + 2, mid + 1, r, nodes);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
void update(int i, Node node) {
update(0, 0, n - 1, i, node);
}
private void update(int id, int l, int r, int i, Node node) {
if (i < l || r < i) {
return;
}
if (l == r) {
st[id] = node;
return;
}
int mid = (l + r) >> 1;
update((id << 1) + 1, l, mid, i, node);
update((id << 1) + 2, mid + 1, r, i, node);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
Node get(int x, int y) {
return get(0, 0, n - 1, x, y);
}
private Node get(int id, int l, int r, int x, int y) {
if (x > r || y < l) {
return new Node();
}
if (x <= l && r <= y) {
return st[id];
}
int mid = (l + r) >> 1;
return comb(get((id << 1) + 1, l, mid, x, y), get((id << 1) + 2, mid + 1, r, x, y));
}
Node comb(Node a, Node b) {
if (a == null) {
return b;
}
if (b == null) {
return a;
}
int min = Math.min(a.open, b.close);
return new Node(a.data + b.data + min * 2, a.open + b.open - min, a.close + b.close - min);
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | be9c1cdab3812b7bfc825610f9c30e7a | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Locale;
import java.util.StringTokenizer;
public class Solution implements Runnable {
private PrintStream out;
private BufferedReader in;
private StringTokenizer st;
public void solve() throws IOException {
long time0 = System.currentTimeMillis();
int t = nextInt();
for (int test = 1; test <= t; test++) {
int x = nextInt();
int y = nextInt();
long ans = solve(x, y);
if ((ans % x) != (y % ans)) {
throw new RuntimeException();
}
out.println(ans);
}
System.err.println("time: " + (System.currentTimeMillis() - time0));
}
private long solve(int x, int y) {
if (x == y) {
return x;
}
if (x > y) {
return x + y;
}
int xx = (y / x) * x;
int n = (xx + y) / 2;
return n;
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public String next() throws IOException {
while (!st.hasMoreTokens()) {
String line = in.readLine();
if (line == null) {
return null;
}
st = new StringTokenizer(line);
}
return st.nextToken();
}
@Override
public void run() {
try {
solve();
out.close();
} catch (Throwable e) {
throw new RuntimeException(e);
}
}
public Solution(String name) throws IOException {
Locale.setDefault(Locale.US);
if (name == null) {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintStream(new BufferedOutputStream(System.out));
} else {
in = new BufferedReader(new InputStreamReader(new FileInputStream(name + ".in")));
out = new PrintStream(new BufferedOutputStream(new FileOutputStream(name + ".out")));
}
st = new StringTokenizer("");
}
public static void main(String[] args) throws IOException {
new Thread(new Solution(null)).start();
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | cc3d7983d06f9ce96f1edc9c89c232ab | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class B {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(in.readLine());
for (int ti = 0; ti < t; ti++) {
String[] ss = in.readLine().split(" ");
long x=Long.parseLong(ss[0]), y=Long.parseLong(ss[1]);
if (y<x) {
System.out.println(x+y);
} else {
// n=kx+b
// y=l(kx+b)+b = (lk)x+(l+1)b
// y mod x = (l+1)b
// l=1
long l1b = y%x;
long b=l1b/2;
long n=(y-b);
System.out.println(n);
}
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 1ee1a46fb091fd3f5aaae8e506a4a69b | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.lang.*;
// StringBuilder uses java.lang
public class mC {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
StringBuilder st = new StringBuilder();
int t = sc.nextInt();
for (int test = 0; test < t; test++) {
int n = sc.nextInt();
int m = sc.nextInt();
if (n == m) {
st.append(n+"\n");
} else if (n > m) {
st.append(n+m+"\n");
} else {
st.append(m-((m % n)/2) +"\n");
}
}
System.out.print(st.toString());
}
public static int firstLargerAb(int val,ArrayList<Integer> ok,int left,int right) {
if (Math.abs(ok.get(right))<=val) {
return -1;
}
if (left==right) {
return left;
} else if (left+1==right) {
if (val<Math.abs(ok.get(left))) {
return left;
} else {
return right;
}
} else {
int mid = (left+right)/2;
if (Math.abs(ok.get(mid))>val) {
return firstLargerAb(val,ok,left,mid);
} else {
return firstLargerAb(val,ok,mid+1,right);
}
}
}
public static int findNthInArray(int[] arr,int val,int start,int o) {
if (o==0) {
return start-1;
} else if (arr[start] == val) {
return findNthInArray(arr,val,start+1,o-1);
} else {
return findNthInArray(arr,val,start+1,o);
}
}
public static ArrayList<Integer> dfs(int at,ArrayList<Integer> went,ArrayList<ArrayList<Integer>> connect) {
for (int i=0;i<connect.get(at).size();i++) {
if (!(went.contains(connect.get(at).get(i)))) {
went.add(connect.get(at).get(i));
went=dfs(connect.get(at).get(i),went,connect);
}
}
return went;
} public static int[] bfs (int at, int[] went, ArrayList<ArrayList<Integer>> queue, int numNodes, ArrayList<ArrayList<Integer>> connect) {
if (went[at]==0) {
went[at]=queue.get(numNodes).get(1);
for (int i=0;i<connect.get(at).size();i++) {
if (went[connect.get(at).get(i)]==0) {
ArrayList<Integer> temp = new ArrayList<>();
temp.add(connect.get(at).get(i));
temp.add(queue.get(numNodes).get(1)+1);
queue.add(temp);
}
}
}
if (queue.size()==numNodes+1) {
return went;
} else {
return bfs(queue.get(numNodes+1).get(0),went, queue, numNodes+1, connect);
}
}
public static long fastPow(long base,long exp,long mod) {
if (exp==0) {
return 1;
} else {
if (exp % 2 == 1) {
long z = fastPow(base,(exp-1)/2,mod);
return ((((z*base) % mod) * z) % mod);
} else {
long z = fastPow(base,exp/2,mod);
return ((z*z) % mod);
}
}
}
public static int fastPow(int base,long exp) {
if (exp==0) {
return 1;
} else {
if (exp % 2 == 1) {
int z = fastPow(base,(exp-1)/2);
return ((((z*base)) * z));
} else {
int z = fastPow(base,exp/2);
return ((z*z));
}
}
}
public static int firstLarger(int val,ArrayList<Integer> ok,int left,int right) {
if (ok.get(right)<=val) {
return -1;
}
if (left==right) {
return left;
} else if (left+1==right) {
if (val<ok.get(left)) {
return left;
} else {
return right;
}
} else {
int mid = (left+right)/2;
if (ok.get(mid)>val) {
return firstLarger(val,ok,left,mid);
} else {
return firstLarger(val,ok,mid+1,right);
}
}
}
public static int binSearchArr(long val,ArrayList<Integer> ok,long[] arr,int left,int right) {
if (arr[ok.get(right)]<=val) {
return -1;
}
if (left==right) {
return left;
} else if (left+1==right) {
if (val<arr[ok.get(left)]) {
return left;
} else {
return right;
}
} else {
int mid = (left+right)/2;
if (arr[ok.get(mid)]>val) {
return binSearchArr(val,ok,arr,left,mid);
} else {
return binSearchArr(val,ok,arr,mid+1,right);
}
}
}
public static long gcd(long a, long b) {
if (b>a) {
return gcd(b,a);
}
if (b==0) {
return a;
}
if (a%b==0) {
return b;
} else {
return gcd(b,a%b);
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 8 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | eaab29cc1aae52bba0c8cc516af65889 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | //package CodeForces.CodeForcesRounds.src.main.java.aarkay.codeforcesrounds.round752;
import java.io.*;
import java.util.*;
public class ModerateModularMode_B {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
// Start writing your solution here. -------------------------------------
/*
int n = sc.nextInt(); // read input as integer
long k = sc.nextLong(); // read input as long
double d = sc.nextDouble(); // read input as double
String str = sc.next(); // read input as String
String s = sc.nextLine(); // read whole line as String
int result = 3*n;
out.println(result); // print via PrintWriter
*/
int totalCases = sc.nextInt();
while(totalCases != 0) {
int x = sc.nextInt();
int y = sc.nextInt();
out.println(findModerateModularForInput(x, y));
totalCases--;
}
// Stop writing your solution here. -------------------------------------
out.close();
}
// private static long findModerateModularForInput(long x, long y) {
// int i = 1, j = 1; //
// long numerator = ((x * i) + y);
// long denominator = (j + 1);
// long n = (numerator / denominator);
// while(true) {
// for(;; i++) {
// for(j=i;j>0;j-=1) {
// numerator = ((x * i) + y);
// denominator = (j + 1);
// n = (numerator / denominator);
// //System.out.println("i : " + i + " j : " + j + " n : " + n);
// if((n % 1 == 0) && ((n % x) == (y % n))) {
// //System.out.println("ANS to this "+"i : " + i + " j : " + j + " n : " + n);
// return n;
// }
// }
// }
// }
// }
private static int findModerateModularForInput(int x, int y) {
if(x == y) return x;
else if(y < x) return x+y;
else {
int d = y%x;
return (y-d/2);
}
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
//--------------------------------------------------------
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | d25a20ff237204c97f178b49c04848ec | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
public class P1603B {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = s.nextInt();
while (t-- > 0) {
long a = s.nextLong();
long b = s.nextLong();
if (a <= b) {
if (b % a == 0)
out.println(a);
else
out.println(b - (b % a) / 2);
} else
out.println(a + b);
}
out.flush();
s.close();
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | cd5a08b67a21f0ab289a48480de42a49 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.Scanner;
import java.util.*;
public class Solution {
static int mod = (int) 1e9 + 7;
static Scanner sc = new Scanner(System.in);
static StringBuilder sb = new StringBuilder();
public static void main(String[] args) {
int cases = Integer.parseInt(sc.next());
for (int i = 0; i < cases; i++) {
solve();
}
System.out.print(sb);
}
public static void solve() {
long x = Long.parseLong(sc.next());
long y = Long.parseLong(sc.next());
long n = -1;
if (x == y) {
n = x;
} else if (x < y) {
n = y - (y % x) / 2;
} else {
n = x + y;
}
sb.append(n + "\n");
}
public static int countbit(long n) {
int count = 0;
for (int i = 0; i < 64; i++) {
if (((1L << i) & n) != 0) count++;
}
return count;
}
public static int firstbit(int n) {
for (int i = 31; i >= 0; i--) {
if (((1 << i) & n) != 0) return i;
}
return -1;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 5d5039e1fa5a7fc349497503f6727318 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.Scanner;
import java.util.*;
public class Solution {
static int mod = (int) 1e9 + 7;
static Scanner sc = new Scanner(System.in);
static StringBuilder sb = new StringBuilder();
public static void main(String[] args) {
int cases = Integer.parseInt(sc.next());
for (int i = 0; i < cases; i++) {
solve();
}
System.out.print(sb);
}
public static void solve() {
long x = Long.parseLong(sc.next());
long y = Long.parseLong(sc.next());
long n = -1;
if (x == y) {
n = x;
} else if (x < y) {
long p = y / x * x;
n = (p + y) / 2;
} else {
n = x + y;
}
sb.append(n + "\n");
}
public static int countbit(long n) {
int count = 0;
for (int i = 0; i < 64; i++) {
if (((1L << i) & n) != 0) count++;
}
return count;
}
public static int firstbit(int n) {
for (int i = 31; i >= 0; i--) {
if (((1 << i) & n) != 0) return i;
}
return -1;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 7b3e58cdc19309acfd9990ef8a1b1d0e | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
public class pariNa {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
StringBuilder finalAnswer=new StringBuilder();
//finalAnswer.append().append('\n');
int t=sc.nextInt();
outer:
while(t-->0){
// long n=sc.nextLong();
long a=sc.nextInt();
long b=sc.nextInt();
if(a==b || b%a==0) {
// System.out.println(a);
finalAnswer.append(a).append('\n');
}
else if(a>b) {
finalAnswer.append(a+b).append('\n');
// System.out.println(a+b);
}
else{
// System.out.println(18);
finalAnswer.append(((b/a)*a)+((b%a)/2)).append('\n');
// System.out.println(a+((b-a)/2));
}
}
System.out.println(finalAnswer);
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 5c29a00cf544f7d3c55475fed395936a | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
void solve() {
int a = in.nextInt();
int b = in.nextInt();
long res = 0L + a + b;
if (a == b)res = a;
if (a < b) {
res = b - b % a / 2;
}
out.append(res + endl);
println(res % a == b % res);
}
int gcd(int a, int b) {
if (b == 0)return a;
return gcd(b, a % b);
}
public static void main (String[] args) {
// It happens - Syed Mizbahuddin
Main sol = new Main();
int t = 1;
t = in.nextInt();
while (t-- != 0) {
sol.solve();
}
System.out.print(out);
}
<T> void println(T[] s) {
if (err == null)return;
err.println(Arrays.toString(s));
}
<T> void println(T s) {
if (err == null)return;
err.println(s);
}
void println(int s) {
if (err == null)return;
err.println(s);
}
void println(int[] a) {
if (err == null)return;
println(Arrays.toString(a));
}
int[] array(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
return a;
}
int[] array1(int n) {
int[] a = new int[n + 1];
for (int i = 1; i <= n; i++) {
a[i] = in.nextInt();
}
return a;
}
int max(int[] a) {
int max = a[0];
for (int i = 0; i < a.length; i++)max = Math.max(max, a[i]);
return max;
}
int min(int[] a) {
int min = a[0];
for (int i = 0; i < a.length; i++)min = Math.min(min, a[i]);
return min;
}
int count(int[] a, int x) {
int count = 0;
for (int i = 0; i < a.length; i++)if (x == a[i])count++;
return count;
}
void printArray(int[] a) {
for (int ele : a)out.append(ele + " ");
out.append("\n");
}
static {
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
err = new PrintStream(new FileOutputStream("error.txt"));
} catch (Exception e) {}
}
static FastReader in;
static StringBuilder out;
static PrintStream err;
static String yes , no , endl;
final int MAX;
final int MIN;
int mod ;
Main() {
in = new FastReader();
out = new StringBuilder();
MAX = Integer.MAX_VALUE;
MIN = Integer.MIN_VALUE;
mod = (int)1e9 + 7;
yes = "YES\n";
no = "NO\n";
endl = "\n";
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
class Pair implements Comparable<Pair> {
int first , second;
Pair(int first, int second) {
this.first = first;
this.second = second;
}
public int compareTo(Pair b) {
return this.first - b.first;
}
public String toString() {
String s = "{ " + Integer.toString(first) + " , " + Integer.toString(second) + " }";
return s;
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 7e715dbeaefaed643a95cd7fcbd902aa | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
void solve() {
int a = in.nextInt();
int b = in.nextInt();
long res = 0L + a + b;
if (a == b)res = a;
if (a < b) {
res = b - b % a / 2;
}
out.append(res + endl);
println(res % a == b % res);
}
int gcd(int a, int b) {
if (b == 0)return a;
return gcd(b, a % b);
}
public static void main (String[] args) {
// It happens - Syed Mizbahuddin
Main sol = new Main();
int t = 1;
t = in.nextInt();
while (t-- != 0) {
sol.solve();
}
System.out.print(out);
}
<T> void println(T[] s) {
if (err == null)return;
err.println(Arrays.toString(s));
}
<T> void println(T s) {
if (err == null)return;
err.println(s);
}
void println(int s) {
if (err == null)return;
err.println(s);
}
void println(int[] a) {
if (err == null)return;
println(Arrays.toString(a));
}
int[] array(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
return a;
}
int[] array1(int n) {
int[] a = new int[n + 1];
for (int i = 1; i <= n; i++) {
a[i] = in.nextInt();
}
return a;
}
int max(int[] a) {
int max = a[0];
for (int i = 0; i < a.length; i++)max = Math.max(max, a[i]);
return max;
}
int min(int[] a) {
int min = a[0];
for (int i = 0; i < a.length; i++)min = Math.min(min, a[i]);
return min;
}
int count(int[] a, int x) {
int count = 0;
for (int i = 0; i < a.length; i++)if (x == a[i])count++;
return count;
}
void printArray(int[] a) {
for (int ele : a)out.append(ele + " ");
out.append("\n");
}
static {
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
err = new PrintStream(new FileOutputStream("error.txt"));
} catch (Exception e) {}
}
static FastReader in;
static StringBuilder out;
static PrintStream err;
static String yes , no , endl;
final int MAX;
final int MIN;
int mod ;
Main() {
in = new FastReader();
out = new StringBuilder();
MAX = Integer.MAX_VALUE;
MIN = Integer.MIN_VALUE;
mod = (int)1e9 + 7;
yes = "YES\n";
no = "NO\n";
endl = "\n";
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
class Pair implements Comparable<Pair> {
int first , second;
Pair(int first, int second) {
this.first = first;
this.second = second;
}
public int compareTo(Pair b) {
return this.first - b.first;
}
public String toString() {
String s = "{ " + Integer.toString(first) + " , " + Integer.toString(second) + " }";
return s;
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | f373fd0bc525607224117eee0bf65728 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main {
static int globalVariable = 123456789;
static String author = "pl728 on codeforces";
public static void main(String[] args) {
FastReader sc = new FastReader();
PrintWriter pw = new PrintWriter(System.out);
MathUtils mathUtils = new MathUtils();
ArrayUtils arrayUtils = new ArrayUtils();
int tc = sc.ni();
while (tc-- != 0) {
int x = sc.ni(); int y = sc.ni();
// n % x == y % n
if(x == y) {
System.out.println(x);
continue;
}
if(x > y) {
// find a number m greater than y such that m % x == 0
// return m + y
long ans = x + y;
System.out.println(ans);
continue;
}
// if(y % x == 0) {
// System.out.println(x + x/2);
// continue;
// }
System.out.println(y - (y % x) / 2);
}
}
static class FastReader {
/**
* Uses BufferedReader and StringTokenizer for quick java I/O
* get next int, long, double, string
* get int, long, double, string arrays, both primitive and wrapped object when array contains all elements
* on one line, and we know the array size (n)
* next: gets next space separated item
* nextLine: returns entire line as space
*/
BufferedReader br;
StringTokenizer st;
public FastReader() {
this.br = new BufferedReader(new InputStreamReader(System.in));
}
public String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
// to parse something else:
// T x = T.parseT(fastReader.next());
public int ni() {
return Integer.parseInt(next());
}
public String ns() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public String[] readStringArrayLine(int n) {
String line = "";
try {
line = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return line.split(" ");
}
public String[] readStringArrayLines(int n) {
String[] result = new String[n];
for (int i = 0; i < n; i++) {
result[i] = this.next();
}
return result;
}
public int[] readIntArray(int n) {
int[] result = new int[n];
for (int i = 0; i < n; i++) {
result[i] = this.ni();
}
return result;
}
public long[] readLongArray(int n) {
long[] result = new long[n];
for (int i = 0; i < n; i++) {
result[i] = this.nl();
}
return result;
}
public Integer[] readIntArrayObject(int n) {
Integer[] result = new Integer[n];
for (int i = 0; i < n; i++) {
result[i] = this.ni();
}
return result;
}
public long nl() {
return Long.parseLong(next());
}
public char[] readCharArray(int n) {
return this.ns().toCharArray();
}
}
static class MathUtils {
public MathUtils() {
}
public long gcdLong(long a, long b) {
if (a % b == 0)
return b;
else
return gcdLong(b, a % b);
}
public long lcmLong(long a, long b) {
return a * b / gcdLong(a, b);
}
}
static class ArrayUtils {
public ArrayUtils() {
}
public static int[] reverse(int[] a) {
int n = a.length;
int[] b = new int[n];
int j = n;
for (int i = 0; i < n; i++) {
b[j - 1] = a[i];
j = j - 1;
}
return b;
}
public int sumIntArrayInt(int[] a) {
int ans = 0;
for (int i = 0; i < a.length; i++) {
ans += a[i];
}
return ans;
}
public long sumLongArrayLong(int[] a) {
long ans = 0;
for (int i = 0; i < a.length; i++) {
ans += a[i];
}
return ans;
}
}
public static int lowercaseToIndex(char c) {
return (int) c - 97;
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 89fc1c2518c0fd4fab42c27e8fa97c33 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.io.*;
import java.util.*;
public class Codeforces{
static long mod = 1000000007L;
// map.put(a[i],map.getOrDefault(a[i],0)+1);
// map.putIfAbsent;
static MyScanner sc = new MyScanner();
//<----------------------------------------------WRITE HERE------------------------------------------->
static void solve(){
long n = sc.nextInt();
long m = sc.nextLong();
long ans;
if(n<=m) {
if(m%n == 0) ans = n;
else {
ans = m-((m%n)/2);
}
}
else ans = n+m;
out.println(ans);
}
static long phi(long n) {
long res = n;
for(int i=2;i*i<=n;i++) {
if(n%i == 0) {
while(n%i == 0) n /= i;
res -= res/i;
}
}
if(n>1) {
res -= res/n;
}
return res;
}
//<----------------------------------------------WRITE HERE------------------------------------------->
static void swap(int[] a,int i,int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void priArr(int[] a) {
for(int i=0;i<a.length;i++) {
out.print(a[i] + " ");
}
out.println();
}
static long gcd(long a,long b){
if(b==0) return a;
return gcd(b,a%b);
}
static class Pair implements Comparable<Pair>{
Integer val;
Integer ind;
Pair(int v,int f){
val = v;
ind = f;
}
public int compareTo(Pair p){
int s1 = this.ind-this.val;
int s2 = p.ind-p.val;
if(s1 != s2) return s1-s2;
return this.val - p.val;
}
}
public static void main(String[] args) {
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
// int t= 1;
while(t-- >0){
solve();
}
// Stop writing your solution here. -------------------------------------
out.close();
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int[] readIntArray(int n){
int arr[] = new int[n];
for(int i = 0;i<n;i++){
arr[i] = Integer.parseInt(next());
}
return arr;
}
int[] reverse(int arr[]){
int n= arr.length;
int i = 0;
int j = n-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
j--;i++;
}
return arr;
}
long[] readLongArray(int n){
long arr[] = new long[n];
for(int i = 0;i<n;i++){
arr[i] = Long.parseLong(next());
}
return arr;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
private static void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 5d61b6e04b2a64e12015c5f1f45049ce | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.lang.*;
import java.util.*;
public class ComdeFormces {
public static boolean gl;
public static ArrayList<Integer> anss;
public static int ans;
public static void main(String[] args) throws Exception{
// TODO Auto-generated method stub
// Reader.init(System.in);
FastReader sc=new FastReader();
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
// OutputStream out = new BufferedOutputStream ( System.out );
int t=sc.nextInt();
int tc=1;
while(t--!=0) {
long n=sc.nextLong();
long m=sc.nextLong();
if(n<=m) {
long nm=n*(m/n)+(m%n)/2;
log.write(nm+"\n");
}
else log.write((n+m)+"\n");
log.flush();
}
}
public static class Node{
int data;
Node left,right;
public Node(int data) {
this.data=data;
this.left=this.right=null;
}
}
public static int dfs(ArrayList<ArrayList<Integer>> ar,int src, boolean vis[],int cnt[]) {
vis[src]=true;
for(int k:ar.get(src)) {
if(!vis[k]) {
cnt[src]+=dfs(ar,k,vis,cnt);
}
}
cnt[src]+=1;
return cnt[src];
}
public static int eval(ArrayList<ArrayList<Integer>> ar,int cnt[],int src,boolean vis[]) {
vis[src]=true;
int vl=-1;
int nd=-1;
int ch[]= {-1,-1};
int p=0;
for(int k:ar.get(src)) {
if(!vis[k]) {
ch[p++]=k;
}
}
if(ch[0]==-1 && ch[1]==-1)return 0;
int ans=0;
for(int i=0;i<2;i++) {
if(i==0) {
if(ch[1]==-1) {
ans+=cnt[ch[0]]-1;
return ans;
}
else {
ans=Math.max(ans, cnt[ch[1]]-1+eval(ar,cnt,ch[0],vis));
}
}
else {
ans=Math.max(ans, cnt[ch[0]]-1+eval(ar,cnt,ch[1],vis));
}
}
return ans;
}
public static void build(long a[][],long b[]) {
for(int i=0;i<b.length;i++) {
a[i][0]=b[i];
}
int jmp=2;
while(jmp<=b.length) {
for(int j=0;j<b.length;j++) {
int ind=(int)(Math.log(jmp/2)/Math.log(2));
int ind2=(int)(Math.log(jmp)/Math.log(2));
if(j+jmp-1<b.length) {
a[j][ind2]=Math.max(a[j][ind],a[j+(jmp/2)][ind]);
}
}
jmp*=2;
}
// for(int i=0;i<a.length;i++) {
// for(int j=0;j<33;j++) {
// System.out.print(a[i][j]+" ");
// }
// System.out.println();
// }
}
public static long serst(long a[][],int i,int j) {
int len=j-i+1;
int hp=1;
int tlen=len>>=1;
// System.out.println(tlen);
while(tlen!=0) {
tlen>>=1;
hp<<=1;
}
// System.out.println(hp);
int ind=(int)(Math.log(hp)/Math.log(2));
int i2=j+1-hp;
return Math.max(a[i][ind], a[i2][ind]);
}
static void update(int f[],int upd,int ind) {
int vl=ind;
while(vl<f.length) {
f[vl]+=upd;
int tp=~vl;
tp++;
tp&=vl;
vl+=tp;
}
}
static long ser(int f[],int ind) {
int vl=ind;
long sm=0;
while(vl!=0) {
sm+=f[vl];
int tp=~vl;
tp++;
tp&=vl;
vl-=tp;
}
return sm;
}
static int find(int el,int p[]) {
if(p[el]<0)return el;
return p[el]=find(p[el],p);
}
static boolean union(int a,int b,int p[]) {
int p1=find(a,p);
int p2=find(b,p);
if(p1>=0 && p1==p2)return false;
else {
if(p[p1]<p[p2]) {
p[p1]+=p[p2];
p[p2]=p1;
}
else {
p[p2]+=p[p1];
p[p1]=p2;
}
return true;
}
}
public static void radixSort(int a[]) {
int n=a.length;
int res[]=new int[n];
int p=1;
for(int i=0;i<=8;i++) {
int cnt[]=new int[10];
for(int j=0;j<n;j++) {
a[j]=res[j];
cnt[(a[j]/p)%10]++;
}
for(int j=1;j<=9;j++) {
cnt[j]+=cnt[j-1];
}
for(int j=n-1;j>=0;j--) {
res[cnt[(a[j]/p)%10]-1]=a[j];
cnt[(a[j]/p)%10]--;
}
p*=10;
}
}
static int bits(long n) {
int ans=0;
while(n!=0) {
if((n&1)==1)ans++;
n>>=1;
}
return ans;
}
public static int kadane(int a[]) {
int sum=0,mx=Integer.MIN_VALUE;
for(int i=0;i<a.length;i++) {
sum+=a[i];
mx=Math.max(mx, sum);
if(sum<0) sum=0;
}
return mx;
}
public static int m=(int)(1e9+7);
public static int mul(int a, int b) {
return ((a%m)*(b%m))%m;
}
public static long mul(long a, long b) {
return ((a%md)*(b%md))%md;
}
public static int add(int a, int b) {
return ((a%m)+(b%m))%m;
}
public static long add(long a, long b) {
return ((a%m)+(b%m))%m;
}
//debug
public static <E> void p(E[][] a,String s) {
System.out.println(s);
for(int i=0;i<a.length;i++) {
for(int j=0;j<a[0].length;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
public static void p(int[] a,String s) {
System.out.print(s+"=");
for(int i=0;i<a.length;i++)System.out.print(a[i]+" ");
System.out.println();
}
public static void p(long[] a,String s) {
System.out.print(s+"=");
for(int i=0;i<a.length;i++)System.out.print(a[i]+" ");
System.out.println();
}
public static <E> void p(E a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(ArrayList<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(LinkedList<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(HashSet<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(Stack<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(Queue<E> a,String s){
System.out.println(s+"="+a);
}
//utils
static ArrayList<Integer> divisors(int n){
ArrayList<Integer> ar=new ArrayList<>();
for (int i=2; i<=Math.sqrt(n); i++){
if (n%i == 0){
if (n/i == i) {
ar.add(i);
}
else {
ar.add(i);
ar.add(n/i);
}
}
}
return ar;
}
static ArrayList<Integer> prime(int n){
ArrayList<Integer> ar=new ArrayList<>();
int cnt=0;
boolean pr=false;
while(n%2==0) {
ar.add(2);
n/=2;
}
for(int i=3;i*i<=n;i+=2) {
pr=false;
while(n%i==0) {
n/=i;
ar.add(i);
pr=true;
}
}
if(n>2) ar.add(n);
return ar;
}
static long gcd(long a,long b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static int gcd(int a,int b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static long factmod(long n,long mod,long img) {
if(n==0)return 1;
long ans=1;
long temp=1;
while(n--!=0) {
if(temp!=img) {
ans=((ans%mod)*((temp)%mod))%mod;
}
temp++;
}
return ans%mod;
}
static int ncr(int n, int r){
if(r>n-r)r=n-r;
int ans=1;
for(int i=0;i<r;i++){
ans*=(n-i);
ans/=(i+1);
}
return ans;
}
public static class trip{
double a;
int b;
int c;
public trip(double a,int b,int c) {
this.a=a;
this.b=b;
this.c=c;
}
// public int compareTo(trip q) {
// return this.b-q.b;
// }
}
static void mergesort(long[] a,int start,int end) {
if(start>=end)return ;
int mid=start+(end-start)/2;
mergesort(a,start,mid);
mergesort(a,mid+1,end);
merge(a,start,mid,end);
}
static void merge(long[] a, int start,int mid,int end) {
int ptr1=start;
int ptr2=mid+1;
long b[]=new long[end-start+1];
int i=0;
while(ptr1<=mid && ptr2<=end) {
if(a[ptr1]<a[ptr2]) {
b[i]=a[ptr1];
ptr1++;
i++;
}
else {
b[i]=a[ptr2];
ptr2++;
i++;
}
}
while(ptr1<=mid) {
b[i]=a[ptr1];
ptr1++;
i++;
}
while(ptr2<=end) {
b[i]=a[ptr2];
ptr2++;
i++;
}
for(int j=start;j<=end;j++) {
a[j]=b[j-start];
}
}
public static class FastReader {
BufferedReader b;
StringTokenizer s;
public FastReader() {
b=new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(s==null ||!s.hasMoreElements()) {
try {
s=new StringTokenizer(b.readLine());
}
catch(IOException e) {
e.printStackTrace();
}
}
return s.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str="";
try {
str=b.readLine();
}
catch(IOException e) {
e.printStackTrace();
}
return str;
}
boolean hasNext() {
if (s != null && s.hasMoreTokens()) {
return true;
}
String tmp;
try {
b.mark(1000);
tmp = b.readLine();
if (tmp == null) {
return false;
}
b.reset();
} catch (IOException e) {
return false;
}
return true;
}
}
public static class pair{
int a;
int b;
public pair(int a,int b) {
this.a=a;
this.b=b;
}
// public int compareTo(pair b) {
// return this.a-b.a;
//
// }
// // public int compareToo(pair b) {
// return this.b-b.b;
// }
@Override
public String toString() {
return "{"+this.a+" "+this.b+"}";
}
}
static long pow(long a, long pw) {
long temp;
if(pw==0)return 1;
temp=pow(a,pw/2);
if(pw%2==0)return temp*temp;
return a*temp*temp;
}
public static int md=998244353;
static long mpow(long a, long pw) {
long temp;
if(pw==0)return 1;
temp=pow(a,pw/2)%md;
if(pw%2==0)return mul(temp,temp);
return mul(a,mul(temp,temp));
}
static int pow(int a, int pw) {
int temp;
if(pw==0)return 1;
temp=pow(a,pw/2);
if(pw%2==0)return temp*temp;
return a*temp*temp;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | b8909a4c66fc14ee7334526aba74acd9 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import static java.lang.Math.*;
import static java.lang.Math.ceil;
import static java.util.Arrays.sort;
public class Round9 {
public static void main(String[] args) {
FastReader fastReader = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = fastReader.nextInt();
while (t-- > 0) {
long a = fastReader.nextLong();
long b = fastReader.nextLong();
//
if ( a == b) {
out.println(Math.min(a, b));
} else {
if (a > b) {
out.println(a + b);
} else {
long div = b / a;
long a1 = div * a;
if (b % a == 0) {
a1 -= a;
}
long mid = (Math.max(a1, a) + b) / 2;
out.println(mid);
}
}
}
out.close();
}
// constants
static final int IBIG = 1000000007;
static final int IMAX = 2147483647;
static final long LMAX = 9223372036854775807L;
static Random __r = new Random();
// math util
static int minof(int a, int b, int c) {
return min(a, min(b, c));
}
static int minof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
int min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static long minof(long a, long b, long c) {
return min(a, min(b, c));
}
static long minof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
long min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static int maxof(int a, int b, int c) {
return max(a, max(b, c));
}
static int maxof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
int max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static long maxof(long a, long b, long c) {
return max(a, max(b, c));
}
static long maxof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
long max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static int powi(int a, int b) {
if (a == 0) return 0;
int ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static long powl(long a, int b) {
if (a == 0) return 0;
long ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static int fli(double d) {
return (int) d;
}
static int cei(double d) {
return (int) ceil(d);
}
static long fll(double d) {
return (long) d;
}
static long cel(double d) {
return (long) ceil(d);
}
static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
static int[] exgcd(int a, int b) {
if (b == 0) return new int[]{1, 0};
int[] y = exgcd(b, a % b);
return new int[]{y[1], y[0] - y[1] * (a / b)};
}
static long[] exgcd(long a, long b) {
if (b == 0) return new long[]{1, 0};
long[] y = exgcd(b, a % b);
return new long[]{y[1], y[0] - y[1] * (a / b)};
}
static int randInt(int min, int max) {
return __r.nextInt(max - min + 1) + min;
}
static long mix(long x) {
x += 0x9e3779b97f4a7c15L;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L;
x = (x ^ (x >> 27)) * 0x94d049bb133111ebL;
return x ^ (x >> 31);
}
public static boolean[] findPrimes(int limit) {
assert limit >= 2;
final boolean[] nonPrimes = new boolean[limit];
nonPrimes[0] = true;
nonPrimes[1] = true;
int sqrt = (int) Math.sqrt(limit);
for (int i = 2; i <= sqrt; i++) {
if (nonPrimes[i]) continue;
for (int j = i; j < limit; j += i) {
if (!nonPrimes[j] && i != j) nonPrimes[j] = true;
}
}
return nonPrimes;
}
// array util
static void reverse(int[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
int swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(long[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
long swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(double[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
double swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
char swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void shuffle(int[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
int swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(long[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
long swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(double[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
double swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void rsort(int[] a) {
shuffle(a);
sort(a);
}
static void rsort(long[] a) {
shuffle(a);
sort(a);
}
static void rsort(double[] a) {
shuffle(a);
sort(a);
}
static int[] copy(int[] a) {
int[] ans = new int[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static long[] copy(long[] a) {
long[] ans = new long[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static double[] copy(double[] a) {
double[] ans = new double[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static char[] copy(char[] a) {
char[] ans = new char[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 06f878014d490f1fdae886992fbaab87 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main{
static final Random random=new Random();
static long mod=1000000007L;
static HashMap<String,Integer>map=new HashMap<>();
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while(st==null || !st.hasMoreTokens()){
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
int[] readIntArray(int n){
int[] res=new int[n];
for(int i=0;i<n;i++)res[i]=nextInt();
return res;
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
public static void main(String[] args) {
try {
FastReader in=new FastReader();
FastWriter out = new FastWriter();
int testCases=in.nextInt();
//int testCases=1;
while(testCases-- > 0){
solve(in);
}
out.close();
} catch (Exception e) {
return;
}
}
public static void solve( FastReader in){
//int n=in.nextInt();
//int w=in.nextInt();
//int h=in.nextInt();
long n=in.nextLong();
//int k=in.nextInt();
long k=in.nextLong();
StringBuilder res=new StringBuilder();
if(n<=k){
long y=k-(k%n)/2;
//a=n+a;
res.append(""+(y));
}
else{
long a=n+k;
res.append(""+(a));
}
System.out.println(res.toString());
}
static int gcd(int a,int b){
if(b==0){
return a;
}
return gcd(b,a%b);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static void debug(String x){
System.out.println(x);
}
static < E > void print(E res)
{
System.out.println(res);
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 3cef847819a98d54efff61c65b71681f | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.Scanner;
public class E1603B {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
StringBuilder sb = new StringBuilder();
while (t-- > 0) {
int x = scn.nextInt();
int y = scn.nextInt();
long ans;
if (x > y) {
ans = x + y;
} else {
ans = y - (y % x) / 2;
}
sb.append(ans);
sb.append("\n");
}
System.out.println(sb);
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 5140b854e52bcde862bb0b4f1a3a9976 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /*
_oo0oo_
o8888888o
88" . "88
(| -_- |)
0\ = /0
___/`---'\___
.' \\| |// '.
/ \\||| : |||// \
/ _||||| -:- |||||- \
| | \\\ - /// | |
| \_| ''\---/'' |_/ |
\ .-\__ '-' ___/-. /
___'. .' /--.--\ `. .'___
."" '< `.___\_<|>_/___.' >' "".
| | : `- \`.;`\ _ /`;.`/ - ` : | |
\ \ `_. \_ __\ /__ _/ .-` / /
=====`-.____`.___ \_____/___.-`___.-'=====
`=---='
*/
import java.util.*;
import java.math.*;
import java.io.*;
import java.lang.Math.*;
public class KickStart2020 {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
float nextFloat() {
return Float.parseFloat(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long lcm(long a, long b) {
return a / gcd(a, b) * b;
}
public static class Pair implements Comparable<Pair> {
public int index;
public int value;
public Pair(int index,int value) {
this.index = index;
this.value = value;
}
@Override
public int compareTo(Pair other) {
// multiplied to -3 as the author need descending sort order
if(other.index < this.index) return 3;
if(other.index > this.index) return -3;
else return 0;
}
@Override
public String toString() {
return this.index + " " + this.value;
}
}
static boolean isPrime(long d) {
if (d == 3)
return false;
for (int i = 2; i <= (long) Math.sqrt(d); i++) {
if (d % i == 0)
return false;
}
return true;
}
static void decimalTob(int n, int k, Stack<Integer> ss) {
int x = n % k;
int y = n / k;
ss.push(x);
if(y > 0) {
decimalTob(y, k, ss);
}
}
static long powermod(long x, long y, long mod) {
long ans = 1;
x = x % mod;
if (x == 0)
return 0;
int i = 1;
while (y > 0) {
if ((y & 1) != 0)
ans = (ans * x) % mod;
y = y >> 1;
x = (x * x) % mod;
}
return ans;
}
static long power(long x, long y)
{
long res = 1;
while (y > 0)
{
if ((y & 1) != 0)
res = res * x;
y = y >> 1;
x = x * x;
}
return res;
}
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
outerloop:
while(t-- > 0) {
long a = sc.nextLong();
long b = sc.nextLong();
if(b % a == 0) {
out.println(a);
continue outerloop;
}
if(b < a) out.println(a + b);
else {
out.println(b - (b % a) / 2);
}
}
out.close();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | de3c1ccf3d933bacbe136c984ad2d388 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
static AReader scan = new AReader();
static int gcd(int a,int b){
return b == 0 ? a : gcd(b,a%b);
}
static void solve() {
int x = scan.nextInt();
int y = scan.nextInt();
if(y < x){
System.out.println(x + y);
}else if(x == y){
System.out.println(x);
}else{
if(y % x == 0) System.out.println(y);
// y > x?
else{
System.out.println(y - (y % x / 2));
}
}
}
public static void main(String[] args) {
int T = scan.nextInt();
while (T-- > 0) {
solve();
}
}
}
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
class Pair {
int x,y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
class Node{
int l,r;
int val;
int lazy;
int cnt = 0,lnum,rnum;
public Node(int l, int r) {
this.l = l;
this.r = r;
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 423e3510d7a3a57129405ffd84240bdc | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class B1603 {
public static void main(String[] args) throws IOException, FileNotFoundException {
// Scanner in = new Scanner(new File("test.in"));
Kattio in = new Kattio();
int T = in.nextInt();
while(T > 0){
T--;
long X = in.nextLong();
long Y = in.nextLong();
if(X > Y){
System.out.println(X + Y);
} else {
System.out.println(Y - (Y%X)/2);
}
}
}
static long gcd (long a, long b) {
if (b == 0){
return a;
} else {
return gcd (b, a % b);
}
}
static long lcm (long a, long b){
return a / gcd(a, b) * b;
}
static class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio() { this(System.in, System.out); }
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio(String problemName) throws IOException {
super(problemName + ".out");
r = new BufferedReader(new FileReader(problemName + ".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) { }
return null;
}
public int nextInt() { return Integer.parseInt(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
public long nextLong() { return Long.parseLong(next()); }
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 806227d219821db30e15ddb09e78657d | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class A1603 {
public static void main(String[] args) throws IOException {
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int numTests = Integer.parseInt(rd.readLine());
for (int t = 0; t < numTests; t++) {
StringTokenizer st = new StringTokenizer(rd.readLine());
long x = Long.parseLong(st.nextToken());
long y = Long.parseLong(st.nextToken());
long ans;
if (y % x == 0) {
ans = x;
} else {
// find m s.t. (2m-1)x < y < (2m+1)y.
if (y < x) {
ans = x + y;
} else {
// find largest m s.t. (2m-1)x<y.
long div = (y - 1) / x;
long m = div;
if (m % 2 == 0) {
m--;
}
ans = (y + m * x) / 2;
}
}
pw.println(ans);
}
pw.flush();
pw.close();
rd.close();
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 8d9fb773fabe54b09f59ba82bbb6661c | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
public class cf1603b {
public static void main(String[] args) throws IOException {
int t = ri();
next: while (t --> 0) {
int x = rni(), y = ni();
if (y < x) {
prln(x + y);
} else if (y % x == 0) {
prln(x);
} else {
prln((y + ((y + x - 1) / x - 1) * x) / 2);
}
}
close();
}
@FunctionalInterface
interface IntCondition {
boolean check(int x);
}
static int maxst(int min, int max, IntCondition condition) {
int ans = min - 1;
for (int l = min, r = max, m; l <= r; ) {
if (condition.check(m = l + (r - l) / 2)) l = (ans = m) + 1;
else r = m - 1;
}
return ans;
}
static int minst(int min, int max, IntCondition condition) {
int ans = max + 1;
for (int l = min, r = max, m; l <= r; ) {
if (condition.check(m = l + (r - l) / 2)) r = (ans = m) - 1;
else l = m + 1;
}
return ans;
}
static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
static StringTokenizer input;
static Random __r = new Random();
// references
// IBIG = 1e9 + 7
// IMAX ~= 2e9
// LMAX ~= 9e18
// constants
static final int IBIG = 1000000007;
static final int IMAX = 2147483647;
static final long LMAX = 9223372036854775807L;
// math util
static int minof(int a, int b, int c) {return min(a, min(b, c));}
static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
static long minof(long a, long b, long c) {return min(a, min(b, c));}
static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
static int maxof(int a, int b, int c) {return max(a, max(b, c));}
static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
static long maxof(long a, long b, long c) {return max(a, max(b, c));}
static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
static int fli(double d) {return (int) d;}
static int cei(double d) {return (int) ceil(d);}
static long fll(double d) {return (long) d;}
static long cel(double d) {return (long) ceil(d);}
static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
static long gcd(long a, long b) {return b == 0 ? a : gcd(b, a % b);}
static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b)};}
static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b)};}
static int randInt(int min, int max) {return __r.nextInt(max - min + 1) + min;}
static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
// array util
static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void rsort(int[] a) {shuffle(a); sort(a);}
static void rsort(long[] a) {shuffle(a); sort(a);}
static void rsort(double[] a) {shuffle(a); sort(a);}
static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
// input
static void r() throws IOException {input = new StringTokenizer(rline());}
static int ri() throws IOException {return Integer.parseInt(rline());}
static long rl() throws IOException {return Long.parseLong(rline());}
static double rd() throws IOException {return Double.parseDouble(rline());}
static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
static void ria(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni();}
static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
static void riam1(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1;}
static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
static void rla(long[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nl();}
static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
static void rda(double[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nd();}
static char[] rcha() throws IOException {return rline().toCharArray();}
static void rcha(char[] a) throws IOException {int n = a.length, i = 0; for (char c : rline().toCharArray()) a[i++] = c;}
static String rline() throws IOException {return __i.readLine();}
static String n() {return input.nextToken();}
static int rni() throws IOException {r(); return ni();}
static int ni() {return Integer.parseInt(n());}
static long rnl() throws IOException {r(); return nl();}
static long nl() {return Long.parseLong(n());}
static double rnd() throws IOException {r(); return nd();}
static double nd() {return Double.parseDouble(n());}
// output
static void pr(int i) {__o.print(i);}
static void prln(int i) {__o.println(i);}
static void pr(long l) {__o.print(l);}
static void prln(long l) {__o.println(l);}
static void pr(double d) {__o.print(d);}
static void prln(double d) {__o.println(d);}
static void pr(char c) {__o.print(c);}
static void prln(char c) {__o.println(c);}
static void pr(char[] s) {__o.print(new String(s));}
static void prln(char[] s) {__o.println(new String(s));}
static void pr(String s) {__o.print(s);}
static void prln(String s) {__o.println(s);}
static void pr(Object o) {__o.print(o);}
static void prln(Object o) {__o.println(o);}
static void prln() {__o.println();}
static void pryes() {prln("yes");}
static void pry() {prln("Yes");}
static void prY() {prln("YES");}
static void prno() {prln("no");}
static void prn() {prln("No");}
static void prN() {prln("NO");}
static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
static void h() {prln("hlfd"); flush();}
static void flush() {__o.flush();}
static void close() {__o.close();}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 9482ee31b654dd37467d7fedb905f780 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class mmm {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
while (t-- > 0) {
st = new StringTokenizer(br.readLine());
long x = Long.parseLong(st.nextToken());
long y = Long.parseLong(st.nextToken());
if (y % x == 0) {
pw.println(x);
} else if (x > y) {
pw.println(x + y);
} else {
long ceil = y - 1;
long xv = ceil % x;
long yv = y % ceil;
pw.println(ceil - (xv - yv) / 2);
}
}
pw.close();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 4cabc8aad5b88d0d7d2b5da583e72070 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | // import java.io.BufferedReader;
// import java.io.InputStreamReader;
// public class Main{
// static int m=(int)1e9+7;
// public static void main(String[] args) throws Exception{
// BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// int t=Integer.parseInt(br.readLine());
// while(t-->0){
// String[] in=br.readLine().split(" ");
// int a=Integer.parseInt(in[0]);
// int b=Integer.parseInt(in[1]);
// int c=Integer.parseInt(in[2]);
// // System.out.println(expo(b,c,m-1));
// System.out.println(expo(a,expo(b,c,m-1),m));
// }
// }
// static int expo(int a,int b,int c){
// if(a==0 && b==0) return 1;
// long res=1;
// long x=a;
// while(b>0){
// if((b&1)==1){
// res=(x*res)%c;
// }
// x=(x*x)%c;
// b=b>>1;
// }
// return (int)(res%c);
// }
// }
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class Main{
static long mod=(long)1e9+7;
public static void main(String[] args) throws Exception{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
// long[] dp=new long[(int)(2*1e6+1)];
// dp[1]=0;
// dp[2]=1;
// for(int i=3;i<=(int)(2*1e6);i++){
// dp[i]=((i-1)*(dp[i-1]%mod+dp[i-2]%mod))%mod;
// }
while(t-->0){
// int n=Integer.parseInt(br.readLine());
String[] in=br.readLine().split(" ");
long a=Long.parseLong(in[0]);
long b=Long.parseLong(in[1]);
// long c=(dp[(int)a])*(modular_inverse((dp[(int)b]*dp[(int)(a-b)])%mod))%mod;
// System.out.println(a*b%mod);
if(a<=b){
if(b%a==0) System.out.println(a);
else{
long p=b-b%a;
System.out.println((p+b)/2);
}
}
else{
System.out.println(a+b);
}
}
}
static long n_of_factors(long[][] a){
long ans=1;
for(int i=0;i<a.length;i++){
ans=ans*(a[i][1]+1)%mod;
}
return ans%mod;
}
static long sum_of_factors(long[][] a){
long ans=1;
for(int i=0;i<a.length;i++){
long res=(((expo(a[i][0],a[i][1]+1)-1)%mod)*(modular_inverse(a[i][0]-1))%mod)%mod;
ans=((res%mod)*(ans%mod))%mod;
}
return (ans%mod);
}
static long prod_of_divisors(long[][] a){
long ans=1;
long d=1;
for(int i=0;i<a.length;i++){
long res=expo(a[i][0],(a[i][1]*(a[i][1]+1)/2));
ans=((expo(ans,a[i][1]+1)%mod)*(expo(res,d)%mod))%mod;
d=(d*(a[i][1]+1))%(mod-1);
}
return ans%mod;
}
static long expo(long a,long b){
long ans=1;
a=a%mod;
while(b>0){
if((b&1)==1){
ans=ans*a%mod;
}
a=a*a%mod;
b>>=1;
}
return ans%mod;
}
static long modular_inverse(long a){
return expo(a,mod-2)%mod;
}
static long phin(long a){
if(isPrime(a)) return a-1;
long res=a;
for(int i=2;i*i<=(int)a;i++){
if(a%i==0){
while(a%i==0){
a=a/i;
}
res-=res/i;
}
}
if(a>1){
res-=res/a;
}
return res;
}
static boolean isPrime(long a){
if(a<2) return false;
for(int i=2;i*i<=(int)a;i++){
if(a%i==0) return false;
}
return true;
}
// }
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | bf004dbdcb8b886e7db5656a33c598d1 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Moderate_Modular_Mode {
static FastScanner fs;
static FastWriter fw;
static boolean checkOnlineJudge = System.getProperty("ONLINE_JUDGE") == null;
private static final int[][] kdir = new int[][]{{-1, 2}, {-2, 1}, {-2, -1}, {-1, -2}, {1, -2}, {2, -1}, {2, 1}, {1, 2}};
private static final int[][] dir = new int[][]{{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
private static final int iMax = (int) (1e9 + 100), iMin = (int) (-1e9 - 100);
private static final long lMax = (long) (1e18 + 100), lMin = (long) (-1e18 - 100);
private static final int mod1 = (int) (1e9 + 7);
private static final int mod2 = 998244353;
public static void main(String[] args) throws IOException {
fs = new FastScanner();
fw = new FastWriter();
int t = 1;
t = fs.nextInt();
while (t-- > 0) {
solve();
}
fw.out.close();
}
private static void solve() {
long x = fs.nextLong(), y = fs.nextLong();
if (x == y) {
fw.out.println(x);
}else if (x > y) {
long n = x + y;
fw.out.println(n);
} else if ((3 * x) > y) {
long n = (x + y) / 2;
fw.out.println(n);
} else {
long k1 = ceilDiv(y - (2 * x), x);
if ((y - (2 * x)) % x == 0) k1++;
long n = ((k1 * x) + y) / 2;
fw.out.println(n);
}
}
private static class UnionFind {
private final int[] parent;
private final int[] rank;
UnionFind(int n) {
parent = new int[n + 5];
rank = new int[n + 5];
for (int i = 0; i <= n; i++) {
parent[i] = i;
rank[i] = 0;
}
}
private int find(int i) {
if (parent[i] == i)
return i;
return parent[i] = find(parent[i]);
}
private void union(int a, int b) {
a = find(a);
b = find(b);
if (a != b) {
if (rank[a] < rank[b]) {
int temp = a;
a = b;
b = temp;
}
parent[b] = a;
if (rank[a] == rank[b])
rank[a]++;
}
}
}
private static class Calc_nCr {
private final long[] fact;
private final long[] invfact;
private final int p;
Calc_nCr(int n, int prime) {
fact = new long[n + 5];
invfact = new long[n + 5];
p = prime;
fact[0] = 1;
for (int i = 1; i <= n; i++) {
fact[i] = (i * fact[i - 1]) % p;
}
invfact[n] = pow(fact[n], p - 2, p);
for (int i = n - 1; i >= 0; i--) {
invfact[i] = (invfact[i + 1] * (i + 1)) % p;
}
}
private long nCr(int n, int r) {
if (r > n || n < 0 || r < 0) return 0;
return (((fact[n] * invfact[r]) % p) * invfact[n - r]) % p;
}
}
private static long gcd(long a, long b) {
return (b == 0 ? a : gcd(b, a % b));
}
private static long lcm(long a, long b) {
return ((a * b) / gcd(a, b));
}
private static long pow(long a, long b, int mod) {
long result = 1;
while (b > 0) {
if ((b & 1L) == 1) {
result = (result * a) % mod;
}
a = (a * a) % mod;
b >>= 1;
}
return result;
}
private static long ceilDiv(long a, long b) {
return ((a + b - 1) / b);
}
private static long getMin(long... args) {
long min = lMax;
for (long arg : args)
min = Math.min(min, arg);
return min;
}
private static long getMax(long... args) {
long max = lMin;
for (long arg : args)
max = Math.max(max, arg);
return max;
}
private static boolean isPalindrome(String s, int l, int r) {
int i = l, j = r;
while (j - i >= 1) {
if (s.charAt(i) != s.charAt(j))
return false;
i++;
j--;
}
return true;
}
private static List<Integer> primes(int n) {
boolean[] primeArr = new boolean[n + 5];
Arrays.fill(primeArr, true);
for (int i = 2; (i * i) <= n; i++) {
if (primeArr[i]) {
for (int j = i * i; j <= n; j += i) {
primeArr[j] = false;
}
}
}
List<Integer> primeList = new ArrayList<>();
for (int i = 2; i <= n; i++) {
if (primeArr[i])
primeList.add(i);
}
return primeList;
}
private static int noOfSetBits(long x) {
int cnt = 0;
while (x != 0) {
x = x & (x - 1);
cnt++;
}
return cnt;
}
private static boolean isPerfectSquare(long num) {
long sqrt = (long) Math.sqrt(num);
return ((sqrt * sqrt) == num);
}
private static class Pair<U, V> {
private final U first;
private final V second;
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return first.equals(pair.first) && second.equals(pair.second);
}
@Override
public int hashCode() {
return Objects.hash(first, second);
}
@Override
public String toString() {
return "(" + first + ", " + second + ")";
}
private Pair(U ff, V ss) {
this.first = ff;
this.second = ss;
}
}
private static void randomizeIntArr(int[] arr, int n) {
Random r = new Random();
for (int i = (n - 1); i > 0; i--) {
int j = r.nextInt(i + 1);
swapInIntArr(arr, i, j);
}
}
private static void randomizeLongArr(long[] arr, int n) {
Random r = new Random();
for (int i = (n - 1); i > 0; i--) {
int j = r.nextInt(i + 1);
swapInLongArr(arr, i, j);
}
}
private static void swapInIntArr(int[] arr, int a, int b) {
int temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
}
private static void swapInLongArr(long[] arr, int a, int b) {
long temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
}
private static int[] readIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = fs.nextInt();
return arr;
}
private static long[] readLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = fs.nextLong();
return arr;
}
private static List<Integer> readIntList(int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++)
list.add(fs.nextInt());
return list;
}
private static List<Long> readLongList(int n) {
List<Long> list = new ArrayList<>();
for (int i = 0; i < n; i++)
list.add(fs.nextLong());
return list;
}
private static class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner() throws IOException {
if (checkOnlineJudge)
this.br = new BufferedReader(new FileReader("src/input.txt"));
else
this.br = new BufferedReader(new InputStreamReader(System.in));
this.st = new StringTokenizer("");
}
public String next() {
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException err) {
err.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine() {
if (st.hasMoreTokens()) {
return st.nextToken("").trim();
}
try {
return br.readLine().trim();
} catch (IOException err) {
err.printStackTrace();
}
return "";
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
private static class FastWriter {
PrintWriter out;
FastWriter() throws IOException {
if (checkOnlineJudge)
out = new PrintWriter(new FileWriter("src/output.txt"));
else
out = new PrintWriter(System.out);
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 46f0841e1a025918827b614443887e3b | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws Throwable {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
for (int tc = 0; tc < t; ++tc) {
st = new StringTokenizer(br.readLine());
int x = Integer.parseInt(st.nextToken());
int y = Integer.parseInt(st.nextToken());
System.out.println(solve(x, y));
}
}
static int solve(int x, int y) {
if (x == y) {
return x;
}
if (x > y) {
return x + y;
}
return (y / x * x + y) / 2;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | fa4edf8a9388da9ac28d5b8f3b8ba28b | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
public class A {
static class fast {
BufferedReader br;
StringTokenizer st;
public fast() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
public static void main(String[] args)throws IOException {
if (System.getProperty("ONLINE_JUDGE") == null) {
try {
System.setIn(new FileInputStream(new File("i.txt")));
System.setOut(new PrintStream(new File("o.txt")));
} catch (Exception e) {}
}
fast f = new fast();
int t = f.nextInt();
StringBuilder sb = new StringBuilder();
while (t-- > 0) {
long x = f.nextInt();
long y = f.nextInt();
if (x == y)sb.append(x).append('\n');
else
if (x > y)sb.append(x + y).append('\n');
else {
long m = y / x;
long X = x * m;
sb.append((X + y) / 2).append('\n');
}
}
System.out.println(sb);
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 157127eadd7a971b0d103d401aa759a3 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
//import java.lang.*;
import java.io.*;
public class Solution {
static long[] fac;
static int m = (int)1e9+7;
static int c = 1;
// static int[] x = {1,-1,0,0};
// static int[] y = {0,0,1,-1};
// static int cycle_node;
public static void main(String[] args) throws IOException {
Reader.init(System.in);
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
// Do not delete this //
//Uncomment this before using nCrModPFermat
// fac = new long[200000 + 1];
// fac[0] = 1;
//
// for (int i = 1; i <= 200000; i++)
// fac[i] = (fac[i - 1] * i % 1000000007);
// long[] fact = factorial((int)1e6);
int te = Reader.nextInt();
// int te = 1;
while(te-->0) {
long a = Reader.nextLong();
long b = Reader.nextLong();
if(a>b){
output.write(a+b+"\n");
}
else if(b%a==0){
output.write(a+"\n");
}
else{
if(b<a*2){
output.write(b-(b-a)/2+"\n");
}
else {
long n = b - b % a - (a - b % a) / 2;
output.write(n + "\n");
}
}
}
output.close();
}
// Recursive function to return (x ^ n) % m
static long modexp(long x, long n)
{
if (n == 0) {
return 1;
}
else if (n % 2 == 0) {
return modexp((x * x) % m, n / 2);
}
else {
return (x * modexp((x * x) % m, (n - 1) / 2) % m);
}
}
// Function to return the fraction modulo mod
static long getFractionModulo(long a, long b)
{
long c = gcd(a, b);
a = a / c;
b = b / c;
// (b ^ m-2) % m
long d = modexp(b, m - 2);
// Final answer
long ans = ((a % m) * (d % m)) % m;
return ans;
}
public static boolean isP(String n){
StringBuilder s1 = new StringBuilder(n);
StringBuilder s2 = new StringBuilder(n);
s2.reverse();
for(int i = 0;i<s1.length();i++){
if(s1.charAt(i)!=s2.charAt(i)) return false;
}
return true;
}
public static long[] factorial(int n){
long[] factorials = new long[n+1];
factorials[0] = 1;
factorials[1] = 1;
for(int i = 2;i<=n;i++){
factorials[i] = (factorials[i-1]*i)%1000000007;
}
return factorials;
}
public static long numOfBits(long n){
long ans = 0;
while(n>0){
n = n & (n-1);
ans++;
}
return ans;
}
public static long ceilOfFraction(long x, long y){
// ceil using integer division: ceil(x/y) = (x+y-1)/y
// using double may go out of range.
return (x+y-1)/y;
}
public static long power(long x, long y, long p) {
long res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// Returns n^(-1) mod p
public static long modInverse(long n, long p) {
return power(n, p - 2, p);
}
// Returns nCr % p using Fermat's
// little theorem.
public static long nCrModPFermat(int n, int r, int p) {
if (n<r) return 0;
if (r == 0) return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;
}
public static long ncr(long n, long r) {
long p = 1, k = 1;
if (n - r < r) {
r = n - r;
}
if (r != 0) {
while (r > 0) {
p *= n;
k *= r;
long m = gcd(p, k);
p /= m;
k /= m;
n--;
r--;
}
} else {
p = 1;
}
return p;
}
public static boolean isPrime(long n){
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; (long) i * i <= n; i = i + 6){
if (n % i == 0 || n % (i + 2) == 0) {
return false;
}
}
return true;
}
public static int powOf2JustSmallerThanN(int n) {
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
return n ^ (n >> 1);
}
public static long merge(int[] arr, int[] aux, int low, int mid, int high)
{
int k = low, i = low, j = mid + 1;
long inversionCount = 0;
// while there are elements in the left and right runs
while (i <= mid && j <= high)
{
if (arr[i] <= arr[j]) {
aux[k++] = arr[i++];
}
else {
aux[k++] = arr[j++];
inversionCount += (mid - i + 1); // NOTE
}
}
// copy remaining elements
while (i <= mid) {
aux[k++] = arr[i++];
}
/* no need to copy the second half (since the remaining items
are already in their correct position in the temporary array) */
// copy back to the original array to reflect sorted order
for (i = low; i <= high; i++) {
arr[i] = aux[i];
}
return inversionCount;
}
public static long mergesort(int[] arr, int[] aux, int low, int high)
{
if (high <= low) { // if run size <= 1
return 0;
}
int mid = (low + ((high - low) >> 1));
long inversionCount = 0;
// recursively split runs into two halves until run size <= 1,
// then merges them and return up the call chain
// split/merge left half
inversionCount += mergesort(arr, aux, low, mid);
// split/merge right half
inversionCount += mergesort(arr, aux, mid + 1, high);
// merge the two half runs
inversionCount += merge(arr, aux, low, mid, high);
return inversionCount;
}
public static void reverseArray(int[] arr,int start, int end) {
int temp;
while (start < end) {
temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
public static long gcd(long a, long b){
if(a==0){
return b;
}
return gcd(b%a,a);
}
public static long lcm(long a, long b){
if(a>b) return a/gcd(b,a) * b;
return b/gcd(a,b) * a;
}
public static long largeExponentMod(long x,long y,long mod){
// computing (x^y) % mod
x%=mod;
long ans = 1;
while(y>0){
if((y&1)==1){
ans = (ans*x)%mod;
}
x = (x*x)%mod;
y = y >> 1;
}
return ans;
}
public static boolean[] numOfPrimesInRange(long L, long R){
boolean[] isPrime = new boolean[(int) (R-L+1)];
Arrays.fill(isPrime,true);
long lim = (long) Math.sqrt(R);
for (long i = 2; i <= lim; i++){
for (long j = Math.max(i * i, (L + i - 1) / i * i); j <= R; j += i){
isPrime[(int) (j - L)] = false;
}
}
if (L == 1) isPrime[0] = false;
return isPrime;
}
public static ArrayList<Long> primeFactors(long n){
ArrayList<Long> factorization = new ArrayList<>();
if(n%2==0){
factorization.add(2L);
}
while(n%2==0){
n/=2;
}
if(n%3==0){
factorization.add(3L);
}
while(n%3==0){
n/=3;
}
if(n%5==0){
factorization.add(5L);
}
while(n%5==0){
// factorization.add(5L);
n/=5;
}
int[] increments = {4, 2, 4, 2, 4, 6, 2, 6};
int i = 0;
for (long d = 7; d * d <= n; d += increments[i++]) {
if(n%d==0){
factorization.add(d);
}
while (n % d == 0) {
// factorization.add(d);
n /= d;
}
if (i == 8)
i = 0;
}
if (n > 1)
factorization.add(n);
return factorization;
}
}
class DSU {
int[] size, parent;
int n;
public DSU(int n){
this.n = n;
size = new int[n];
parent = new int[n];
for(int i = 0;i<n;i++){
parent[i] = i;
size[i] = 1;
}
}
public int find(int u){
if(parent[u]==u){
return u;
}
return parent[u] = find(parent[u]);
}
public void merge(int u, int v){
u = find(u);
v = find(v);
if(u!=v){
if(size[u]>size[v]){
parent[v] = u;
size[u] += size[v];
}
else{
parent[u] = v;
size[v] += size[u];
}
}
}
}
class Reader {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(
new InputStreamReader(input) );
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() throws IOException {
while ( ! tokenizer.hasMoreTokens() ) {
//TODO add check for eof if necessary
tokenizer = new StringTokenizer(
reader.readLine() );
}
return tokenizer.nextToken();
}
static String nextLine() throws IOException {
return reader.readLine();
}
static int nextInt() throws IOException {
return Integer.parseInt( next() );
}
static long nextLong() throws IOException{
return Long.parseLong(next() );
}
static double nextDouble() throws IOException {
return Double.parseDouble( next() );
}
}
class SegTree{
int n; // array size
// Max size of tree
public int[] tree;
public SegTree(int n){
this.n = n;
tree = new int[4*n];
}
// function to build the tree
void update(int pos, int val, int s, int e, int treeIdx){
if(pos < s || pos > e){
return;
}
if(s == e){
tree[treeIdx] = val;
return;
}
int mid = s + (e - s) / 2;
update(pos, val, s, mid, 2 * treeIdx);
update(pos, val, mid + 1, e, 2 * treeIdx + 1);
tree[treeIdx] = tree[2 * treeIdx] + tree[2 * treeIdx + 1];
}
void update(int pos, int val){
update(pos, val, 1, n, 1);
}
int query(int qs, int qe, int s, int e, int treeIdx){
if(qs <= s && qe >= e){
return tree[treeIdx];
}
if(qs > e || qe < s){
return 0;
}
int mid = s + (e - s) / 2;
int subQuery1 = query(qs, qe, s, mid, 2 * treeIdx);
int subQuery2 = query(qs, qe, mid + 1, e, 2 * treeIdx + 1);
int res = subQuery1 + subQuery2;
return res;
}
int query(int l, int r){
return query(l, r, 1, n, 1);
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 17f32c65d42739c27b30761960f36ff2 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /*
*** author: cypher70
*/
import java.io.*;
import java.util.*;
public class sol {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() { return Double.parseDouble(next()); }
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
}
static class pair<E, T, K> {
E val1;
T val2;
K val3;
public pair(E d1, T d2, K d3) {
val1 = d1;
val2 = d2;
val3 = d3;
}
}
static String reverse(String s) {
StringBuilder rev = new StringBuilder();
for (int i = s.length()-1; i >= 0; i--) {
rev.append(s.charAt(i));
}
return rev.toString();
}
static void reverse(int[] arr, int l, int r) {
while (l < r) {
int h = arr[l];
arr[l] = arr[r];
arr[r] = h;
l++; r--;
}
}
static void debug(int[] arr) {
for (int e: arr) {
System.out.print(e + " ");
}
System.out.println();
}
static void sort(int[] arr, int l, int r) {
ArrayList<Integer> a = new ArrayList<>();
for (int i = l; i <= r; i++) {
a.add(arr[i]);
}
Collections.sort(a);
for (int i = l; i <= r; i++) {
arr[i] = a.get(i-l);
}
}
/*
------------------------------- Implementation Begins ------------------------------
*/
public static void main(String[] args) {
FastReader fr = new FastReader();
int t = fr.nextInt();
while (t-->0) {
long a = fr.nextLong();
long b = fr.nextLong();
if (b >= a && (b % a == 0 || a == b)) {
System.out.println(b);
} else if (b > a) {
a *= (b / a);
System.out.println((a + b) / 2);
} else {
System.out.println(a + b);
}
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 21fb0cdb66385081008b983967b8d337 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
// when can't think of anything -->>
// 1. In sorting questions try to think about all possibilities like starting from start, end, middle.
// 2. Two pointers, brute force.
// 3. In graph query questions try to solve it reversely or try to process all the queries in a single parse.
// 4. If order does not matter then just sort the data if constraints allow. It'll never harm.
// 5. In greedy problems if you are just overwhelmed by the possibilities and stuck, try to code whatever you come up with.
// 6. Try to solve it from back or reversely.
// 7. When can't prove something take help of contradiction.
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
PrintWriter writer = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-->0) {
long a = sc.nextLong();
long b = sc.nextLong();
if(b>a)writer.println((long)(b-0.5*(b%a)));
else if(a==b)writer.println(a);
else writer.println(a+b);
}
writer.flush();
writer.close();
}
private static int findVal(String a, String b, int n) {
int diff1 = 0;
for(int i = 0; i < n; i++) {
if(a.charAt(i) != b.charAt(i))diff1++;
}
return diff1;
}
private static long power (long a, long n, long p) {
long res = 1;
while(n!=0) {
if(n%2==1) {
res=(res*a)%p;
n--;
}else {
a= (a*a)%p;
n/=2;
}
}
return res;
}
private static boolean isPrime(int c) {
for (int i = 2; i*i <= c; i++) {
if(c%i==0)return false;
}
return true;
}
private static int find(int a , int arr[]) {
if(arr[a] != a) return arr[a] = find(arr[a], arr);
return arr[a];
}
private static void union(int a, int b, int arr[]) {
int aa = find(a,arr);
int bb = find(b, arr);
arr[aa] = bb;
}
private static int gcd(int a, int b) {
if(a==0)return b;
return gcd(b%a, a);
}
public static int[] readIntArray(int size, FastReader s) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = s.nextInt();
}
return array;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
class Pair implements Comparable<Pair>{
int a;
int b;
Pair(int a, int b){
this.a = a;
this.b = b;
}
@Override
public boolean equals(Object obj) {
if(this == obj) return true;
if(obj == null || obj.getClass()!= this.getClass()) return false;
Pair pair = (Pair) obj;
return (pair.a == this.a && pair.b == this.b);
}
@Override
public int hashCode()
{
return Objects.hash(a,b);
}
@Override
public int compareTo(Pair o) {
if(this.a == o.a) {
return Integer.compare(this.b, o.b);
}else {
return Integer.compare(this.a, o.a);
}
}
@Override
public String toString() {
return this.a + " " + this.b;
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 35011f1afe4f207b5977d59b8cac6c7c | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.sql.Statement;
import java.text.DecimalFormat;
import java.util.*;
public class Main {
static class Pair
{
long val,ind;
public Pair(long val,long ind)
{
this.val=val;
this.ind=ind;
}
public String toString() {
return val +" "+ind;
}
}
static final int INF = 1 << 30;
static final long INFL = 1L << 60;
static final long NINF = INFL * -1;
static final long mod = (long) 1e9 + 7;
static final long mod2 = 998244353;
static DecimalFormat df = new DecimalFormat("0.00000000000000");
public static final double PI = 3.141592653589793d, eps = 1e-9;
public static void main(String[] args) throws IOException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
static class TaskC
{
public static void solve(int testNumber, InputReader in, OutputWriter out)
{
int t1=in.readInt();
while (t1-->0)
{
long a=in.nextLong();
long b=in.nextLong();
if(a<=b)
{
out.printLine(b-((b%a)/2));
}
else
{
out.printLine(a+b);
}
}
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int arraySize) {
int[] array = new int[arraySize];
for (int i = 0; i < arraySize; i++) {
array[i] = readInt();
}
return array;
}
private int skip() {
int b;
while ((b = read()) != -1 && isSpaceChar(b)) ;
return b;
}
private String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = read();
}
return sb.toString();
}
private char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while (p < n && !(isSpaceChar(b))) {
buf[p++] = (char) b;
b = read();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isNewLine(int c) {
return c == '\n';
}
public String nextLine() {
int c = read();
StringBuilder result = new StringBuilder();
do {
result.appendCodePoint(c);
c = read();
} while (!isNewLine(c));
return result.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sign = 1;
if (c == '-') {
sign = -1;
c = read();
}
long result = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
result *= 10;
result += c & 15;
c = read();
} while (!isSpaceChar(c));
return result * sign;
}
public long[] nextLongArray(int arraySize) {
long array[] = new long[arraySize];
for (int i = 0; i < arraySize; i++) {
array[i] = nextLong();
}
return array;
}
public double nextDouble() {
double ret = 0, div = 1;
byte c = (byte) read();
while (c <= ' ') {
c = (byte) read();
}
boolean neg = (c == '-');
if (neg) {
c = (byte) read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = (byte) read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = (byte) read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg) {
return -ret;
}
return ret;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
public static class CP {
static boolean isPrime(long n) {
if (n <= 1)
return false;
if (n == 2 || n == 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; (long) i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
public static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long pow(long a, long n, long mod) {
// a %= mod;
long ret = 1;
int x = 63 - Long.numberOfLeadingZeros(n);
for (; x >= 0; x--) {
ret = ret * ret % mod;
if (n << 63 - x < 0)
ret = ret * a % mod;
}
return ret;
}
public static boolean isComposite(long n) {
if (n < 2)
return true;
if (n == 2 || n == 3)
return false;
if (n % 2 == 0 || n % 3 == 0)
return true;
for (long i = 6L; i * i <= n; i += 6)
if (n % (i - 1) == 0 || n % (i + 1) == 0)
return true;
return false;
}
static int ifnotPrime(int[] prime, int x) {
return (prime[x / 64] & (1 << ((x >> 1) & 31)));
}
static int log2(long n) {
return (int) (Math.log10(n) / Math.log10(2));
}
static void makeComposite(int[] prime, int x) {
prime[x / 64] |= (1 << ((x >> 1) & 31));
}
public static String swap(String a, int i, int j) {
char temp;
char[] charArray = a.toCharArray();
temp = charArray[i];
charArray[i] = charArray[j];
charArray[j] = temp;
return String.valueOf(charArray);
}
static void reverse(long arr[]){
int l = 0, r = arr.length-1;
while(l<r){
long temp = arr[l];
arr[l] = arr[r];
arr[r] = temp;
l++;
r--;
}
}
static void reverse(int arr[]){
int l = 0, r = arr.length-1;
while(l<r){
int temp = arr[l];
arr[l] = arr[r];
arr[r] = temp;
l++;
r--;
}
}
public static int[] sieveEratosthenes(int n) {
if (n <= 32) {
int[] primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31};
for (int i = 0; i < primes.length; i++) {
if (n < primes[i]) {
return Arrays.copyOf(primes, i);
}
}
return primes;
}
int u = n + 32;
double lu = Math.log(u);
int[] ret = new int[(int) (u / lu + u / lu / lu * 1.5)];
ret[0] = 2;
int pos = 1;
int[] isnp = new int[(n + 1) / 32 / 2 + 1];
int sup = (n + 1) / 32 / 2 + 1;
int[] tprimes = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31};
for (int tp : tprimes) {
ret[pos++] = tp;
int[] ptn = new int[tp];
for (int i = (tp - 3) / 2; i < tp << 5; i += tp)
ptn[i >> 5] |= 1 << (i & 31);
for (int j = 0; j < sup; j += tp) {
for (int i = 0; i < tp && i + j < sup; i++) {
isnp[j + i] |= ptn[i];
}
}
}
// 3,5,7
// 2x+3=n
int[] magic = {0, 1, 23, 2, 29, 24, 19, 3, 30, 27, 25, 11, 20, 8, 4,
13, 31, 22, 28, 18, 26, 10, 7, 12, 21, 17, 9, 6, 16, 5, 15, 14};
int h = n / 2;
for (int i = 0; i < sup; i++) {
for (int j = ~isnp[i]; j != 0; j &= j - 1) {
int pp = i << 5 | magic[(j & -j) * 0x076be629 >>> 27];
int p = 2 * pp + 3;
if (p > n)
break;
ret[pos++] = p;
if ((long) p * p > n)
continue;
for (int q = (p * p - 3) / 2; q <= h; q += p)
isnp[q >> 5] |= 1 << q;
}
}
return Arrays.copyOf(ret, pos);
}
static int digit(long s) {
int brute = 0;
while (s > 0) {
s /= 10;
brute++;
}
return brute;
}
public static int[] primefacs(int n, int[] primes) {
int[] ret = new int[15];
int rp = 0;
for (int p : primes) {
if (p * p > n) break;
int i;
for (i = 0; n % p == 0; n /= p, i++) ;
if (i > 0) ret[rp++] = p;
}
if (n != 1) ret[rp++] = n;
return Arrays.copyOf(ret, rp);
}
static ArrayList<Integer> bitWiseSieve(int n) {
ArrayList<Integer> al = new ArrayList<>();
int prime[] = new int[n / 64 + 1];
for (int i = 3; i * i <= n; i += 2) {
if (ifnotPrime(prime, i) == 0)
for (int j = i * i, k = i << 1;
j < n; j += k)
makeComposite(prime, j);
}
al.add(2);
for (int i = 3; i <= n; i += 2)
if (ifnotPrime(prime, i) == 0)
al.add(i);
return al;
}
public static long[] sort(long arr[]) {
List<Long> list = new ArrayList<>();
for (long n : arr) {
list.add(n);
}
Collections.sort(list);
for (int i = 0; i < arr.length; i++) {
arr[i] = list.get(i);
}
return arr;
}
public static long[] revsort(long[] arr) {
List<Long> list = new ArrayList<>();
for (long n : arr) {
list.add(n);
}
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < arr.length; i++) {
arr[i] = list.get(i);
}
return arr;
}
public static int[] revsort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int n : arr) {
list.add(n);
}
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < arr.length; i++) {
arr[i] = list.get(i);
}
return arr;
}
public static ArrayList<Integer> reverse(
ArrayList<Integer> data,
int left, int right)
{
// Reverse the sub-array
while (left < right)
{
int temp = data.get(left);
data.set(left++,
data.get(right));
data.set(right--, temp);
}
// Return the updated array
return data;
}
static ArrayList<Integer> sieve(long size) {
ArrayList<Integer> pr = new ArrayList<Integer>();
boolean prime[] = new boolean[(int) size];
for (int i = 2; i < prime.length; i++) prime[i] = true;
for (int i = 2; i * i < prime.length; i++) {
if (prime[i]) {
for (int j = i * i; j < prime.length; j += i) {
prime[j] = false;
}
}
}
for (int i = 2; i < prime.length; i++) if (prime[i]) pr.add(i);
return pr;
}
static ArrayList<Integer> segmented_sieve(int l, int r, ArrayList<Integer> primes) {
ArrayList<Integer> al = new ArrayList<>();
if (l == 1) ++l;
int max = r - l + 1;
int arr[] = new int[max];
for (int p : primes) {
if (p * p <= r) {
int i = (l / p) * p;
if (i < l) i += p;
for (; i <= r; i += p) {
if (i != p) {
arr[i - l] = 1;
}
}
}
}
for (int i = 0; i < max; ++i) {
if (arr[i] == 0) {
al.add(l + i);
}
}
return al;
}
static boolean isfPrime(long n, int iteration) {
if (n == 0 || n == 1)
return false;
if (n == 2)
return true;
if (n % 2 == 0)
return false;
Random rand = new Random();
for (int i = 0; i < iteration; i++) {
long r = Math.abs(rand.nextLong());
long a = r % (n - 1) + 1;
if (modPow(a, n - 1, n) != 1)
return false;
}
return true;
}
static long modPow(long a, long b, long c) {
long res = 1;
for (int i = 0; i < b; i++) {
res *= a;
res %= c;
}
return res % c;
}
private static long binPower(long a, long l, long mod) {
long res = 0;
while (l > 0) {
if ((l & 1) == 1) {
res = mulmod(res, a, mod);
l >>= 1;
}
a = mulmod(a, a, mod);
}
return res;
}
private static long mulmod(long a, long b, long c) {
long x = 0, y = a % c;
while (b > 0) {
if (b % 2 == 1) {
x = (x + y) % c;
}
y = (y * 2L) % c;
b /= 2;
}
return x % c;
}
static long binary_Expo(long a, long b) {
long res = 1;
while (b != 0) {
if ((b & 1) == 1) {
res *= a;
--b;
}
a *= a;
b /= 2;
}
return res;
}
static void swap(int a, int b) {
int tp = b;
b = a;
a = tp;
}
static long Modular_Expo(long a, long b, long mod) {
long res = 1;
while (b != 0) {
if ((b & 1) == 1) {
res = (res * a) % mod;
--b;
}
a = (a * a) % mod;
b /= 2;
}
return res % mod;
}
static int i_gcd(int a, int b) {
while (true) {
if (b == 0)
return a;
int c = a;
a = b;
b = c % b;
}
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long ceil_div(long a, long b) {
return (a + b - 1) / b;
}
static int getIthBitFromInt(int bits, int i) {
return (bits >> (i - 1)) & 1;
}
private static TreeMap<Long, Long> primeFactorize(long n) {
TreeMap<Long, Long> pf = new TreeMap<>(Collections.reverseOrder());
long cnt = 0;
long total = 1;
for (long i = 2; (long) i * i <= n; ++i) {
if (n % i == 0) {
cnt = 0;
while (n % i == 0) {
++cnt;
n /= i;
}
pf.put(cnt, i);
//total*=(cnt+1);
}
}
if (n > 1) {
pf.put(1L, n);
//total*=2;
}
return pf;
}
//less than or equal
private static int lower_bound(List<Integer> list, int val) {
int ans = -1, lo = 0, hi = list.size() - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(int[] arr, int val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(long[] arr, long val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
// greater than or equal
private static int upper_bound(List<Integer> list, int val) {
int ans = list.size(), lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(int[] arr, int val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(long[] arr, long val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
// ==================== LIS & LNDS ================================
private static int[] LIS(long arr[], int n) {
List<Long> list = new ArrayList<>();
int[] cnt=new int[n];
for (int i = 0; i < n; i++) {
int idx = find1(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
cnt[i]=list.size();
}
return cnt;
}
private static int find1(List<Long> list, long val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid)>=val) {
ret = mid;
j = mid - 1;
} else {
i = mid + 1;
}
}
return ret;
}
private static int LNDS(int[] arr, int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
int idx = find2(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
}
return list.size();
}
private static int find2(List<Integer> list, int val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid) <= val) {
i = mid + 1;
} else {
ret = mid;
j = mid - 1;
}
}
return ret;
}
private static long nCr(long n, long r,long mod) {
if (n - r > r)
r = n - r;
long ans = 1L;
for (long i = r + 1; i <= n; i++)
ans = (ans%mod*i%mod)%mod;
for (long i = 2; i <= n - r; i++)
ans /= i;
return ans%mod;
}
private static boolean isPalindrome(String str) {
int i = 0, j = str.length() - 1;
while (i < j)
if (str.charAt(i++) != str.charAt(j--))
return false;
return true;
}
private static String reverseString(String str) {
StringBuilder sb = new StringBuilder(str);
return sb.reverse().toString();
}
private static String sortString(String str) {
int[] arr = new int[256];
for (char ch : str.toCharArray())
arr[ch]++;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < 256; i++)
while (arr[i]-- > 0)
sb.append((char) i);
return sb.toString();
}
static boolean isSquarefactor(int x, int factor, int target) {
int s = (int) Math.round(Math.sqrt(x));
return factor * s * s == target;
}
static boolean isSquare(int x) {
int s = (int) Math.round(Math.sqrt(x));
return x * x == s;
}
static int bs(ArrayList<Integer> al, int val) {
int l = 0, h = al.size() - 1, mid = 0, ans = -1;
while (l <= h) {
mid = (l + h) >> 1;
if (al.get(mid) == val) {
return mid;
} else if (al.get(mid) > val) {
h = mid - 1;
} else {
l = mid + 1;
}
}
return ans;
}
static void sort(int a[]) // heap sort
{
PriorityQueue<Integer> q = new PriorityQueue<>();
for (int i = 0; i < a.length; i++)
q.add(a[i]);
for (int i = 0; i < a.length; i++)
a[i] = q.poll();
}
static void shuffle(int[] in) {
for (int i = 0; i < in.length; i++) {
int idx = (int) (Math.random() * in.length);
fast_swap(in, idx, i);
}
}
public static int[] radixSort2(int[] a) {
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for (int v : a) {
c0[(v & 0xff) + 1]++;
c1[(v >>> 8 & 0xff) + 1]++;
c2[(v >>> 16 & 0xff) + 1]++;
c3[(v >>> 24 ^ 0x80) + 1]++;
}
for (int i = 0; i < 0xff; i++) {
c0[i + 1] += c0[i];
c1[i + 1] += c1[i];
c2[i + 1] += c2[i];
c3[i + 1] += c3[i];
}
int[] t = new int[n];
for (int v : a) t[c0[v & 0xff]++] = v;
for (int v : t) a[c1[v >>> 8 & 0xff]++] = v;
for (int v : a) t[c2[v >>> 16 & 0xff]++] = v;
for (int v : t) a[c3[v >>> 24 ^ 0x80]++] = v;
return a;
}
static int[] computeLps(String pat) {
int len = 0, i = 1, m = pat.length();
int lps[] = new int[m];
lps[0] = 0;
while (i < m) {
if (pat.charAt(i) == pat.charAt(len)) {
++len;
lps[i] = len;
++i;
} else {
if (len != 0) {
len = lps[len - 1];
} else {
lps[i] = len;
++i;
}
}
}
return lps;
}
static ArrayList<Integer> kmp(String s, String pat) {
ArrayList<Integer> al = new ArrayList<>();
int n = s.length(), m = pat.length();
int lps[] = computeLps(pat);
int i = 0, j = 0;
while (i < n) {
if (s.charAt(i) == pat.charAt(j)) {
i++;
j++;
if (j == m) {
al.add(i - j);
j = lps[j - 1];
}
} else {
if (j != 0) {
j = lps[j - 1];
} else {
i++;
}
}
}
return al;
}
static void reverse_ruffle_sort(int a[]) {
shuffle(a);
Arrays.sort(a);
for (int l = 0, r = a.length - 1; l < r; ++l, --r)
fast_swap(a, l, r);
}
static void ruffle_sort(int a[]) {
shuffle(a);
Arrays.sort(a);
}
static int getMax(int arr[], int n) {
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return mx;
}
static ArrayList<Long> primeFactors(long n) {
ArrayList<Long> al = new ArrayList<>();
al.add(1L);
while (n % 2 == 0) {
if (!al.contains(2L)) {
al.add(2L);
}
n /= 2L;
}
for (long i = 3; (long) i * i <= n; i += 2) {
while ((n % i == 0)) {
if (!al.contains((long) i)) {
al.add((long) i);
}
n /= i;
}
}
if (n > 2) {
if (!al.contains(n)) {
al.add(n);
}
}
return al;
}
public static long totFactors(long n) {
long cnt = 0, tot = 1;
while (n % 2 == 0) {
n /= 2;
++cnt;
}
tot *= (cnt + 1);
for (int i = 3; i <= Math.sqrt(n); i += 2) {
cnt = 0;
while (n % i == 0) {
n /= i;
++cnt;
}
tot *= (cnt + 1);
}
if (n > 2) {
tot *= 2;
}
return tot;
}
static int[] z_function(String s) {
int n = s.length(), z[] = new int[n];
for (int i = 1, l = 0, r = 0; i < n; ++i) {
if (i <= r)
z[i] = Math.min(z[i - l], r - i + 1);
while (i + z[i] < n && s.charAt(z[i]) == s.charAt(i + z[i]))
++z[i];
if (i + z[i] - 1 > r) {
l = i;
r = i + z[i] - 1;
}
}
return z;
}
static void fast_swap(int[] a, int idx1, int idx2) {
if (a[idx1] == a[idx2])
return;
a[idx1] ^= a[idx2];
a[idx2] ^= a[idx1];
a[idx1] ^= a[idx2];
}
static void fast_swap(long[] a, int idx1, int idx2) {
if (a[idx1] == a[idx2])
return;
a[idx1] ^= a[idx2];
a[idx2] ^= a[idx1];
a[idx1] ^= a[idx2];
}
public static void fast_sort(long[] array) {
ArrayList<Long> copy = new ArrayList<>();
for (long i : array)
copy.add(i);
Collections.sort(copy);
for (int i = 0; i < array.length; i++)
array[i] = copy.get(i);
}
static int factorsCount(int n) {
boolean hash[] = new boolean[n + 1];
Arrays.fill(hash, true);
for (int p = 2; p * p < n; p++)
if (hash[p] == true)
for (int i = p * 2; i < n; i += p)
hash[i] = false;
int total = 1;
for (int p = 2; p <= n; p++) {
if (hash[p]) {
int count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
static long binomialCoeff(long n, long k) {
long res = 1;
if (k > n - k)
k = n - k;
for (int i = 0; i < k; ++i) {
res = (res * (n - i));
res /= (i + 1);
}
return res;
}
static long nck(long fact[], long inv[], long n, long k) {
if (k > n) return 0;
long res = fact[(int) n]%mod;
res = (int) ((res%mod* inv[(int) k]%mod))%mod;
res = (int) ((res%mod*inv[(int) (n - k)]%mod))%mod;
return res % mod;
}
public static long fact(long x) {
long fact = 1;
for (int i = 2; i <= x; ++i) {
fact = fact * i;
}
return fact;
}
public static ArrayList<Long> getFact(long x) {
ArrayList<Long> facts = new ArrayList<>();
for (long i = 2; i * i <= x; ++i) {
if (x % i == 0) {
facts.add(i);
if (i != x / i) {
facts.add(x / i);
}
}
}
return facts;
}
static void matrix_ex(long n, long[][] A, long[][] I) {
while (n > 0) {
if (n % 2 == 0) {
Multiply(A, A);
n /= 2;
} else {
Multiply(I, A);
n--;
}
}
}
static void Multiply(long[][] A, long[][] B) {
int n = A.length, m = A[0].length, p = B[0].length;
long[][] C = new long[n][p];
for (int i = 0; i < n; i++) {
for (int j = 0; j < p; j++) {
for (int k = 0; k < m; k++) {
C[i][j] += ((A[i][k] % mod) * (B[k][j] % mod)) % mod;
C[i][j] = C[i][j] % mod;
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < p; j++) {
A[i][j] = C[i][j];
}
}
}
public static HashMap<Character, Integer> sortMapDesc(HashMap<Character, Integer> map) {
List<Map.Entry<Character, Integer>> list = new LinkedList<>(map.entrySet());
Collections.sort(list, (o1, o2) -> o2.getValue() - o1.getValue());
HashMap<Character, Integer> temp = new LinkedHashMap<>();
for (Map.Entry<Character, Integer> i : list) {
temp.put(i.getKey(), i.getValue());
}
return temp;
}
public static HashMap<Integer, Integer> sortMapAsc(HashMap<Integer, Integer> map) {
List<Map.Entry<Integer, Integer>> list = new LinkedList<>(map.entrySet());
Collections.sort(list, (o1, o2) -> o1.getValue() - o2.getValue());
HashMap<Integer, Integer> temp = new LinkedHashMap<>();
for (Map.Entry<Integer, Integer> i : list) {
temp.put(i.getKey(), i.getValue());
}
return temp;
}
public static long lcm(long l, long l2) {
long val = gcd(l, l2);
return (l * l2) / val;
}
public static int isSubsequence(String s, String t) {
int n = s.length();
int m = t.length();
if (m > n) {
return Integer.MAX_VALUE;
}
int i = 0, j = 0, skip = 0;
while (i < n && j < m) {
if (s.charAt(i) == t.charAt(j)) {
--skip;
++j;
}
++skip;
++i;
}
while (i < n) {
++i;
++skip;
}
if (j != m) {
skip = Integer.MAX_VALUE;
}
return skip;
}
public static long[][] combination(int l, int r) {
long[][] pascal = new long[l + 1][r + 1];
pascal[0][0] = 1;
for (int i = 1; i <= l; ++i) {
pascal[i][0] = 1;
for (int j = 1; j <= r; ++j) {
pascal[i][j] = pascal[i - 1][j - 1] + pascal[i - 1][j];
}
}
return pascal;
}
public static long gcd(long... array) {
long ret = array[0];
for (int i = 1; i < array.length; ++i) ret = gcd(ret, array[i]);
return ret;
}
public static long lcm(int... array) {
long ret = array[0];
for (int i = 1; i < array.length; ++i) ret = lcm(ret, array[i]);
return ret;
}
public static int min(int a, int b) {
return a < b ? a : b;
}
public static int min(int... array) {
int ret = array[0];
for (int i = 1; i < array.length; ++i) ret = min(ret, array[i]);
return ret;
}
public static long min(long a, long b) {
return a < b ? a : b;
}
public static long min(long... array) {
long ret = array[0];
for (int i = 1; i < array.length; ++i) ret = min(ret, array[i]);
return ret;
}
public static int max(int a, int b) {
return a > b ? a : b;
}
public static int max(int... array) {
int ret = array[0];
for (int i = 1; i < array.length; ++i) ret = max(ret, array[i]);
return ret;
}
public static long max(long a, long b) {
return a > b ? a : b;
}
public static long max(long... array) {
long ret = array[0];
for (int i = 1; i < array.length; ++i) ret = max(ret, array[i]);
return ret;
}
public static long sum(int... array) {
long ret = 0;
for (int i : array) ret += i;
return ret;
}
public static long sum(long... array) {
long ret = 0;
for (long i : array) ret += i;
return ret;
}
public static long[] facts(int n)
{
long[] fact=new long[1005];
fact[0]=1;
fact[1]=1;
for(int i=2;i<n;i++)
{
fact[i]=(long)(i*fact[i-1])%mod;
}
return fact;
}
public static long[] inv(long[] fact,int n)
{
long[] inv=new long[n];
inv[n-1]= CP.Modular_Expo(fact[n-1],mod-2,mod)%mod;
for(int i=n-2;i>=0;--i)
{
inv[i]=(inv[i+1]*(i+1))%mod;
}
return inv;
}
public static int modinv(long x, long mod) {
return (int) (CP.Modular_Expo(x, mod - 2, mod) % mod);
}
public static int lcs(String s, String t) {
int n = s.length(), m = t.length();
int dp[][] = new int[n + 1][m + 1];
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
if (i == 0 || j == 0) {
dp[i][j] = 0;
} else if (s.charAt(i - 1) == t.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[n][m];
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(int[] array) {
for (int i = 0; i < array.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(array[i]);
}
}
public void printLine(int[] array) {
print(array);
writer.println();
}
public void close() {
writer.close();
}
public void printLine(long i) {
writer.println(i);
}
public void printLine(int i) {
writer.println(i);
}
public void printLine(char c) {
writer.println(c);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
void flush() {
writer.flush();
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 12f7420368410d609b1a054f1a72465c | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Mod {
public static void main(String[] args) throws Exception {
FastIO in = new FastIO();
int t = in.nextInt();
for (int tc=0; tc<t; tc++) {
long x = in.nextLong();
long y = in.nextLong();
if (x>y) {
System.out.println(x+y);
}
else {
if (x==y) System.out.println(x);
else {
System.out.println(y-(y%x)/2);
}
}
}
}
static class FastIO {
BufferedReader br;
StringTokenizer st;
PrintWriter pr;
public FastIO() throws IOException
{
br = new BufferedReader(new InputStreamReader(System.in));
}
public String next() throws IOException
{
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
}
public int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(next()); }
public long nextLong() throws NumberFormatException, IOException { return Long.parseLong(next()); }
public double nextDouble() throws NumberFormatException, IOException
{
return Double.parseDouble(next());
}
public String nextLine() throws IOException
{
String str = br.readLine();
return str;
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 44f6d1f65f3fb5715078cbc5ebb0a2a5 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | /*
Rating: 1367
Date: 17-01-2022
Time: 16-37-01
Author: Kartik Papney
Linkedin: https://www.linkedin.com/in/kartik-papney-4951161a6/
Leetcode: https://leetcode.com/kartikpapney/
Codechef: https://www.codechef.com/users/kartikpapney
*/
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class B_Moderate_Modular_Mode {
public static void s() {
// (x+y)%x = y%(x+y)
int x = sc.nextInt(), y = sc.nextInt();
if(x > y) {
p.writeln(x + y);
} else {
p.writeln(y - ((y%x)/2));
}
}
public static void main(String[] args) {
int t = 1;
t = sc.nextInt();
while (t-- != 0) {
s();
}
p.print();
}
static final Integer MOD = (int) 1e9 + 7;
static final FastReader sc = new FastReader();
static final Print p = new Print();
static class Functions {
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static void sort(long[] a) {
ArrayList<Long> l = new ArrayList<>();
for (long i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static int max(int[] a) {
int max = Integer.MIN_VALUE;
for (int val : a) max = Math.max(val, max);
return max;
}
static int min(int[] a) {
int min = Integer.MAX_VALUE;
for (int val : a) min = Math.min(val, min);
return min;
}
static long min(long[] a) {
long min = Long.MAX_VALUE;
for (long val : a) min = Math.min(val, min);
return min;
}
static long max(long[] a) {
long max = Long.MIN_VALUE;
for (long val : a) max = Math.max(val, max);
return max;
}
static long sum(long[] a) {
long sum = 0;
for (long val : a) sum += val;
return sum;
}
static int sum(int[] a) {
int sum = 0;
for (int val : a) sum += val;
return sum;
}
public static long mod_add(long a, long b) {
return (a % MOD + b % MOD + MOD) % MOD;
}
public static long pow(long a, long b) {
long res = 1;
while (b > 0) {
if ((b & 1) != 0)
res = mod_mul(res, a);
a = mod_mul(a, a);
b >>= 1;
}
return res;
}
public static long mod_mul(long a, long b) {
long res = 0;
a %= MOD;
while (b > 0) {
if ((b & 1) > 0) {
res = mod_add(res, a);
}
a = (2 * a) % MOD;
b >>= 1;
}
return res;
}
public static long gcd(long a, long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
public static long factorial(long n) {
long res = 1;
for (int i = 1; i <= n; i++) {
res = (i % MOD * res % MOD) % MOD;
}
return res;
}
public static int count(int[] arr, int x) {
int count = 0;
for (int val : arr) if (val == x) count++;
return count;
}
public static ArrayList<Integer> generatePrimes(int n) {
boolean[] primes = new boolean[n];
for (int i = 2; i < primes.length; i++) primes[i] = true;
for (int i = 2; i < primes.length; i++) {
if (primes[i]) {
for (int j = i * i; j < primes.length; j += i) {
primes[j] = false;
}
}
}
ArrayList<Integer> arr = new ArrayList<>();
for (int i = 0; i < primes.length; i++) {
if (primes[i]) arr.add(i);
}
return arr;
}
}
static class Print {
StringBuffer strb = new StringBuffer();
public void write(Object str) {
strb.append(str);
}
public void writes(Object str) {
char c = ' ';
strb.append(str).append(c);
}
public void writeln(Object str) {
char c = '\n';
strb.append(str).append(c);
}
public void writeln() {
char c = '\n';
strb.append(c);
}
public void writes(int[] arr) {
for (int val : arr) {
write(val);
write(' ');
}
}
public void writes(long[] arr) {
for (long val : arr) {
write(val);
write(' ');
}
}
public void writeln(int[] arr) {
for (int val : arr) {
writeln(val);
}
}
public void print() {
System.out.print(strb);
}
public void println() {
System.out.println(strb);
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
double[] readArrayDouble(int n) {
double[] a = new double[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
String nextLine() {
String str = new String();
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | d4fc930596b3edb7305f52c2b66652ac | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.Scanner;
import java.util.Arrays;
import java.util.Comparator;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[]args){
FastReader at = new FastReader();
int t = at.nextInt();
while(t-->0){
long x = at.nextLong();
long y = at.nextLong();
if(x == y){
System.out.println(x);
}
else if(x > y){
System.out.println(x+y);
}
else{
long temp = y;
temp -= (y%x)/2;
System.out.println(temp);
}
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 6fc822a057e93ec9497ef9a48b6388bd | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
public class b {
public static void main(String[] args) {
// in.nextInt()
// out.println()
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
long t=in.nextLong();
while(t-- > 0) {
long x=in.nextLong();
long y=in.nextLong();
if(x>y)out.println(x+y);
else if(x==y)out.println(x);
else {
long z=y-(y%x)/2;
out.println(z);
}
}
out.close();
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 278d62c15fbd6df4e4af37322eb00650 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
static PrintWriter out=new PrintWriter((System.out));
static Reader sc=new Reader();
public static void main(String args[])throws IOException {
int t=sc.nextInt();
for (int i = 1;i <= t;i++) {
solve();
}
out.close();
}
public static void solve() {
long x = sc.nextInt();
long y = sc.nextInt();
long n;
if (x > y) {
n = x + y;
} else {
n = (y + (y/x) * x) / 2;
}
out.println(n);
}
static class Reader {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
public String next()
{
while(!st.hasMoreTokens())
{
try
{
st=new StringTokenizer(br.readLine());
}
catch(Exception e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt()
{
return Integer.parseInt(next());
}
public long nextLong()
{
return Long.parseLong(next());
}
public double nextDouble()
{
return Double.parseDouble(next());
}
public String nextLine()
{
try
{
return br.readLine();
}
catch(Exception e)
{
e.printStackTrace();
}
return null;
}
public boolean hasNext()
{
String next=null;
try
{
next=br.readLine();
}
catch(Exception e)
{
}
if(next==null)
{
return false;
}
st=new StringTokenizer(next);
return true;
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | a232a42db914a0d1c0e7c9fe201ae1b9 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class B{
static void print(long x, long y) {
for (int i = 1; i <= y; i++) {
out.print(i%x + " ");
}
out.println();
for (int i = 1; i <= y; i++) {
out.print(y%i + " ");
}
out.println();
}
public static void main(String[] args) throws IOException{
// br = new BufferedReader(new FileReader(".in"));
// out = new PrintWriter(new FileWriter(".out"));
// new Thread(null, new (), "fisa balls", 1<<28).start();
int t = readInt();
while(t-->0) {
long x= readLong(), y = readLong();
if (x==y) out.println(x);
else if (x>=y) {
out.println(x + y);
}
else if (y%x==0) out.println(x);
else {
// Answer is < y
// Answer exists after y/2
// Stupid guess, ternary search on min(i%x,y%i)
// for the last "stretch" of things
long l = max(y/2+1, y-y%x);
long r = y;
long i = (l+r)/2;
//out.println(i%x + " " + y%i);
out.println(i);
}
}
out.close();
}
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
static StringTokenizer st = new StringTokenizer("");
static String read() throws IOException{
while (!st.hasMoreElements()) st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public static int readInt() throws IOException{return Integer.parseInt(read());}
public static long readLong() throws IOException{return Long.parseLong(read());}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 15ca6c2075b72bcb253fc47fab63dd2a | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
import java.text.*;
public class CF_1603_B{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
long X = nl(), Y = nl();
long N = f(X, Y);
hold(N%X == Y%N);
pn(N);
}
long f(long X, long Y){
if(X == Y)return X;
else if(X > Y)return X+Y;
else if(Y%X == 0)return X;
else {
long f = Y-Y%X;
return (f+Y)/2;
}
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
void exit(boolean b){if(!b)System.exit(0);}
static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
final long IINF = (long)1e17;
final int INF = (int)1e9+2;
DecimalFormat df = new DecimalFormat("0.00000000000");
double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
static boolean multipleTC = true, memory = true, fileIO = false;
FastReader in;PrintWriter out;
void run() throws Exception{
long ct = System.currentTimeMillis();
if (fileIO) {
in = new FastReader("");
out = new PrintWriter("");
} else {
in = new FastReader();
out = new PrintWriter(System.out);
}
//Solution Credits: Taranpreet Singh
int T = multipleTC? ni():1;
pre();
for (int t = 1; t <= T; t++) solve(t);
out.flush();
out.close();
System.err.println(System.currentTimeMillis() - ct);
}
public static void main(String[] args) throws Exception{
if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1603_B().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
else new CF_1603_B().run();
}
int[][] make(int n, int e, int[] from, int[] to, boolean f){
int[][] g = new int[n][];int[]cnt = new int[n];
for(int i = 0; i< e; i++){
cnt[from[i]]++;
if(f)cnt[to[i]]++;
}
for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
for(int i = 0; i< e; i++){
g[from[i]][--cnt[from[i]]] = to[i];
if(f)g[to[i]][--cnt[to[i]]] = from[i];
}
return g;
}
int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
int[][][] g = new int[n][][];int[]cnt = new int[n];
for(int i = 0; i< e; i++){
cnt[from[i]]++;
if(f)cnt[to[i]]++;
}
for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
for(int i = 0; i< e; i++){
g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
}
return g;
}
int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object... o){for(Object oo:o)out.print(oo+" ");}
void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str;
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | be8618b1ccd01cb4ceab3cd00f28888e | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;
public class weird_algrithm {
static BufferedWriter output = new BufferedWriter(
new OutputStreamWriter(System.out));
static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
static int mod = 1000000007;
static String toReturn = "";
static int steps = Integer.MAX_VALUE;
static int maxlen = 1000005;
/*MATHEMATICS FUNCTIONS START HERE
MATHS
MATHS
MATHS
MATHS*/
static long gcd(long a, long b) {
if(b == 0) return a;
else return gcd(b, a % b);
}
static long powerMod(long x, long y, int mod) {
if(y == 0) return 1;
long temp = powerMod(x, y / 2, mod);
temp = ((temp % mod) * (temp % mod)) % mod;
if(y % 2 == 0) return temp;
else return ((x % mod) * (temp % mod)) % mod;
}
static long modInverse(long n, int p) {
return powerMod(n, p - 2, p);
}
static long nCr(int n, int r, int mod, long [] fact, long [] ifact) {
return ((fact[n] % mod) * ((ifact[r] * ifact[n - r]) % mod)) % mod;
}
static boolean isPrime(long a) {
if(a == 1) return false;
else if(a == 2 || a == 3 || a== 5) return true;
else if(a % 2 == 0 || a % 3 == 0) return false;
for(int i = 5; i * i <= a; i = i + 6) {
if(a % i == 0 || a % (i + 2) == 0) return false;
}
return true;
}
static int [] seive(int a) {
int [] toReturn = new int [a + 1];
for(int i = 0; i < a; i++) toReturn[i] = 1;
toReturn[0] = 0;
toReturn[1] = 0;
toReturn[2] = 1;
for(int i = 2; i * i <= a; i++) {
if(toReturn[i] == 0) continue;
for(int j = 2 * i; j <= a; j += i) toReturn[j] = 0;
}
return toReturn;
}
static long [] fact(int a) {
long [] arr = new long[a + 1];
arr[0] = 1;
for(int i = 1; i < a + 1; i++) {
arr[i] = (arr[i - 1] * i) % mod;
}
return arr;
}
static int divisors(int n, HashMap<Integer, Integer> map) {
int max = 1;
for(int i = 1; i * i <= n; i++) {
if(n % i == 0) {
if(map.containsKey(i)) {
max = Math.max(max, i);
}else map.put(i, 1);
if(n / i == i) continue;
if(map.containsKey(n / i)) {
max = Math.max(max, n / i);
}else map.put(n / i, 1);
}
}
return max;
}
static int euler(int n) {
int ans = n;
for(int i = 2; i * i <= n; i++) {
if(n % i == 0) {
while(n % i == 0) {
n /= i;
}
ans -= ans / i;
}
}
if(n > 1) ans -= ans / n;
return ans;
}
/*MATHS
MATHS
MATHS
MATHS
MATHEMATICS FUNCTIONS END HERE */
/*SWAP FUNCTION START HERE
SWAP
SWAP
SWAP
SWAP
*/
static void swap(int i, int j, long[] arr) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, int[] arr) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, String [] arr) {
String temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, char [] arr) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
/*SWAP
SWAP
SWAP
SWAP
SWAP FUNCTION END HERE*/
/*BINARY SEARCH METHODS START HERE
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
*/
static boolean BinaryCheck(long test, long [] arr, long health) {
for(int i = 0; i <= arr.length - 1; i++) {
if(i == arr.length - 1) health -= test;
else if(arr[i + 1] - arr[i] > test) {
health = health - test;
}else {
health = health - (arr[i + 1] - arr[i]);
}
if(health <= 0) return true;
}
return false;
}
static long binarySearchModified(long start1, long n, ArrayList<Long> arr, int a, long r) {
long start = start1, end = n, ans = -1;
while(start < end) {
long mid = (start + end) / 2;
//System.out.println(mid);
if(arr.get((int)mid) + arr.get(a) <= r && mid != start1) {
ans = mid;
start = mid + 1;
}else{
end = mid;
}
}
//System.out.println();
return ans;
}
static int upper(int start, int end, ArrayList<Integer> pairs, long val) {
while(start < end) {
int mid = (start + end) / 2;
if(pairs.get(mid) <= val) start = mid + 1;
else end = mid;
}
return start;
}
static int lower(int start, int end, ArrayList<Long> arr, long val) {
while(start < end) {
int mid = (start + end) / 2;
if(arr.get(mid) >= val) end = mid;
else start = mid + 1;
}
return start;
}
/*BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
BINARY SEARCH METHODS END HERE*/
/*RECURSIVE FUNCTION START HERE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
*/
static int recurse(int x, int y, int n, int steps1, Integer [][] dp) {
if(x > n || y > n) return 0;
if(dp[x][y] != null) {
return dp[x][y];
}
else if(x == n || y == n) {
return steps1;
}
return dp[x][y] = Math.max(recurse(x + y, y, n, steps1 + 1, dp), recurse(x, x + y, n, steps1 + 1, dp));
}
/*RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
RECURSIVE FUNCTION END HERE*/
/*GRAPH FUNCTIONS START HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
* */
static class edge{
int from, to;
long weight;
public edge(int x, int y, long weight2) {
this.from = x;
this.to = y;
this.weight = weight2;
}
}
static class sort implements Comparator<pair>{
@Override
public int compare(pair o1, pair o2) {
// TODO Auto-generated method stub
return (int)o1.a - (int)o2.a;
}
}
static void addEdge(ArrayList<ArrayList<edge>> graph, int from, int to, long weight) {
edge temp = new edge(from, to, weight);
edge temp1 = new edge(to, from, weight);
graph.get(from).add(temp);
//graph.get(to).add(temp1);
}
static int ans = 0;
static void topoSort(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, ArrayList<Integer> toReturn) {
if(visited[vertex]) return;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(!visited[graph.get(vertex).get(i)]) topoSort(graph, graph.get(vertex).get(i), visited, toReturn);
}
toReturn.add(vertex);
}
static boolean isCyclicDirected(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, boolean [] reStack) {
if(reStack[vertex]) return true;
if(visited[vertex]) return false;
reStack[vertex] = true;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(isCyclicDirected(graph, graph.get(vertex).get(i), visited, reStack)) return true;
}
reStack[vertex] = false;
return false;
}
static int e = 0;
static long mst(PriorityQueue<edge> pq, int nodes) {
long weight = 0;
int [] size = new int[nodes + 1];
Arrays.fill(size, 1);
while(!pq.isEmpty()) {
edge temp = pq.poll();
int x = parent(parent, temp.to);
int y = parent(parent, temp.from);
if(x != y) {
//System.out.println(temp.weight);
union(x, y, rank, parent, size);
weight += temp.weight;
e++;
}
}
return weight;
}
static void floyd(long [][] dist) { // to find min distance between two nodes
for(int k = 0; k < dist.length; k++) {
for(int i = 0; i < dist.length; i++) {
for(int j = 0; j < dist.length; j++) {
if(dist[i][j] > dist[i][k] + dist[k][j]) {
dist[i][j] = dist[i][k] + dist[k][j];
}
}
}
}
}
static void dijkstra(ArrayList<ArrayList<edge>> graph, long [] dist, int src) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MAX_VALUE / 2;
dist[src] = 0;
boolean visited[] = new boolean[dist.length];
PriorityQueue<pair> pq = new PriorityQueue<>(new sort());
pq.add(new pair(src, 0));
while(!pq.isEmpty()) {
pair temp = pq.poll();
int index = (int)temp.a;
for(int i = 0; i < graph.get(index).size(); i++) {
if(dist[graph.get(index).get(i).to] > dist[index] + graph.get(index).get(i).weight) {
dist[graph.get(index).get(i).to] = dist[index] + graph.get(index).get(i).weight;
pq.add(new pair(graph.get(index).get(i).to, graph.get(index).get(i).weight));
}
}
}
}
static int parent1 = -1;
static boolean ford(ArrayList<ArrayList<edge>> graph1, ArrayList<edge> graph, long [] dist, int src, int [] parent) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MIN_VALUE / 2;
dist[src] = 0;
boolean hasNeg = false;
for(int i = 0; i < dist.length - 1; i++) {
for(int j = 0; j < graph.size(); j++) {
int from = graph.get(j).from;
int to = graph.get(j).to;
long weight = graph.get(j).weight;
if(dist[to] < dist[from] + weight) {
dist[to] = dist[from] + weight;
parent[to] = from;
}
}
}
for(int i = 0; i < graph.size(); i++) {
int from = graph.get(i).from;
int to = graph.get(i).to;
long weight = graph.get(i).weight;
if(dist[to] < dist[from] + weight) {
parent1 = from;
hasNeg = true;
/*
* dfs(graph1, parent1, new boolean[dist.length], dist.length - 1);
* //System.out.println(ans); dfs(graph1, 0, new boolean[dist.length], parent1);
*/
//System.out.println(ans);
if(ans == 2) break;
else ans = 0;
}
}
return hasNeg;
}
/*GRAPH FUNCTIONS END HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
*/
/*disjoint Set START HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
static int [] rank;
static int [] parent;
static int parent(int [] parent, int x) {
if(parent[x] == x) return x;
else return parent[x] = parent(parent, parent[x]);
}
static boolean union(int x, int y, int [] rank, int [] parent, int [] setSize) {
if(parent(parent, x) == parent(parent, y)) {
return true;
}
if (rank[x] > rank[y]) {
parent[y] = x;
setSize[x] += setSize[y];
} else {
parent[x] = y;
setSize[y] += setSize[x];
if (rank[x] == rank[y]) rank[y]++;
}
return false;
}
/*disjoint Set END HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
/*INPUT START HERE
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT
*/
static int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(sc.readLine());
}
static long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(sc.readLine());
}
static long [] inputLongArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
long [] toReturn = new long[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Long.parseLong(s[i]);
}
return toReturn;
}
static int max = 0;
static int [] inputIntArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
//System.out.println(s.length);
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Integer.parseInt(s[i]);
}
return toReturn;
}
/*INPUT
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT END HERE
*/
static long [] preCompute(int level) {
long [] toReturn = new long[level];
toReturn[0] = 1;
toReturn[1] = 16;
for(int i = 2; i < level; i++) {
toReturn[i] = ((toReturn[i - 1] % mod) * (toReturn[i - 1] % mod)) % mod;
}
return toReturn;
}
static class pair{
long a;
long b;
long d;
public pair(long in, long y) {
this.a = in;
this.b = y;
this.d = 0;
}
}
static int [] nextGreaterBack(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int [] nextGreaterFront(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = s.length - 1; i >= 0; i--) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int [] lps(String s) {
int [] lps = new int[s.length()];
lps[0] = 0;
int j = 0;
for(int i = 1; i < lps.length; i++) {
j = lps[i - 1];
while(j > 0 && s.charAt(i) != s.charAt(j)) j = lps[j - 1];
if(s.charAt(i) == s.charAt(j)) {
lps[i] = j + 1;
}
}
return lps;
}
static int [][] vectors = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
static String dir = "DRUL";
static boolean check(int i, int j, boolean [][] visited) {
if(i >= visited.length || j >= visited[0].length) return false;
if(i < 0 || j < 0) return false;
return true;
}
static void selectionSort(long arr[], long [] arr1, ArrayList<ArrayList<Integer>> ans)
{
int n = arr.length;
for (int i = 0; i < n-1; i++)
{
int min_idx = i;
for (int j = i+1; j < n; j++)
if (arr[j] < arr[min_idx])
min_idx = j;
else if(arr[j] == arr[min_idx]) {
if(arr1[j] < arr1[min_idx]) min_idx = j;
}
if(i == min_idx) {
continue;
}
ArrayList<Integer> p = new ArrayList<Integer>();
p.add(min_idx + 1);
p.add(i + 1);
ans.add(new ArrayList<Integer>(p));
swap(i, min_idx, arr);
swap(i, min_idx, arr1);
}
}
static int saved = Integer.MAX_VALUE;
static String ans1 = "";
public static boolean isValid(int x, int y, String [] mat) {
if(x >= mat.length || x < 0) return false;
if(y >= mat[0].length() || y < 0) return false;
return true;
}
public static void recurse3(ArrayList<Character> arr, int index, String s, int max, ArrayList<String> toReturn) {
if(s.length() == max) {
toReturn.add(s);
return;
}
if(index == arr.size()) return;
recurse3(arr, index + 1, s + arr.get(index), max, toReturn);
recurse3(arr, index + 1, s, max, toReturn);
}
/*
if(arr[i] > q) return Math.max(f(i + 1, q - 1) + 1, f(i + 1, q);
else return f(i + 1, q) + 1
*/
static void dfsDP(ArrayList<ArrayList<Integer>> graph, int src, int [] dp1, int [] dp2, int parent) {
int sum1 = 0; int sum2 = 0;
for(int x : graph.get(src)) {
if(x == parent) continue;
dfsDP(graph, x, dp1, dp2, src);
sum1 += Math.min(dp1[x], dp2[x]);
sum2 += dp1[x];
}
dp1[src] = 1 + sum1;
dp2[src] = sum2;
System.out.println(src + " " + dp1[src] + " " + dp2[src]);
}
static int balanced = 0;
static void dfs(ArrayList<ArrayList<ArrayList<Long>>> graph, long src, int [] dist, long sum1, long sum2, long parent, ArrayList<Long> arr, int index) {
index = 0;//binarySearch(index, arr.size() - 1, arr, sum1);
if(index < arr.size() && arr.get(index) <= sum1) {
dist[(int)src] = index + 1;
}
else dist[(int)src] = index;
for(ArrayList<Long> x : graph.get((int)src)) {
if(x.get(0) == parent) continue;
if(arr.size() != 0) arr.add(arr.get(arr.size() - 1) + x.get(2));
else arr.add(x.get(2));
dfs(graph, x.get(0), dist, sum1 + x.get(1), sum2, src, arr, index);
arr.remove(arr.size() - 1);
}
}
static int compare(String s1, String s2) {
Queue<Character> q1 = new LinkedList<>();
Queue<Character> q2 = new LinkedList<Character>();
for(int i = 0; i < s1.length(); i++) {
q1.add(s1.charAt(i));
q2.add(s2.charAt(i));
}
int k = 0;
while(k < s1.length()) {
if(q1.equals(q2)) {
break;
}
q2.add(q2.poll());
k++;
}
return k;
}
static long pro = 0;
public static int len(ArrayList<ArrayList<Integer>> graph, int src, boolean [] visited
) {
visited[src] = true;
int max = 0;
for(int x : graph.get(src)) {
if(!visited[x]) {
visited[x] = true;
int len = len(graph, x, visited) + 1;
//System.out.println(len);
pro = Math.max(max * (len - 1), pro);
max = Math.max(len, max);
}
}
return max;
}
public static void recurse(int l, int [] ans) {
if(l < 0) return;
int r = (int)Math.sqrt(l * 2);
int s = r * r;
r = s - l;
recurse(r - 1, ans);
while(r <= l) {
ans[r] = l;
ans[l] = r;
r++;
l--;
}
}
static boolean isSmaller(String str1, String str2)
{
// Calculate lengths of both string
int n1 = str1.length(), n2 = str2.length();
if (n1 < n2)
return true;
if (n2 < n1)
return false;
for (int i = 0; i < n1; i++)
if (str1.charAt(i) < str2.charAt(i))
return true;
else if (str1.charAt(i) > str2.charAt(i))
return false;
return false;
}
// Function for find difference of larger numbers
static String findDiff(String str1, String str2)
{
// Before proceeding further, make sure str1
// is not smaller
if (isSmaller(str1, str2)) {
String t = str1;
str1 = str2;
str2 = t;
}
// Take an empty string for storing result
String str = "";
// Calculate length of both string
int n1 = str1.length(), n2 = str2.length();
// Reverse both of strings
str1 = new StringBuilder(str1).reverse().toString();
str2 = new StringBuilder(str2).reverse().toString();
int carry = 0;
// Run loop till small string length
// and subtract digit of str1 to str2
for (int i = 0; i < n2; i++) {
// Do school mathematics, compute difference of
// current digits
int sub
= ((int)(str1.charAt(i) - '0')
- (int)(str2.charAt(i) - '0') - carry);
// If subtraction is less than zero
// we add then we add 10 into sub and
// take carry as 1 for calculating next step
if (sub < 0) {
sub = sub + 10;
carry = 1;
}
else
carry = 0;
str += (char)(sub + '0');
}
// subtract remaining digits of larger number
for (int i = n2; i < n1; i++) {
int sub = ((int)(str1.charAt(i) - '0') - carry);
// if the sub value is -ve, then make it
// positive
if (sub < 0) {
sub = sub + 10;
carry = 1;
}
else
carry = 0;
str += (char)(sub + '0');
}
// reverse resultant string
return new StringBuilder(str).reverse().toString();
}
static void solve() throws IOException {
long [] in = inputLongArr();
if(in[0] > in[1]) {
output.write(in[0] + in[1] + "\n");
}else if(in[0] == in[1]) {
output.write(in[0] + "\n");
}else {
long p = in[1] - in[1] % in[0];
long t = p + (in[1] - p) / 2;
output.write(t + "\n");
}
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
int t = Integer.parseInt(sc.readLine()); for(int i = 0; i < t; i++)
solve();
output.flush();
}
}
class TreeNode {
int val; int start; int end;
public TreeNode(int val, int x, int y) {
this.val = val;
this.start = x;
this.end = y;
}
public String toString() {
return Integer.toString(this.val);
}
}
/*
1
10
6 10 7 9 11 99 45 20 88 31
*/
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 502fca1b034ad3f6ecca7fda608c3de9 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.StringTokenizer;
/**
* Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
*
* To modify the template, go to Preferences -> Editor -> File and Code Templates -> Other
*/
public class Main {
private static int avg(int x, int y) {
return Math.min(x,y) + (Math.max(x,y)-Math.min(x,y))/2;
}
public static void main(String[] args) {
int t = fs.nextInt();
while (t-- > 0) {
int x = fs.nextInt(), y = fs.nextInt();
int N;
if (x <= y) {
boolean averageWorks = y < 3*x;
if (averageWorks) {
// the "easy" case
// because avg%x equals y%avg
// this is the case for x = 10, y = 12
// where avg = 11 and 12%11 = 11%10
// but it is NOT the case for x = 4, y = 12
// where avg = 8 and 12%8 is NOT 8%4
// the average "works" if avg is not too much bigger
// than x
// because if it is not too much bigger,
// we have avg%x = (y-x)/2
// specifically avg = x + (y-x)/2 < 2x
// iff (y-x)/2 < x
// iff y < 3x
N = avg(x,y);
} else {
// can be proven
N = y-avg(x,y)%x;
}
} else {
// x > y
N = x + y;
// since then y%(x+y) = y
// and (x+y)%x = y%x = y (last equality due to y < x)
}
o.println(N);
}
o.flush();
}
static FastScanner fs = new FastScanner();
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
static PrintWriter o = new PrintWriter(new OutputStreamWriter(System.out));
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | e1d2a3bb009a295ee331f10386a438bc | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.io.*;
import java.util.*;
public class cp {
static int mod=(int)1e9+7;
// static Reader sc=new Reader();
static FastReader sc=new FastReader(System.in);
static int[] sp;
static int size=(int)1e6;
static int[] arInt;
static long[] arLong;
public static void main(String[] args) throws IOException {
long tc=sc.nextLong();
// Scanner sc=new Scanner(System.in);
// int tc=1;
// print_int(pre);
// primeSet=new HashSet<>();
// primeCnt=new int[(int)1e9];
// sieveOfEratosthenes((int)1e9);
while(tc-->0)
{
long x=sc.nextLong();
long y=sc.nextLong();
if(x>y)
{
out.println(x+y);
}
else {
long ans=y-(y%x)/2;
out.println(ans);
}
}
// System.out.flush();
out.flush();
out.close();
System.gc();
}
/*
...SOLUTION ENDS HERE...........SOLUTION ENDS HERE...
*/
//
static boolean util(long a[],long b[],long mid,long k)
{
long cnt=0;
for (int i = 0; i < b.length; i++) {
cnt+=Math.max(0, a[i]*mid-b[i]);
if(cnt>k)
return false;
}
return true;
}
static void mapping(int a[])
{
Pair[] temp=new Pair[a.length];
for (int i = 0; i < temp.length; i++) {
temp[i]=new Pair(a[i], i);
}
Arrays.sort(temp);
int k=0;
for (int i = 0; i < temp.length; i++) {
a[temp[i].y]=k++;
}
}
static boolean palin(String s)
{
for(int i=0;i<s.length()/2;i++)
if(s.charAt(i)!=s.charAt(s.length()-i-1))
return false;
return true;
}
static class temp implements Comparable<temp>{
int x;
int y;
int sec;
public temp(int x,int y,int l) {
// TODO Auto-generated constructor stub
this.x=x;
this.y=y;
this.sec=l;
}
@Override
public int compareTo(cp.temp o) {
// TODO Auto-generated method stub
return this.sec-o.sec;
}
}
static class Node{
int x;
int y;
ArrayList<Integer> edges;
public Node(int x,int y) {
// TODO Auto-generated constructor stub
this.x=x;
this.y=y;
this.edges=new ArrayList<>();
}
}
static int lis(int arr[],int n)
{
int ans=0;
int dp[]=new int[n+1];
Arrays.fill(dp, int_max);
dp[0]=int_min;
for(int i=0;i<n;i++)
{
int j=UpperBound(dp,arr[i]);
if(dp[j-1]<=arr[i] && arr[i]<dp[j])
dp[j]=arr[i];
}
for(int i=0;i<=n;i++)
{
if(dp[i]<int_max)
ans=i;
}
return ans;
}
static long get(long n)
{
return n*(n+1)/2L;
}
static boolean go(ArrayList<Pair> caves,int k)
{
for(Pair each:caves)
{
if(k<=each.x)
return false;
k+=each.y;
}
return true;
}
static String revString(String s)
{
char arr[]=s.toCharArray();
int n=s.length();
for(int i=0;i<n/2;i++)
{
char temp=arr[i];
arr[i]=arr[n-i-1];
arr[n-i-1]=temp;
}
return String.valueOf(arr);
}
// Fuction return the number of set bits in n
static int SetBits(int n)
{
int cnt=0;
while(n>0)
{
if((n&1)==1)
{
cnt++;
}
n=n>>1;
}
return cnt;
}
static boolean isPowerOfTwo(int n)
{
return (int)(Math.ceil((Math.log(n) / Math.log(2))))
== (int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
static void arrInt(int n) throws IOException
{
arInt=new int[n];
for (int i = 0; i < arInt.length; i++) {
arInt[i]=sc.nextInt();
}
}
static void arrLong(int n) throws IOException
{
arLong=new long[n];
for (int i = 0; i < arLong.length; i++) {
arLong[i]=sc.nextLong();
}
}
static ArrayList<Integer> add(int id,int c)
{
ArrayList<Integer> newArr=new ArrayList<>();
for(int i=0;i<id;i++)
newArr.add(arInt[i]);
newArr.add(c);
for(int i=id;i<arInt.length;i++)
{
newArr.add(arInt[i]);
}
return newArr;
}
// function to find first index >= y
static int upper(ArrayList<Integer> arr, int n, int x)
{
int l = 0, h = n - 1;
while (h-l>1)
{
int mid = (l + h) / 2;
if (arr.get(mid) <= x)
l=mid+1;
else
{
h=mid;
}
}
if(arr.get(l)>x)
{
return l;
}
if(arr.get(h)>x)
return h;
return -1;
}
static int upper(ArrayList<Long> arr, int n, long x)
{
int l = 0, h = n - 1;
while (h-l>1)
{
int mid = (l + h) / 2;
if (arr.get(mid) <= x)
l=mid+1;
else
{
h=mid;
}
}
if(arr.get(l)>x)
{
return l;
}
if(arr.get(h)>x)
return h;
return -1;
}
static int lower(ArrayList<Integer> arr, int n, int x)
{
int l = 0, h = n - 1;
while (h-l>1)
{
int mid = (l + h) / 2;
if (arr.get(mid) < x)
l=mid+1;
else
{
h=mid;
}
}
if(arr.get(l)>=x)
{
return l;
}
if(arr.get(h)>=x)
return h;
return -1;
}
static int N = 501;
// Array to store inverse of 1 to N
static long[] factorialNumInverse = new long[N + 1];
// Array to precompute inverse of 1! to N!
static long[] naturalNumInverse = new long[N + 1];
// Array to store factorial of first N numbers
static long[] fact = new long[N + 1];
// Function to precompute inverse of numbers
public static void InverseofNumber(int p)
{
naturalNumInverse[0] = naturalNumInverse[1] = 1;
for(int i = 2; i <= N; i++)
naturalNumInverse[i] = naturalNumInverse[p % i] *
(long)(p - p / i) % p;
}
// Function to precompute inverse of factorials
public static void InverseofFactorial(int p)
{
factorialNumInverse[0] = factorialNumInverse[1] = 1;
// Precompute inverse of natural numbers
for(int i = 2; i <= N; i++)
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p;
}
// Function to calculate factorial of 1 to N
public static void factorial(int p)
{
fact[0] = 1;
// Precompute factorials
for(int i = 1; i <= N; i++)
{
fact[i] = (fact[i - 1] * (long)i) % p;
}
}
// Function to return nCr % p in O(1) time
public static long Binomial(int N, int R, int p)
{
// n C r = n!*inverse(r!)*inverse((n-r)!)
long ans = ((fact[N] * factorialNumInverse[R]) %
p * factorialNumInverse[N - R]) % p;
return ans;
}
static String tr(String s)
{
int now = 0;
while (now + 1 < s.length() && s.charAt(now)== '0')
++now;
return s.substring(now);
}
static ArrayList<Integer> ans;
static void dfs(int node,Graph gg,int cnt,int k,ArrayList<Integer> temp)
{
if(cnt==k)
return;
for(Integer each:gg.list[node])
{
if(each==0)
{
temp.add(each);
ans=new ArrayList<>(temp);
temp.remove(temp.size()-1);
continue;
}
temp.add(each);
dfs(each,gg,cnt+1,k,temp);
temp.remove(temp.size()-1);
}
return;
}
static boolean isPrime(long n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static ArrayList<Integer> commDiv(int a, int b)
{
// find gcd of a, b
int n = gcd(a, b);
// Count divisors of n.
ArrayList<Integer> Div=new ArrayList<>();
for (int i = 1; i <= Math.sqrt(n); i++) {
// if 'i' is factor of n
if (n % i == 0) {
// check if divisors are equal
if (n / i == i)
Div.add(i);
else
{
Div.add(i);
Div.add(n/i);
}
}
}
return Div;
}
static HashSet<Integer> factors(int x)
{
HashSet<Integer> a=new HashSet<Integer>();
for(int i=2;i*i<=x;i++)
{
if(x%i==0)
{
a.add(i);
a.add(x/i);
}
}
return a;
}
static void primeFactors(int n,HashSet<Integer> factors)
{
// Print the number of 2s that divide n
while (n%2==0)
{
factors.add(2);
n /= 2;
}
// n must be odd at this point. So we can
// skip one element (Note i = i +2)
for (int i = 3; i <= Math.sqrt(n); i+= 2)
{
// While i divides n, print i and divide n
while (n%i == 0)
{
factors.add(i);
n /= i;
}
}
// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
factors.add(n);
}
// static class Node
// {
// int vertex;
// HashSet<Node> adj;
// int deg;
// Node(int ver)
// {
// vertex=ver;
// deg=0;
// adj=new HashSet<Node>();
// }
// @Override
// public String toString()
// {
// return vertex+" ";
// }
// }
static class Tuple{
int a;
int b;
int c;
public Tuple(int a,int b,int c) {
this.a=a;
this.b=b;
this.c=c;
}
}
//function to find prime factors of n
static HashMap<Long,Long> findFactors(long n2)
{
HashMap<Long,Long> ans=new HashMap<>();
if(n2%2==0)
{
ans.put(2L, 0L);
// cnt++;
while((n2&1)==0)
{
n2=n2>>1;
ans.put(2L, ans.get(2L)+1);
//
}
}
for(long i=3;i*i<=n2;i+=2)
{
if(n2%i==0)
{
ans.put((long)i, 0L);
// cnt++;
while(n2%i==0)
{
n2=n2/i;
ans.put((long)i, ans.get((long)i)+1);
}
}
}
if(n2!=1)
{
ans.put(n2, ans.getOrDefault(n2, (long) 0)+1);
}
return ans;
}
//fenwick tree implementaion
static class fwt
{
int n;
long BITree[];
fwt(int n)
{
this.n=n;
BITree=new long[n+1];
}
fwt(int arr[], int n)
{
this.n=n;
BITree=new long[n+1];
for(int i = 0; i < n; i++)
updateBIT(n, i, arr[i]);
}
long getSum(int index)
{
long sum = 0;
index = index + 1;
while(index>0)
{
sum += BITree[index];
index -= index & (-index);
}
return sum;
}
void updateBIT(int n, int index,int val)
{
index = index + 1;
while(index <= n)
{
BITree[index] += val;
index += index & (-index);
}
}
long sum(int l, int r) {
return getSum(r) - getSum(l - 1);
}
void print()
{
for(int i=0;i<n;i++)
out.print(getSum(i)+" ");
out.println();
}
}
static class sparseTable{
int n;
long[][]dp;
int log2[];
int P;
void buildTable(long[] arr)
{
n=arr.length;
P=(int)Math.floor(Math.log(n)/Math.log(2));
log2=new int[n+1];
log2[0]=log2[1]=0;
for(int i=2;i<=n;i++)
{
log2[i]=log2[i/2]+1;
}
dp=new long[P+1][n];
for(int i=0;i<n;i++)
{
dp[0][i]=arr[i];
}
for(int p=1;p<=P;p++)
{
for(int i=0;i+(1<<p)<=n;i++)
{
long left=dp[p-1][i];
long right=dp[p-1][i+(1<<(p-1))];
dp[p][i]=gcd(left, right);
}
}
}
long maxQuery(int l,int r)
{
int len=r-l+1;
int p=(int)Math.floor(log2[len]);
long left=dp[p][l];
long right=dp[p][r-(1<<p)+1];
return gcd(left, right);
}
}
//Function to find number of set bits
static int setBitNumber(long n)
{
if (n == 0)
return 0;
int msb = 0;
n = n / 2;
while (n != 0) {
n = n / 2;
msb++;
}
return msb;
}
static int getFirstSetBitPos(long n)
{
return (int)((Math.log10(n & -n)) / Math.log10(2)) + 1;
}
static ArrayList<Integer> primes;
static HashSet<Integer> primeSet;
static boolean prime[];
static int primeCnt[];
static void sieveOfEratosthenes(int n)
{
prime= new boolean[n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
primeCnt[i]=primeCnt[i-1]+1;
}
}
static long mod(long a, long b) {
long c = a % b;
return (c < 0) ? c + b : c;
}
static void swap(long arr[],int i,int j)
{
long temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
static boolean util(int a,int b,int c)
{
if(b>a)util(b, a, c);
while(c>=a)
{
c-=a;
if(c%b==0)
return true;
}
return (c%b==0);
}
static void flag(boolean flag)
{
out.println(flag ? "YES" : "NO");
out.flush();
}
static void print(int a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print(long a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print_int(ArrayList<Integer> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static void print_long(ArrayList<Long> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static void printYesNo(boolean condition)
{
if (condition) {
out.println("YES");
}
else {
out.println("NO");
}
}
static int LowerBound(int a[], int x)
{ // x is the target value or key
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
static int lowerIndex(int arr[], int n, int x)
{
int l = 0, h = n - 1;
while (l<=h)
{
int mid = (l + h) / 2;
if (arr[mid] >= x)
h = mid -1 ;
else
l = mid + 1;
}
return l;
}
// function to find last index <= y
static int upperIndex(int arr[], int n, int y)
{
int l = 0, h = n - 1;
while (l <= h)
{
int mid = (l + h) / 2;
if (arr[mid] <= y)
l = mid + 1;
else
h = mid - 1;
}
return h;
}
static int upperIndex(long arr[], int n, long y)
{
int l = 0, h = n - 1;
while (l <= h)
{
int mid = (l + h) / 2;
if (arr[mid] <= y)
l = mid + 1;
else
h = mid - 1;
}
return h;
}
static int UpperBound(int a[], int x)
{// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
static int UpperBound(long a[], long x)
{// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
static class DisjointUnionSets
{
int[] rank, parent;
int n;
// Constructor
public DisjointUnionSets(int n)
{
rank = new int[n];
parent = new int[n];
this.n = n;
makeSet();
}
// Creates n sets with single item in each
void makeSet()
{
for (int i = 0; i < n; i++)
parent[i] = i;
}
int find(int x)
{
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}
// Unites the set that includes x and the set
// that includes x
void union(int x, int y)
{
int xRoot = find(x), yRoot = find(y);
if (xRoot == yRoot)
return;
if (rank[xRoot] < rank[yRoot])
parent[xRoot] = yRoot;
else if (rank[yRoot] < rank[xRoot])
parent[yRoot] = xRoot;
else // if ranks are the same
{
parent[yRoot] = xRoot;
rank[xRoot] = rank[xRoot] + 1;
}
// if(xRoot!=yRoot)
// parent[y]=x;
}
int connectedComponents()
{
int cnt=0;
for(int i=0;i<n;i++)
{
if(parent[i]==i)
cnt++;
}
return cnt;
}
}
static class Graph
{
int v;
ArrayList<Integer> list[];
Graph(int v)
{
this.v=v;
list=new ArrayList[v+1];
for(int i=1;i<=v;i++)
list[i]=new ArrayList<Integer>();
}
void addEdge(int a, int b)
{
this.list[a].add(b);
}
}
// static class GraphMap{
// Map<String,ArrayList<String>> graph;
// GraphMap() {
// // TODO Auto-generated constructor stub
// graph=new HashMap<String,ArrayList<String>>();
//
// }
// void addEdge(String a,String b)
// {
// if(graph.containsKey(a))
// this.graph.get(a).add(b);
// else {
// this.graph.put(a, new ArrayList<>());
// this.graph.get(a).add(b);
// }
// }
// }
// static void dfsMap(GraphMap g,HashSet<String> vis,String src,int ok)
// {
// vis.add(src);
//
// if(g.graph.get(src)!=null)
// {
// for(String each:g.graph.get(src))
// {
// if(!vis.contains(each))
// {
// dfsMap(g, vis, each, ok+1);
// }
// }
// }
//
// cnt=Math.max(cnt, ok);
// }
static double sum[];
static long cnt;
// static void DFS(Graph g, boolean[] visited, int u)
// {
// visited[u]=true;
//
// for(int i=0;i<g.list[u].size();i++)
// {
// int v=g.list[u].get(i);
//
// if(!visited[v])
// {
// cnt1=cnt1*2;
// DFS(g, visited, v);
//
// }
//
// }
//
//
// }
static class Pair implements Comparable<Pair>
{
int x;
int y;
Pair(int x,int y)
{
this.x=x;
this.y=y;
}
@Override
public int compareTo(Pair o) {
// TODO Auto-generated method stub
return this.x-o.x;
}
}
static long sum_array(int a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static long sum_array(long a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static void sort(int[] a)
{
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void sort(long[] a)
{
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void reverse_array(int a[])
{
int n=a.length;
int i,t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static void reverse_array(long a[])
{
int n=a.length;
int i; long t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
// static long modInverse(long a, long m)
// {
// long g = gcd(a, m);
//
// return power(a, m - 2, m);
//
// }
static long power(long x, long y)
{
long res = 1;
x = x % mod;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % mod;
y = y >> 1; // y = y/2
x = (x * x) % mod;
}
return res;
}
static int power(int x, int y)
{
int res = 1;
x = x % mod;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % mod;
y = y >> 1; // y = y/2
x = (x * x) % mod;
}
return res;
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
//cnt+=a/b;
return gcd(b%a,a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static class FastReader{
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static PrintWriter out=new PrintWriter(System.out);
static int int_max=Integer.MAX_VALUE;
static int int_min=Integer.MIN_VALUE;
static long long_max=Long.MAX_VALUE;
static long long_min=Long.MIN_VALUE;
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | fc737c85772ef4f5b9754578f3db999f | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes |
import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Main {
private static FS sc = new FS();
private static class FS {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
private static class extra {
static int[] intArr(int size) {
int[] a = new int[size];
for(int i = 0; i < size; i++) a[i] = sc.nextInt();
return a;
}
static long[] longArr(int size) {
Scanner scc = new Scanner(System.in);
long[] a = new long[size];
for(int i = 0; i < size; i++) a[i] = sc.nextLong();
return a;
}
static long intSum(int[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static long longSum(long[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static LinkedList[] graphD(int vertices, int edges) {
LinkedList<Integer>[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
}
return temp;
}
static LinkedList[] graphUD(int vertices, int edges) {
LinkedList<Integer>[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
temp[y].add(x);
}
return temp;
}
static void printG(LinkedList[] temp) { for(LinkedList<Integer> aa:temp) System.out.println(aa); }
static long cal(long val, long pow, long mod) {
if(pow == 0) return 1;
long res = cal(val, pow/2, mod);
long ret = (res*res)%mod;
if(pow%2 == 0) return ret;
return (val*ret)%mod;
}
static long gcd(long a, long b) { return b == 0 ? a:gcd(b, a%b); }
}
static int mod = (int) 1e9 + 7;
// static LinkedList<Integer>[] temp, idx;
static long inf = (long) Long.MAX_VALUE;
// static long inf = Long.MAX_VALUE;
// static int max;
// static StringBuilder out;
public static void main(String[] args) {
int t = sc.nextInt();
// int t = 1;
StringBuilder ret = new StringBuilder();
while(t-- > 0) {
int x = sc.nextInt(), y = sc.nextInt();
if(x > y) {
long sum = x + y;
ret.append(sum + "\n");
} else {
long sum = y - (y%x)/2;
ret.append(sum + "\n");
}
}
System.out.println(ret);
}
}
//0110 -> 1000
//1010
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | c4c320954676a1c37d40d6a14acfefe7 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
BModerateModularMode solver = new BModerateModularMode();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class BModerateModularMode {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int x = in.nextInt();
int y = in.nextInt();
if (x <= y) {
out.println(y - y % x / 2);
} else {
out.println(x + y);
}
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return nextString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 1f352024e7bfa3cf0346eec33ee91a88 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class B1603 {
/**
* @param args the command line arguments
*/
public static void main(String[] args) throws IOException {
// TODO code application logic here
// Test later
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(f.readLine());
for(int i = 0; i < T; i++) {
// Let Q = 1
StringTokenizer st = new StringTokenizer(f.readLine());
int x = Integer.parseInt(st.nextToken());
int y = Integer.parseInt(st.nextToken());
int a = y/x;
int b = (y%x)/2;
int n = x*a + b;
if(y < x) {
n = x + y;
}
System.out.println(n);
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 4b5c14de30c9e734595f0c5d479b479a | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
public class Contest1604D
{
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
String next() { // reads in the next string
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() { // reads in the next int
return Integer.parseInt(next());
}
public long nextLong() { // reads in the next long
return Long.parseLong(next());
}
public double nextDouble() { // reads in the next double
return Double.parseDouble(next());
}
}
static InputReader r = new InputReader(System.in);
static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args)
{
//DOES BINARY SEARCH WORK ALWAYS LOOK FOR MONOTONIC STUFF
//LOOK FOR INT OVERFLOW
//LOOK FOR SCOPE ERROR
//Add negatives
//SORT BOTH ARRAYS
//CHECK THE LAST ONE THAT IS OUT LOOP
//SORT BY BOTH VALUES
//DON'T ADD STRINGS,USE STRING BUILDER
int t = r.nextInt();
while (t > 0)
{
t--;
long x = r.nextLong();
long y = r.nextLong();
long ans = 0;
pw.println(x > y? x*y+y:y-(y%x)/2);
}
pw.close();
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 91ea5b4717c21ee666a9d327d2239c7c | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.IOException;
import java.lang.reflect.Field;
import java.nio.charset.StandardCharsets;
import java.io.UncheckedIOException;
import java.io.Closeable;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) throws Exception {
Thread thread = new Thread(null, new TaskAdapter(), "", 1 << 29);
thread.start();
thread.join();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastInput in = new FastInput(inputStream);
FastOutput out = new FastOutput(outputStream);
BModerateModularMode solver = new BModerateModularMode();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
}
static class BModerateModularMode {
public void solve(int testNumber, FastInput in, FastOutput out) {
int x = in.ri();
int y = in.ri();
if (x == y) {
out.println(x);
return;
}
if (x < y) {
int m = (x + y) / 2;
int mod = m % x;
out.println(y - mod);
return;
}
out.println(x + y);
}
}
static class FastInput {
private final InputStream is;
private StringBuilder defaultStringBuf = new StringBuilder(1 << 13);
private byte[] buf = new byte[1 << 13];
private int bufLen;
private int bufOffset;
private int next;
public FastInput(InputStream is) {
this.is = is;
}
private int read() {
while (bufLen == bufOffset) {
bufOffset = 0;
try {
bufLen = is.read(buf);
} catch (IOException e) {
bufLen = -1;
}
if (bufLen == -1) {
return -1;
}
}
return buf[bufOffset++];
}
public void skipBlank() {
while (next >= 0 && next <= 32) {
next = read();
}
}
public String next() {
return readString();
}
public int ri() {
return readInt();
}
public int readInt() {
boolean rev = false;
skipBlank();
if (next == '+' || next == '-') {
rev = next == '-';
next = read();
}
int val = 0;
while (next >= '0' && next <= '9') {
val = val * 10 - next + '0';
next = read();
}
return rev ? val : -val;
}
public String readString(StringBuilder builder) {
skipBlank();
while (next > 32) {
builder.append((char) next);
next = read();
}
return builder.toString();
}
public String readString() {
defaultStringBuf.setLength(0);
return readString(defaultStringBuf);
}
}
static class FastOutput implements AutoCloseable, Closeable, Appendable {
private static final int THRESHOLD = 32 << 10;
private OutputStream writer;
private StringBuilder cache = new StringBuilder(THRESHOLD * 2);
private static Field stringBuilderValueField;
private char[] charBuf = new char[THRESHOLD * 2];
private byte[] byteBuf = new byte[THRESHOLD * 2];
static {
try {
stringBuilderValueField = StringBuilder.class.getSuperclass().getDeclaredField("value");
stringBuilderValueField.setAccessible(true);
} catch (Exception e) {
stringBuilderValueField = null;
}
stringBuilderValueField = null;
}
public FastOutput append(CharSequence csq) {
cache.append(csq);
return this;
}
public FastOutput append(CharSequence csq, int start, int end) {
cache.append(csq, start, end);
return this;
}
private void afterWrite() {
if (cache.length() < THRESHOLD) {
return;
}
flush();
}
public FastOutput(OutputStream writer) {
this.writer = writer;
}
public FastOutput append(char c) {
cache.append(c);
afterWrite();
return this;
}
public FastOutput append(int c) {
cache.append(c);
afterWrite();
return this;
}
public FastOutput println(int c) {
return append(c).println();
}
public FastOutput println() {
return append('\n');
}
public FastOutput flush() {
try {
if (stringBuilderValueField != null) {
try {
byte[] value = (byte[]) stringBuilderValueField.get(cache);
writer.write(value, 0, cache.length());
} catch (Exception e) {
stringBuilderValueField = null;
}
}
if (stringBuilderValueField == null) {
int n = cache.length();
if (n > byteBuf.length) {
//slow
writer.write(cache.toString().getBytes(StandardCharsets.UTF_8));
// writer.append(cache);
} else {
cache.getChars(0, n, charBuf, 0);
for (int i = 0; i < n; i++) {
byteBuf[i] = (byte) charBuf[i];
}
writer.write(byteBuf, 0, n);
}
}
writer.flush();
cache.setLength(0);
} catch (IOException e) {
throw new UncheckedIOException(e);
}
return this;
}
public void close() {
flush();
try {
writer.close();
} catch (IOException e) {
throw new UncheckedIOException(e);
}
}
public String toString() {
return cache.toString();
}
}
}
| Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 250db33f83318283fd9890238afee598 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
import java.awt.Point;
public class Main{
static int mod = (int) (Math.pow(10, 9)+7);
static final int dx[] = { -1, 0, 1, 0 }, dy[] = { 0, -1, 0, 1 };
static final int[] dx8 = { -1, -1, -1, 0, 0, 1, 1, 1 }, dy8 = { -1, 0, 1, -1, 1, -1, 0, 1 };
static final int[] dx9 = { -1, -1, -1, 0, 0, 0, 1, 1, 1 }, dy9 = { -1, 0, 1, -1, 0, 1, -1, 0, 1 };
static final int inf = Integer.MAX_VALUE / 2;
static final long infL = Long.MAX_VALUE / 3;
static final double infD = Double.MAX_VALUE / 3;
static final double eps = 1e-10;
static final double pi = Math.PI;
static List<Integer> primeNumbers = new ArrayList<>();
public static void main(String[] args) throws IOException{
MyScanner sc = new MyScanner();
//changes in this line of code
out = new PrintWriter(new BufferedOutputStream(System.out));
// out = new PrintWriter(new BufferedWriter(new FileWriter("output.out")));
//will this work on codeforces?
//WRITE YOUR CODE IN HERE
int test = sc.nextInt();
while(test -- > 0){
int f = sc.nextInt();
int s = sc.nextInt();
if(f > s){
out.println(f+s);
}else if(f == s){
out.println(f);
}else{
//think
if(s%f == 0){
out.println(f);
continue;
}
//now think for this one
int mod = s%f;
out.println(s-(mod/2));
}
}
out.close();
}
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//-------------------------------------------------------------------
//fast java implementation for sure
public static class Graph{
public int V;
public ArrayList<ArrayList<Integer>> edges;
Graph(int V){
this.V = V;
edges = new ArrayList<>(V+1);
for(int i= 0; i <= V; i++){
edges.add(new ArrayList<>());
}
}
///single sided this for sure
public void addEdge(int from , int to){
edges.get(from).add(to);
}
}
public static class DisjointUnionSets {
int[] rank, parent;
int n;
// Constructor
public DisjointUnionSets(int n)
{
rank = new int[n];
parent = new int[n];
this.n = n;
makeSet();
}
// Creates n sets with single item in each
void makeSet()
{
for (int i = 0; i < n; i++) {
// Initially, all elements are in
// their own set.
parent[i] = i;
}
}
// Returns representative of x's set
int find(int x)
{
// Finds the representative of the set
// that x is an element of
if (parent[x] != x) {
// if x is not the parent of itself
// Then x is not the representative of
// his set,
parent[x] = find(parent[x]);
// so we recursively call Find on its parent
// and move i's node directly under the
// representative of this set
}
return parent[x];
}
// Unites the set that includes x and the set
// that includes x
void union(int x, int y)
{
// Find representatives of two sets
int xRoot = find(x), yRoot = find(y);
// Elements are in the same set, no need
// to unite anything.
if (xRoot == yRoot)
return;
// If x's rank is less than y's rank
if (rank[xRoot] < rank[yRoot])
// Then move x under y so that depth
// of tree remains less
parent[xRoot] = yRoot;
// Else if y's rank is less than x's rank
else if (rank[yRoot] < rank[xRoot])
// Then move y under x so that depth of
// tree remains less
parent[yRoot] = xRoot;
else // if ranks are the same
{
// Then move y under x (doesn't matter
// which one goes where)
parent[yRoot] = xRoot;
// And increment the result tree's
// rank by 1
rank[xRoot] = rank[xRoot] + 1;
}
}
}
//with mod
public static long power(long x, long y)
{
long res = 1L; // Initialize result
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0){
x %= mod;
res %= mod;
res = (res * x)%mod;
}
// y must be even now
y = y >> 1; // y = y/2
x%= mod;
x = (x * x)%mod; // Change x to x^2
}
// 550193677
return res%mod;
}
//without mod
public static long power2(long x, long y)
{
long res = 1L; // Initialize result
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0){
res = (res * x);
}
// y must be even now
y = y >> 1; // y = y/
x = (x * x);
}
// 550193677
return res;
}
public static class segmentTree{
public long[] arr;
public long[] tree;
public long[] lazy;
segmentTree(long[] array){
int n = array.length;
arr = new long[n];
for(int i= 0; i < n; i++) arr[i] = array[i];
tree = new long[4*n + 1];
lazy = new long[4*n + 1];
}
public void build(int[]arr, int s, int e, int[] tree, int index){
if(s == e){
tree[index] = arr[s];
return;
}
//otherwise divide in two parts and fill both sides simply
int mid = (s+e)/2;
build(arr, s, mid, tree, 2*index);
build(arr, mid+1, e, tree, 2*index+1);
//who will build the current position dude
tree[index] = Math.min(tree[2*index], tree[2*index+1]);
}
public int query(int sr, int er, int sc, int ec, int index, int[] tree){
if(lazy[index] != 0){
tree[index] += lazy[index];
if(sc != ec){
lazy[2*index+1] += lazy[index];
lazy[2*index] += lazy[index];
}
lazy[index] = 0;
}
//no overlap
if(sr > ec || sc > er) return Integer.MAX_VALUE;
//found the index baby
if(sr <= sc && ec <= er) return tree[index];
//finding the index on both sides hehehehhe
int mid = (sc + ec)/2;
int left = query(sr, er, sc, mid, 2*index, tree);
int right = query(sr, er, mid+1, ec, 2*index + 1, tree);
return Integer.min(left, right);
}
//now we will do point update implementation
//it should be simple then we expected for sure
public void update(int index, int indexr, int increment, int[] tree, int s, int e){
if(lazy[index] != 0){
tree[index] += lazy[index];
if(s != e){
lazy[2*index+1] = lazy[index];
lazy[2*index] = lazy[index];
}
lazy[index] = 0;
}
//no overlap
if(indexr < s || indexr > e) return;
//found the required index
if(s == e){
tree[index] += increment;
return;
}
//search for the index on both sides
int mid = (s+e)/2;
update(2*index, indexr, increment, tree, s, mid);
update(2*index+1, indexr, increment, tree, mid+1, e);
//now update the current range simply
tree[index] = Math.min(tree[2*index+1], tree[2*index]);
}
public void rangeUpdate(int[] tree , int index, int s, int e, int sr, int er, int increment){
//if not at all in the same range
if(e < sr || er < s) return;
//complete then also move forward
if(s == e){
tree[index] += increment;
return;
}
//otherwise move in both subparts
int mid = (s+e)/2;
rangeUpdate(tree, 2*index, s, mid, sr, er, increment);
rangeUpdate(tree, 2*index + 1, mid+1, e, sr, er, increment);
//update current range too na
//i always forget this step for some reasons hehehe, idiot
tree[index] = Math.min(tree[2*index], tree[2*index + 1]);
}
public void rangeUpdateLazy(int[] tree, int index, int s, int e, int sr, int er, int increment){
//update lazy values
//resolve lazy value before going down
if(lazy[index] != 0){
tree[index] += lazy[index];
if(s != e){
lazy[2*index+1] += lazy[index];
lazy[2*index] += lazy[index];
}
lazy[index] = 0;
}
//no overlap case
if(sr > e || s > er) return;
//complete overlap
if(sr <= s && er >= e){
tree[index] += increment;
if(s != e){
lazy[2*index+1] += increment;
lazy[2*index] += increment;
}
return;
}
//otherwise go on both left and right side and do your shit
int mid = (s + e)/2;
rangeUpdateLazy(tree, 2*index, s, mid, sr, er, increment);
rangeUpdateLazy(tree, 2*index + 1, mid+1, e, sr, er, increment);
tree[index] = Math.min(tree[2*index+1], tree[2*index]);
return;
}
}
//prime sieve
public static void primeSieve(int n){
BitSet bitset = new BitSet(n+1);
for(long i = 0; i < n ; i++){
if (i == 0 || i == 1) {
bitset.set((int) i);
continue;
}
if(bitset.get((int) i)) continue;
primeNumbers.add((int)i);
for(long j = i; j <= n ; j+= i)
bitset.set((int)j);
}
}
//number of divisors
public static int countDivisors(long number){
if(number == 1) return 1;
List<Integer> primeFactors = new ArrayList<>();
int index = 0;
long curr = primeNumbers.get(index);
while(curr * curr <= number){
while(number % curr == 0){
number = number/curr;
primeFactors.add((int) curr);
}
index++;
curr = primeNumbers.get(index);
}
if(number != 1) primeFactors.add((int) number);
int current = primeFactors.get(0);
int totalDivisors = 1;
int currentCount = 2;
for (int i = 1; i < primeFactors.size(); i++) {
if (primeFactors.get(i) == current) {
currentCount++;
} else {
totalDivisors *= currentCount;
currentCount = 2;
current = primeFactors.get(i);
}
}
totalDivisors *= currentCount;
return totalDivisors;
}
//now adding next permutation function to java hehe
public static boolean next_permutation(int[] p) {
for (int a = p.length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (int b = p.length - 1;; --b)
if (p[b] > p[a]) {
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
return false;
}
//finding the value of NCR in O(RlogN) time and O(1) space
public static long getNcR(int n, int r)
{
long p = 1, k = 1;
if (n - r < r) r = n - r;
if (r != 0) {
while (r > 0) {
p *= n;
k *= r;
long m = __gcd(p, k);
p /= m;
k /= m;
n--;
r--;
}
}
else {
p = 1;
}
return p;
}
public static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//for ncr calculator
public static long __gcd(long n1, long n2)
{
long gcd = 1;
for (int i = 1; i <= n1 && i <= n2; ++i) {
// Checks if i is factor of both integers
if (n1 % i == 0 && n2 % i == 0) {
gcd = i;
}
}
return gcd;
}
//is vowel function
public static boolean isVowel(char c)
{
return (c=='a' || c=='A' || c=='e' || c=='E' || c=='i' || c=='I' || c=='o' || c=='O' || c=='u' || c=='U');
}
//two add two big numbers with some mod
public static int add(int a, int b) {
a+=b;
if (a>=mod) return a-mod;
return a;
}
//two sub two numbers
public static int sub(int a, int b) {
a-=b;
if (a<0) a+=mod;
else if (a>=mod) a-=mod;
return a;
}
//to sort the array with better method
public static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//for calculating binomialCoeff
public static int binomialCoeff(int n, int k)
{
int C[] = new int[k + 1];
// nC0 is 1
C[0] = 1;
for (int i = 1; i <= n; i++) {
// Compute next row of pascal
// triangle using the previous row
for (int j = Math.min(i, k); j > 0; j--)
C[j] = C[j] + C[j - 1];
}
return C[k];
}
//Pair with int int
public static class Pair{
public int a;
public int b;
Pair(int a , int b){
this.a = a;
this.b = b;
}
@Override
public String toString(){
return a + " -> " + b;
}
}
//Triplet with int int int
public static class Triplet{
public int a;
public int b;
public int c;
Triplet(int a , int b, int c){
this.a = a;
this.b = b;
this.c = c;
}
@Override
public String toString(){
return a + " -> " + b;
}
}
//Shortcut function
public static long lcm(long a , long b){
return a * (b/gcd(a,b));
}
//let's make one for calculating lcm basically
public static int lcm(int a , int b){
return (a * b)/gcd(a,b);
}
//int version for gcd
public static int gcd(int a, int b){
if(b == 0)
return a;
return gcd(b , a%b);
}
//long version for gcd
public static long gcd(long a, long b){
if(b == 0)
return a;
return gcd(b , a%b);
}
//swapping two elements in an array
public static void swap(int[] arr, int left , int right){
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
//
//for char array
public static void swap(char[] arr, int left , int right){
char temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
//reversing an array
public static void reverse(int[] arr){
int left = 0;
int right = arr.length-1;
while(left <= right){
swap(arr, left,right);
left++;
right--;
}
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
// public MyScanner() throws FileNotFoundException {
// br = new BufferedReader(new FileReader("input.in"));
// }
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
//--------------------------------------------------------
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 7e3c2bfd168c472721bb212610fb92ca | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
/**
*
* @author M1ME
*
*/
public class D {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
StringBuilder st = new StringBuilder();
int t = sc.nextInt();
for (int test = 0; test < t; test++) {
int n = sc.nextInt();
int m = sc.nextInt();
if (n == m) st.append(n + "\n");
else if (n > m) st.append(n + m + "\n");
else st.append(m - ((m % n) / 2) + "\n");
}
System.out.print(st.toString());
}
static void sort(int[] a) {
ArrayList<Integer> al = new ArrayList<>();
for (int i : a)
al.add(i);
Collections.sort(al);
for (int i = 0; i < a.length; i++)
a[i] = al.get(i);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | e02701f0a359735b7f205347c4327bd3 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
public static void main(String[] args) throws Exception {
int T = r.readInt();
for (int t = 0; t < T; t++) {
long x = r.readLong(), y = r.readLong();
long n;
if (y % x == 0) {
n = y;
} else if (x > y) {
n = x + y;
} else {
n = y - (y % x) / 2;
}
System.out.println(n);
}
}
//749999999
static public InputReader r = new InputReader(System.in);
static public OutputWriter w = new OutputWriter(System.out);
static public class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] readIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = this.readInt();
}
return array;
}
public long[] readLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; i++) {
array[i] = this.readLong();
}
return array;
}
public long readLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static public class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | db05b104c568a0b592bac460fef875c1 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class SolutionB extends Thread {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
private static final FastReader scanner = new FastReader();
private static final PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) {
new Thread(null, new SolutionB(), "Main", 1 << 28).start();
}
public void run() {
int t = scanner.nextInt();
for (int i = 0; i < t; i++) {
solve();
}
out.close();
}
private static void solve() {
int x = scanner.nextInt();
int y = scanner.nextInt();
if (y % x == 0) {
out.println(x);
return;
}
if (y < x) {
out.println(x + y);
return;
}
int n = y-1 - (((y-1) % x) - 1) / 2;
out.println(n);
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 77deff7a0b2d3edc1baaba1527013527 | train_110.jsonl | 1635604500 | YouKn0wWho has two even integers $$$x$$$ and $$$y$$$. Help him to find an integer $$$n$$$ such that $$$1 \le n \le 2 \cdot 10^{18}$$$ and $$$n \bmod x = y \bmod n$$$. Here, $$$a \bmod b$$$ denotes the remainder of $$$a$$$ after division by $$$b$$$. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | 256 megabytes | import java.util.*;
import java.io.*;
public class _752 {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
while (t-- > 0) {
long x = sc.nextLong();
long y = sc.nextLong();
if (x == y) {
out.println(x);
} else if (y < x) {
out.println(x + y);
} else if (y % x != 0) {
long mult = (y / x) * x;
out.println((y + mult) / 2);
} else {
out.println(y);
}
}
out.close();
}
static void sort(int[] a) {
ArrayList<Integer> q = new ArrayList<>();
for (int i : a) q.add(i);
Collections.sort(q);
for (int i = 0; i < a.length; i++) a[i] = q.get(i);
}
static void sort(long[] a) {
ArrayList<Long> q = new ArrayList<>();
for (long i : a) q.add(i);
Collections.sort(q);
for (int i = 0; i < a.length; i++) a[i] = q.get(i);
}
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["4\n4 8\n4 2\n420 420\n69420 42068"] | 1 second | ["4\n10\n420\n9969128"] | NoteIn the first test case, $$$4 \bmod 4 = 8 \bmod 4 = 0$$$.In the second test case, $$$10 \bmod 4 = 2 \bmod 10 = 2$$$.In the third test case, $$$420 \bmod 420 = 420 \bmod 420 = 0$$$. | Java 11 | standard input | [
"constructive algorithms",
"math",
"number theory"
] | a24aac9152417527d43b9b422e3d2303 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains two integers $$$x$$$ and $$$y$$$ ($$$2 \le x, y \le 10^9$$$, both are even). | 1,600 | For each test case, print a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{18}$$$) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints. | standard output | |
PASSED | 7809ae91430638183c4caae3704debf1 | train_110.jsonl | 1635604500 | It was October 18, 2017. Shohag, a melancholic soul, made a strong determination that he will pursue Competitive Programming seriously, by heart, because he found it fascinating. Fast forward to 4 years, he is happy that he took this road. He is now creating a contest on Codeforces. He found an astounding problem but has no idea how to solve this. Help him to solve the final problem of the round.You are given three integers $$$n$$$, $$$k$$$ and $$$x$$$. Find the number, modulo $$$998\,244\,353$$$, of integer sequences $$$a_1, a_2, \ldots, a_n$$$ such that the following conditions are satisfied: $$$0 \le a_i \lt 2^k$$$ for each integer $$$i$$$ from $$$1$$$ to $$$n$$$. There is no non-empty subsequence in $$$a$$$ such that the bitwise XOR of the elements of the subsequence is $$$x$$$. A sequence $$$b$$$ is a subsequence of a sequence $$$c$$$ if $$$b$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements. | 512 megabytes | import java.util.*;
import java.io.*;
public class cf {
static long modulo(long a, long b, long c) {
long x=1;
long y=a;
while(b > 0){
if(b%2 == 1){
x=(x*y)%c;
}
y = (y*y)%c; // squaring the base
b /= 2;
}
return x%c;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
long m = 998244353;
while (t-- > 0) {
String[] ss = br.readLine().split(" ");
long n = Long.parseLong(ss[0]), k = Long.parseLong(ss[1]), x = Long.parseLong(ss[2]);
if (x == 0) {
if (n > k) {
System.out.println(0);
} else {
long power = modulo(2, k, m);
long prod = 1;
long exp = 1;
for (int i = 0; i < n; i++) {
prod = prod * (power - exp) % m;
exp = 2 * exp % m;
}
if (prod < 0) {
prod += m;
}
System.out.println(prod);
}
} else {
long coeff = 1;
long sum = 0;
long term = modulo(2, (k - 1) * (n + 1), m);
long pow = modulo(2, k - 1, m);
long inv = modulo(499122177, n + 1, m);
for (int i = 0; i < k; i++) {
sum = (sum + term * coeff) % m;
term = term * inv % m;
coeff = coeff * (1 - pow) % m;
pow = pow * 499122177 % m;
}
if (sum < 0) {
sum += m;
}
System.out.println(sum);
}
}
}
} | Java | ["6\n2 2 0\n2 1 1\n3 2 3\n69 69 69\n2017 10 18\n5 7 0"] | 4 seconds | ["6\n1\n15\n699496932\n892852568\n713939942"] | NoteIn the first test case, the valid sequences are $$$[1, 2]$$$, $$$[1, 3]$$$, $$$[2, 1]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$ and $$$[3, 2]$$$.In the second test case, the only valid sequence is $$$[0, 0]$$$. | Java 11 | standard input | [
"combinatorics",
"dp",
"implementation",
"math"
] | aa08245d8959ed3901eb23f5b97b1b7e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains three space-separated integers $$$n$$$, $$$k$$$, and $$$x$$$ ($$$1 \le n \le 10^9$$$, $$$0 \le k \le 10^7$$$, $$$0 \le x \lt 2^{\operatorname{min}(20, k)}$$$). It is guaranteed that the sum of $$$k$$$ over all test cases does not exceed $$$5 \cdot 10^7$$$. | 2,700 | For each test case, print a single integer — the answer to the problem. | standard output | |
PASSED | f072421ed23a809d10c93fd19596e013 | train_110.jsonl | 1635604500 | It was October 18, 2017. Shohag, a melancholic soul, made a strong determination that he will pursue Competitive Programming seriously, by heart, because he found it fascinating. Fast forward to 4 years, he is happy that he took this road. He is now creating a contest on Codeforces. He found an astounding problem but has no idea how to solve this. Help him to solve the final problem of the round.You are given three integers $$$n$$$, $$$k$$$ and $$$x$$$. Find the number, modulo $$$998\,244\,353$$$, of integer sequences $$$a_1, a_2, \ldots, a_n$$$ such that the following conditions are satisfied: $$$0 \le a_i \lt 2^k$$$ for each integer $$$i$$$ from $$$1$$$ to $$$n$$$. There is no non-empty subsequence in $$$a$$$ such that the bitwise XOR of the elements of the subsequence is $$$x$$$. A sequence $$$b$$$ is a subsequence of a sequence $$$c$$$ if $$$b$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements. | 512 megabytes | import java.util.*;
import java.io.*;
public class cf {
static long modulo(long a, long b, long c) {
long x=1;
long y=a;
while(b > 0){
if(b%2 == 1){
x=(x*y)%c;
}
y = (y*y)%c; // squaring the base
b /= 2;
}
return x%c;
}
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
long m = 998244353;
while (t-- > 0) {
String[] ss = br.readLine().split(" ");
long n = Long.parseLong(ss[0]), k = Long.parseLong(ss[1]), x = Long.parseLong(ss[2]);
if (x == 0) {
if (n > k) {
System.out.println(0);
} else {
long power = modulo(2, k, m);
long prod = 1;
long exp = 1;
for (int i = 0; i < n; i++) {
prod = prod * (power - exp) % m;
exp = 2 * exp % m;
}
if (prod < 0) {
prod += m;
}
System.out.println(prod);
}
} else {
long coeff = 1;
long sum = 0;
long term = modulo(2, (k - 1) * (n + 1), m);
long pow = modulo(2, k - 1, m);
long inv = modulo(499122177, n + 1, m);
for (int i = 0; i < k; i++) {
sum = (sum + term * coeff) % m;
term = term * inv % m;
coeff = coeff * (1 - pow) % m;
pow = pow * 499122177 % m;
}
if (sum < 0) {
sum += m;
}
System.out.println(sum);
}
}
}
} | Java | ["6\n2 2 0\n2 1 1\n3 2 3\n69 69 69\n2017 10 18\n5 7 0"] | 4 seconds | ["6\n1\n15\n699496932\n892852568\n713939942"] | NoteIn the first test case, the valid sequences are $$$[1, 2]$$$, $$$[1, 3]$$$, $$$[2, 1]$$$, $$$[2, 3]$$$, $$$[3, 1]$$$ and $$$[3, 2]$$$.In the second test case, the only valid sequence is $$$[0, 0]$$$. | Java 8 | standard input | [
"combinatorics",
"dp",
"implementation",
"math"
] | aa08245d8959ed3901eb23f5b97b1b7e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The first and only line of each test case contains three space-separated integers $$$n$$$, $$$k$$$, and $$$x$$$ ($$$1 \le n \le 10^9$$$, $$$0 \le k \le 10^7$$$, $$$0 \le x \lt 2^{\operatorname{min}(20, k)}$$$). It is guaranteed that the sum of $$$k$$$ over all test cases does not exceed $$$5 \cdot 10^7$$$. | 2,700 | For each test case, print a single integer — the answer to the problem. | standard output | |
PASSED | 1f990f607164dbf7e048339be901f4e2 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.util.*;
import java.util.function.*;
import java.io.*;
// you can compare with output.txt and expected out
public class Round752Div1C {
MyPrintWriter out;
MyScanner in;
// final static long FIXED_RANDOM;
// static {
// FIXED_RANDOM = System.currentTimeMillis();
// }
final static String IMPOSSIBLE = "IMPOSSIBLE";
final static String POSSIBLE = "POSSIBLE";
final static String YES = "YES";
final static String NO = "NO";
private void initIO(boolean isFileIO) {
if (System.getProperty("ONLINE_JUDGE") == null && isFileIO) {
try{
in = new MyScanner(new FileInputStream("input.txt"));
out = new MyPrintWriter(new FileOutputStream("output.txt"));
}
catch(FileNotFoundException e){
e.printStackTrace();
}
}
else{
in = new MyScanner(System.in);
out = new MyPrintWriter(new BufferedOutputStream(System.out));
}
}
public static void main(String[] args){
// Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
// the code for the deep recursion (more than 100k or 3~400k or so)
// Thread t = new Thread(null, new RoundEdu130F(), "threadName", 1<<28);
// t.start();
// t.join();
Round752Div1C sol = new Round752Div1C();
sol.run();
}
private void run() {
boolean isDebug = false;
boolean isFileIO = true;
boolean hasMultipleTests = true;
initIO(isFileIO);
final int MAX = 100_000;
cnt = new long[2][MAX+1];
int t = hasMultipleTests? in.nextInt() : 1;
for (int i = 1; i <= t; ++i) {
if(isDebug){
out.printf("Test %d\n", i);
}
getInput();
solve();
printOutput();
}
in.close();
out.close();
}
// use suitable one between matrix(2, n), matrix(n, 2), transposedMatrix(2, n) for graph edges, pairs, ...
int n;
int[] a;
void getInput() {
n = in.nextInt();
a = in.nextIntArray(n);
}
void printOutput() {
out.printlnAns(ans);
}
long[][] cnt;
final long MOD = 998244353;
long ans;
void solve(){
// f(l, r) = extreme value for a[l..r]
// we should leave a[r] as is
// a[r-1] <= a[r] -> leave it
// a[r-1] > a[r] -> then what? we should make x as large as possible, while using not so many operations
// a[r-1] -> (a[r]-1)/(k+1), (a[r]-1)/(k+1), ..., a[r]/(k+1)
// if a[r-1] = a[r]*k + r, then k-r floor values and r ceil values
// ex: 100, 23
// 20, 20, 20, 20, 20
// 5 4 3
// 5 2 2 3 -- 1
// 1 2 2 2 2 3 -- 2
// for each i, we have to split it, based on b[i+1] value
// count how many b[i+1] values are possible, and give it to i-1
// there are sqrt(a[i]) different values for a[i]/k
ArrayList<Integer> curr = new ArrayList<>();
// HashMap<Integer, Long> cnt = new HashMap<>();
ans = 0;
for(int i=n-1; i>=0; i--) {
// HashMap<Integer, Long> next = new HashMap<>();
ArrayList<Integer> next = new ArrayList<>();
int bit = i & 1;
int last = 0;
for(int val: curr) {
long multitude = cnt[bit][val];
if(a[i] <= val)
cnt[bit^1][a[i]] += multitude;
else {
int k = a[i]/val;
if(a[i] % val > 0)
k++;
int newVal = a[i]/k;
cnt[bit^1][newVal] += multitude;
if(last != newVal) {
last = newVal;
next.add(last);
}
ans += multitude*(k-1)*(i+1) % MOD; // there are i+1 subarrays that can include this
}
}
if(last != a[i]) {
last = a[i];
next.add(last);
}
cnt[bit^1][a[i]]++;
for(int val: curr)
cnt[bit][val] = 0;
curr = next;
// for(var entry: cnt.entrySet()) {
// int val = entry.getKey();
// long multitude = entry.getValue();
//
// if(a[i] <= val) {
// next.merge(a[i], multitude, (x, y) -> (x+y));
// }
// else { // a[i] > val
// int k = a[i] / val;
// if(a[i] % val > 0)
// k++;
// int newVal = a[i]/k;
// next.merge(newVal, multitude, (x, y) -> (x+y));
// ans += multitude*(k-1)*(i+1) % MOD; // there are i+1 subarrays that can include this
// }
// }
// next.merge(a[i], 1L, (x, y) -> (x+y));
// cnt = next;
}
ans %= MOD;
for(int val: curr)
cnt[1][val] = 0;
}
// Optional<T> solve()
// return Optional.empty();
static class Pair implements Comparable<Pair>{
final static long FIXED_RANDOM = System.currentTimeMillis();
int first, second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
@Override
public int hashCode() {
// http://xorshift.di.unimi.it/splitmix64.c
long x = first;
x <<= 32;
x += second;
x += FIXED_RANDOM;
x += 0x9e3779b97f4a7c15l;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9l;
x = (x ^ (x >> 27)) * 0x94d049bb133111ebl;
return (int)(x ^ (x >> 31));
}
@Override
public boolean equals(Object obj) {
// if (this == obj)
// return true;
// if (obj == null)
// return false;
// if (getClass() != obj.getClass())
// return false;
Pair other = (Pair) obj;
return first == other.first && second == other.second;
}
@Override
public String toString() {
return "[" + first + "," + second + "]";
}
@Override
public int compareTo(Pair o) {
int cmp = Integer.compare(first, o.first);
return cmp != 0? cmp: Integer.compare(second, o.second);
}
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
// 32768?
public MyScanner(InputStream is, int bufferSize) {
br = new BufferedReader(new InputStreamReader(is), bufferSize);
}
public MyScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
// br = new BufferedReader(new InputStreamReader(System.in));
// br = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
}
public void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[][] nextMatrix(int n, int m) {
return nextMatrix(n, m, 0);
}
int[][] nextMatrix(int n, int m, int offset) {
int[][] mat = new int[n][m];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[i][j] = nextInt()+offset;
}
}
return mat;
}
int[][] nextTransposedMatrix(int n, int m){
return nextTransposedMatrix(n, m, 0);
}
int[][] nextTransposedMatrix(int n, int m, int offset){
int[][] mat = new int[m][n];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[j][i] = nextInt()+offset;
}
}
return mat;
}
int[] nextIntArray(int len) {
return nextIntArray(len, 0);
}
int[] nextIntArray(int len, int offset){
int[] a = new int[len];
for(int j=0; j<len; j++)
a[j] = nextInt()+offset;
return a;
}
long[] nextLongArray(int len) {
return nextLongArray(len, 0);
}
long[] nextLongArray(int len, int offset){
long[] a = new long[len];
for(int j=0; j<len; j++)
a[j] = nextLong()+offset;
return a;
}
String[] nextStringArray(int len) {
String[] s = new String[len];
for(int i=0; i<len; i++)
s[i] = next();
return s;
}
}
public static class MyPrintWriter extends PrintWriter{
public MyPrintWriter(OutputStream os) {
super(os);
}
// public <T> void printlnAns(Optional<T> ans) {
// if(ans.isEmpty())
// println(-1);
// else
// printlnAns(ans.get());
// }
public void printlnAns(OptionalInt ans) {
println(ans.orElse(-1));
}
public void printlnAns(long ans) {
println(ans);
}
public void printlnAns(int ans) {
println(ans);
}
public void printlnAns(boolean[] ans) {
for(boolean b: ans)
printlnAns(b);
}
public void printlnAns(boolean ans) {
if(ans)
println(YES);
else
println(NO);
}
public void printAns(long[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(long[] arr){
printAns(arr);
println();
}
public void printAns(int[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(int[] arr){
printAns(arr);
println();
}
public <T> void printAns(ArrayList<T> arr){
if(arr != null && arr.size() > 0){
print(arr.get(0));
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i));
}
}
}
public <T> void printlnAns(ArrayList<T> arr){
printAns(arr);
println();
}
public void printAns(int[] arr, int add){
if(arr != null && arr.length > 0){
print(arr[0]+add);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]+add);
}
}
}
public void printlnAns(int[] arr, int add){
printAns(arr, add);
println();
}
public void printAns(ArrayList<Integer> arr, int add) {
if(arr != null && arr.size() > 0){
print(arr.get(0)+add);
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i)+add);
}
}
}
public void printlnAns(ArrayList<Integer> arr, int add){
printAns(arr, add);
println();
}
public void printlnAnsSplit(long[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public void printlnAnsSplit(int[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public <T> void printlnAnsSplit(ArrayList<T> arr, int split){
if(arr != null && !arr.isEmpty()){
for(int i=0; i<arr.size(); i+=split){
print(arr.get(i));
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr.get(j));
}
println();
}
}
}
}
static private void permutateAndSort(long[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
long temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private void permutateAndSort(int[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
int temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private int[][] constructChildren(int n, int[] parent, int parentRoot){
int[][] childrens = new int[n][];
int[] numChildren = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
numChildren[parent[i]]++;
}
for(int i=0; i<n; i++) {
childrens[i] = new int[numChildren[i]];
}
int[] idx = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
childrens[parent[i]][idx[parent[i]]++] = i;
}
return childrens;
}
static private int[][][] constructDirectedNeighborhood(int n, int[][] e){
int[] inDegree = new int[n];
int[] outDegree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outDegree[u]++;
inDegree[v]++;
}
int[][] inNeighbors = new int[n][];
int[][] outNeighbors = new int[n][];
for(int i=0; i<n; i++) {
inNeighbors[i] = new int[inDegree[i]];
outNeighbors[i] = new int[outDegree[i]];
}
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outNeighbors[u][--outDegree[u]] = v;
inNeighbors[v][--inDegree[v]] = u;
}
return new int[][][] {inNeighbors, outNeighbors};
}
static private int[][] constructNeighborhood(int n, int[][] e) {
int[] degree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
degree[u]++;
degree[v]++;
}
int[][] neighbors = new int[n][];
for(int i=0; i<n; i++)
neighbors[i] = new int[degree[i]];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
neighbors[u][--degree[u]] = v;
neighbors[v][--degree[v]] = u;
}
return neighbors;
}
static private void drawGraph(int[][] e) {
makeDotUndirected(e);
try {
final Process process = new ProcessBuilder("dot", "-Tpng", "graph.dot")
.redirectOutput(new File("graph.png"))
.start();
} catch (IOException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
}
static private void makeDotUndirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict graph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "--" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
static private void makeDotDirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict digraph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "->" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 17 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | 2ac5f964e36134178e7ac8a639ea914b | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that :()
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x1603C
{
static final long MOD = 998244353L;
static final int MAX = 100000;
public static void main(String omkar[]) throws Exception
{
FastScanner infile = new FastScanner();
int T = infile.nextInt();
StringBuilder sb = new StringBuilder();
long[] dp = new long[MAX+1];
long[] next = new long[MAX+1];
int[] keys = new int[1000];
int[] nextKeys = new int[1000];
int size = 0;
for(int q=1; q <= T; q++)
{
int N = infile.nextInt();
int[] arr = infile.nextInts(N);
dp[arr[N-1]] = 1L;
keys[0] = arr[N-1];
size = 1;
long res = 0L;
for(int t=N-2; t >= 0; t--)
{
int nextSize = 0;
int curr = -1;
for(int i=0; i < size; i++)
{
int prev = keys[i];
int sections = (arr[t]+prev-1)/prev;
int val = arr[t]/sections;
if(curr != val)
{
nextKeys[nextSize++] = val;
next[val] = 0L;
curr = val;
}
next[val] += dp[prev];
if(next[val] >= MOD)
next[val] -= MOD;
long temp = (dp[prev]*(sections-1)*(t+1))%MOD;
res += temp;
}
if(curr == arr[t])
{
next[curr]++;
if(next[curr] >= MOD)
next[curr] -= MOD;
}
else
{
next[arr[t]] = 1L;
nextKeys[nextSize++] = arr[t];
}
for(int i=0; i < nextSize; i++)
{
keys[i] = nextKeys[i];
dp[keys[i]] = next[keys[i]];
}
size = nextSize;
}
res %= MOD;
sb.append(res).append('\n');
}
System.out.print(sb);
}
}
/*
23 8 -> 7 8 8 8
24 8 -> 8 8 8 8
25 8 -> 6 6 6 7 8
2
3
5 4 3
2
4 3
*/
class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | 2911159c53600fcfe3addef979c406b5 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that :()
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x1603C
{
static final long MOD = 998244353L;
static final int MAX = 100000;
public static void main(String omkar[]) throws Exception
{
FastScanner infile = new FastScanner();
int T = infile.nextInt();
StringBuilder sb = new StringBuilder();
long[] dp = new long[MAX+1];
long[] next = new long[MAX+1];
int[] keys = new int[1000];
int[] nextKeys = new int[1000];
int size = 0;
for(int q=1; q <= T; q++)
{
int N = infile.nextInt();
int[] arr = infile.nextInts(N);
dp[arr[N-1]] = 1L;
keys[0] = arr[N-1];
size = 1;
long res = 0L;
for(int t=N-2; t >= 0; t--)
{
int nextSize = 0;
int curr = -1;
for(int i=0; i < size; i++)
{
int prev = keys[i];
int sections = (arr[t]+prev-1)/prev;
int val = arr[t]/sections;
if(curr != val)
{
nextKeys[nextSize++] = val;
next[val] = 0L;
curr = val;
}
next[val] += dp[prev];
if(next[val] >= MOD)
next[val] -= MOD;
long temp = (dp[prev]*(sections-1)*(t+1))%MOD;
res += temp;
if(res >= MOD)
res -= MOD;
}
if(curr == arr[t])
{
next[curr]++;
if(next[curr] >= MOD)
next[curr] -= MOD;
}
else
{
next[arr[t]] = 1L;
nextKeys[nextSize++] = arr[t];
}
for(int i=0; i < nextSize; i++)
{
keys[i] = nextKeys[i];
dp[keys[i]] = next[keys[i]];
}
size = nextSize;
}
sb.append(res).append('\n');
}
System.out.print(sb);
}
}
/*
23 8 -> 7 8 8 8
24 8 -> 8 8 8 8
25 8 -> 6 6 6 7 8
2
3
5 4 3
2
4 3
*/
class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | ad00fefef3a9287f6120ef0809b44d17 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that :()
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x1603C
{
static final long MOD = 998244353L;
static final int MAX = 100000;
public static void main(String omkar[]) throws Exception
{
FastScanner infile = new FastScanner();
int T = infile.nextInt();
StringBuilder sb = new StringBuilder();
long[] dp = new long[MAX+1];
long[] next = new long[MAX+1];
int[] keys = new int[1000];
int[] nextKeys = new int[1000];
int size = 0;
for(int q=1; q <= T; q++)
{
int N = infile.nextInt();
int[] arr = infile.nextInts(N);
dp[arr[N-1]] = 1L;
keys[0] = arr[N-1];
size = 1;
long res = 0L;
for(int t=N-2; t >= 0; t--)
{
int nextSize = 0;
int curr = -1;
for(int i=0; i < size; i++)
{
int prev = keys[i];
int sections = (arr[t]+prev-1)/prev;
int val = arr[t]/sections;
if(curr != val)
{
nextKeys[nextSize++] = val;
next[val] = 0L;
curr = val;
}
next[val] += dp[prev];
if(next[val] >= MOD)
next[val] -= MOD;
long temp = (dp[prev]*(sections-1)*(t+1))%MOD;
res += temp;
if(res >= MOD)
res -= MOD;
}
if(curr == arr[t])
{
next[curr]++;
if(next[curr] >= MOD)
next[curr] -= MOD;
}
else
{
next[arr[t]] = 1L;
nextKeys[nextSize++] = arr[t];
}
for(int i=0; i < nextSize; i++)
keys[i] = nextKeys[i];
size = nextSize;
for(int i=0; i < size; i++)
dp[keys[i]] = next[keys[i]];
}
sb.append(res+"\n");
}
System.out.print(sb);
}
}
/*
23 8 -> 7 8 8 8
24 8 -> 8 8 8 8
25 8 -> 6 6 6 7 8
2
3
5 4 3
2
4 3
*/
class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | 5c421b108e3624cffc70fc259f49b3be | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that :()
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x1603C
{
static final long MOD = 998244353L;
static final int MAX = 100000;
public static void main(String omkar[]) throws Exception
{
FastScanner infile = new FastScanner();
int T = infile.nextInt();
StringBuilder sb = new StringBuilder();
long[] dp = new long[MAX+1];
long[] next = new long[MAX+1];
int[] keys = new int[1000];
int[] nextKeys = new int[1000];
int size = 0;
for(int q=1; q <= T; q++)
{
int N = infile.nextInt();
int[] arr = infile.nextInts(N);
dp[arr[N-1]] = 1L;
keys[0] = arr[N-1];
size = 1;
long res = 0L;
for(int t=N-2; t >= 0; t--)
{
int nextSize = 0;
int curr = -1;
for(int i=0; i < size; i++)
{
int prev = keys[i];
int sections = (arr[t]+prev-1)/prev;
int val = arr[t]/sections;
if(curr != val)
{
nextKeys[nextSize++] = val;
next[val] = 0L;
curr = val;
}
next[val] += dp[prev];
if(next[val] >= MOD)
next[val] -= MOD;
long temp = (dp[prev]*(sections-1))%MOD;
temp = (temp*(t+1))%MOD;
res += temp;
if(res >= MOD)
res -= MOD;
}
if(curr == arr[t])
{
next[curr]++;
if(next[curr] >= MOD)
next[curr] -= MOD;
}
else
{
next[arr[t]] = 1L;
nextKeys[nextSize++] = arr[t];
}
for(int i=0; i < nextSize; i++)
keys[i] = nextKeys[i];
size = nextSize;
for(int i=0; i < size; i++)
dp[keys[i]] = next[keys[i]];
}
sb.append(res+"\n");
}
System.out.print(sb);
}
}
/*
23 8 -> 7 8 8 8
24 8 -> 8 8 8 8
25 8 -> 6 6 6 7 8
2
3
5 4 3
2
4 3
*/
class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | fc8fd4e9a408022498efc7db7d6acfd4 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that :()
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class x1603C
{
static final long MOD = 998244353L;
static final int MAX = 100000;
public static void main(String omkar[]) throws Exception
{
FastScanner infile = new FastScanner();
int T = infile.nextInt();
StringBuilder sb = new StringBuilder();
long[] dp = new long[MAX+1];
long[] next = new long[MAX+1];
for(int q=1; q <= T; q++)
{
int N = infile.nextInt();
int[] arr = infile.nextInts(N);
dp[arr[N-1]] = 1L;
ArrayDeque<Integer> keys = new ArrayDeque<Integer>();
keys.add(arr[N-1]);
long res = 0L;
for(int t=N-2; t >= 0; t--)
{
ArrayDeque<Integer> nextKeys = new ArrayDeque<Integer>();
int curr = -1;
for(int prev: keys)
{
int sections = (arr[t]+prev-1)/prev;
int val = arr[t]/sections;
if(curr != val)
{
nextKeys.add(val);
next[val] = 0L;
curr = val;
}
next[val] += dp[prev];
if(next[val] >= MOD)
next[val] -= MOD;
long temp = (dp[prev]*(sections-1))%MOD;
temp = (temp*(t+1))%MOD;
res += temp;
if(res >= MOD)
res -= MOD;
}
if(curr == arr[t])
{
next[curr]++;
if(next[curr] >= MOD)
next[curr] -= MOD;
}
else
{
next[arr[t]] = 1L;
nextKeys.add(arr[t]);
}
for(int x: nextKeys)
dp[x] = next[x];
keys = nextKeys;
}
sb.append(res+"\n");
}
System.out.print(sb);
}
}
/*
23 8 -> 7 8 8 8
24 8 -> 8 8 8 8
25 8 -> 6 6 6 7 8
2
3
5 4 3
2
4 3
*/
class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | 96997eebe21c0d81d9d7c8705f970334 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Locale;
import java.util.StringTokenizer;
public class Solution2 implements Runnable {
public static final long MOD = 998244353;
private PrintStream out;
private BufferedReader in;
private StringTokenizer st;
public void solve() throws IOException {
long time0 = System.currentTimeMillis();
int t = nextInt();
for (int test = 1; test <= t; test++) {
int n = nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
int answer = solve(n, a);
out.println(answer);
}
System.err.println("time: " + (System.currentTimeMillis() - time0));
}
private int solve(int n, int[] a) {
int[] h0 = new int[n];
int[] h1 = new int[n];
int[] w = new int[n];
long sum = 0;
long ans = 0;
for (int i = 0; i < n; i++) {
h0[i] = a[i];
h1[i] = a[i];
w[i] = 0;
for (int j = i - 1; j >= 0; j--) {
if (h1[j] <= h0[j + 1]) {
break;
}
sum = (((sum - 1L * w[j] * (j + 1)) % MOD) + MOD) % MOD;
int r = (a[j] + h0[j + 1] - 1) / h0[j + 1];
h0[j] = a[j] / r;
h1[j] = (a[j] + r - 1) / r;
w[j] = r - 1;
sum = (sum + 1L * w[j] * (j + 1)) % MOD;
}
ans = (ans + sum) % MOD;
}
return (int) ans;
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public String next() throws IOException {
while (!st.hasMoreTokens()) {
String line = in.readLine();
if (line == null) {
return null;
}
st = new StringTokenizer(line);
}
return st.nextToken();
}
@Override
public void run() {
try {
solve();
out.close();
} catch (Throwable e) {
throw new RuntimeException(e);
}
}
public Solution2(String name) throws IOException {
Locale.setDefault(Locale.US);
if (name == null) {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintStream(new BufferedOutputStream(System.out));
} else {
in = new BufferedReader(new InputStreamReader(new FileInputStream(name + ".in")));
out = new PrintStream(new BufferedOutputStream(new FileOutputStream(name + ".out")));
}
st = new StringTokenizer("");
}
public static void main(String[] args) throws IOException {
new Thread(new Solution2(null)).start();
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | cc5868d32c3bff7889016d1636516bd6 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Locale;
import java.util.StringTokenizer;
public class Solution implements Runnable {
public static final long MOD = 998244353;
private PrintStream out;
private BufferedReader in;
private StringTokenizer st;
public void solve() throws IOException {
long time0 = System.currentTimeMillis();
int t = nextInt();
for (int test = 1; test <= t; test++) {
int n = nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
int answer = solve(n, a);
out.println(answer);
}
System.err.println("time: " + (System.currentTimeMillis() - time0));
}
private int solve(int n, int[] a) {
int[] h0 = new int[n];
int[] h1 = new int[n];
int[] w = new int[n];
long sum = 0;
long ans = 0;
for (int i = 0; i < n; i++) {
h0[i] = a[i];
h1[i] = a[i];
w[i] = 0;
for (int j = i - 1; j >= 0; j--) {
if (h1[j] <= h0[j + 1]) {
break;
}
sum = (((sum - 1L * w[j] * (j + 1)) % MOD) + MOD) % MOD;
int l = w[j] + 1;
int r = (a[j] + h0[j + 1] - 1) / h0[j + 1];
while (l + 1 < r) {
int m = (l + r) / 2;
int hh1 = (a[j] + m - 1) / m;
if (hh1 <= h0[j + 1]) {
r = m;
} else {
l = m;
}
}
h0[j] = a[j] / r;
h1[j] = (a[j] + r - 1) / r;
w[j] = r - 1;
sum = (sum + 1L * w[j] * (j + 1)) % MOD;
}
ans = (ans + sum) % MOD;
}
return (int) ans;
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public String next() throws IOException {
while (!st.hasMoreTokens()) {
String line = in.readLine();
if (line == null) {
return null;
}
st = new StringTokenizer(line);
}
return st.nextToken();
}
@Override
public void run() {
try {
solve();
out.close();
} catch (Throwable e) {
throw new RuntimeException(e);
}
}
public Solution(String name) throws IOException {
Locale.setDefault(Locale.US);
if (name == null) {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintStream(new BufferedOutputStream(System.out));
} else {
in = new BufferedReader(new InputStreamReader(new FileInputStream(name + ".in")));
out = new PrintStream(new BufferedOutputStream(new FileOutputStream(name + ".out")));
}
st = new StringTokenizer("");
}
public static void main(String[] args) throws IOException {
new Thread(new Solution(null)).start();
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | bb8ea2dad5c6df70f083973ed8074bc2 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.util.*;
import java.lang.*;
// StringBuilder uses java.lang
public class mC {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
StringBuilder st = new StringBuilder();
int t = sc.nextInt();
long MOD = 998244353;
for (int test = 0; test < t; test++) {
int n = sc.nextInt();
long[] inputList = new long[n];
long[] numDiv = new long[n]; // number of divisions to make, not num of resulting numbers
long indexProdSum = 0;
long ans = 0;
for (int i = 0; i < n; i++) {
inputList[i] = sc.nextInt();
boolean cont = true;
for (int j = i-1; (j>=0 && cont); j--) {
long maxAt = inputList[j+1] / (numDiv[j+1]+1);
long minForcedDivs = (inputList[j]-1) / maxAt; // ceiling minus one
if (minForcedDivs > numDiv[j]) {
indexProdSum += (minForcedDivs - numDiv[j]) * (j+1);
indexProdSum %= MOD;
numDiv[j] = minForcedDivs;
} else {
cont = false;
}
}
ans += indexProdSum;
ans %= MOD;
}
st.append(ans+"\n");
}
System.out.print(st.toString());
}
public static int firstLargerAb(int val,ArrayList<Integer> ok,int left,int right) {
if (Math.abs(ok.get(right))<=val) {
return -1;
}
if (left==right) {
return left;
} else if (left+1==right) {
if (val<Math.abs(ok.get(left))) {
return left;
} else {
return right;
}
} else {
int mid = (left+right)/2;
if (Math.abs(ok.get(mid))>val) {
return firstLargerAb(val,ok,left,mid);
} else {
return firstLargerAb(val,ok,mid+1,right);
}
}
}
public static int findNthInArray(int[] arr,int val,int start,int o) {
if (o==0) {
return start-1;
} else if (arr[start] == val) {
return findNthInArray(arr,val,start+1,o-1);
} else {
return findNthInArray(arr,val,start+1,o);
}
}
public static ArrayList<Integer> dfs(int at,ArrayList<Integer> went,ArrayList<ArrayList<Integer>> connect) {
for (int i=0;i<connect.get(at).size();i++) {
if (!(went.contains(connect.get(at).get(i)))) {
went.add(connect.get(at).get(i));
went=dfs(connect.get(at).get(i),went,connect);
}
}
return went;
} public static int[] bfs (int at, int[] went, ArrayList<ArrayList<Integer>> queue, int numNodes, ArrayList<ArrayList<Integer>> connect) {
if (went[at]==0) {
went[at]=queue.get(numNodes).get(1);
for (int i=0;i<connect.get(at).size();i++) {
if (went[connect.get(at).get(i)]==0) {
ArrayList<Integer> temp = new ArrayList<>();
temp.add(connect.get(at).get(i));
temp.add(queue.get(numNodes).get(1)+1);
queue.add(temp);
}
}
}
if (queue.size()==numNodes+1) {
return went;
} else {
return bfs(queue.get(numNodes+1).get(0),went, queue, numNodes+1, connect);
}
}
public static long fastPow(long base,long exp,long mod) {
if (exp==0) {
return 1;
} else {
if (exp % 2 == 1) {
long z = fastPow(base,(exp-1)/2,mod);
return ((((z*base) % mod) * z) % mod);
} else {
long z = fastPow(base,exp/2,mod);
return ((z*z) % mod);
}
}
}
public static int fastPow(int base,long exp) {
if (exp==0) {
return 1;
} else {
if (exp % 2 == 1) {
int z = fastPow(base,(exp-1)/2);
return ((((z*base)) * z));
} else {
int z = fastPow(base,exp/2);
return ((z*z));
}
}
}
public static int firstLarger(int val,ArrayList<Integer> ok,int left,int right) {
if (ok.get(right)<=val) {
return -1;
}
if (left==right) {
return left;
} else if (left+1==right) {
if (val<ok.get(left)) {
return left;
} else {
return right;
}
} else {
int mid = (left+right)/2;
if (ok.get(mid)>val) {
return firstLarger(val,ok,left,mid);
} else {
return firstLarger(val,ok,mid+1,right);
}
}
}
public static int binSearchArr(long val,ArrayList<Integer> ok,long[] arr,int left,int right) {
if (arr[ok.get(right)]<=val) {
return -1;
}
if (left==right) {
return left;
} else if (left+1==right) {
if (val<arr[ok.get(left)]) {
return left;
} else {
return right;
}
} else {
int mid = (left+right)/2;
if (arr[ok.get(mid)]>val) {
return binSearchArr(val,ok,arr,left,mid);
} else {
return binSearchArr(val,ok,arr,mid+1,right);
}
}
}
public static long gcd(long a, long b) {
if (b>a) {
return gcd(b,a);
}
if (b==0) {
return a;
}
if (a%b==0) {
return b;
} else {
return gcd(b,a%b);
}
}
} | Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 8 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | b551f4d142944d349118fe09171cabce | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.util.*;
import java.io.*;
public class _1604_E {
static final long MOD = 998244353;
public static void main(String[] args) throws IOException {
Reader in = new Reader();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-- > 0) {
int n = in.nextInt();
int[] a = new int[n];
for(int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
int[][] dp = new int[2][100005];
ArrayList<Integer>[] keys = new ArrayList[2];
keys[0] = new ArrayList<Integer>();
keys[1] = new ArrayList<Integer>();
dp[0][a[n - 1]] = 1;
keys[0].add(a[n - 1]);
long res = 0;
int cur = 0;
for(int i = n - 1; i > 0; i--) {
dp[1 - cur][a[i - 1]] += 1;
keys[1 - cur].add(a[i - 1]);
int prev = a[i - 1];
for(int key : keys[cur]) {
int ceil = a[i - 1] / key;
if(a[i - 1] % key != 0) ceil++;
int min = a[i - 1] / ceil;
dp[1 - cur][min] += dp[cur][key];
res = modadd(res, modmult(modmult(dp[cur][key], ceil - 1), i));
if(min != prev) {
keys[1 - cur].add(min);
prev = min;
}
dp[cur][key] = 0;
}
keys[cur].clear();
cur = 1 - cur;
}
out.println(res);
}
in.close();
out.close();
}
static long modadd(long a, long b) {
return (a + b + MOD) % MOD;
}
static long modmult(long a, long b) {
return a * b % MOD;
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
} else {
continue;
}
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg) {
return -ret;
}
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) {
buffer[0] = -1;
}
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) {
fillBuffer();
}
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null) {
return;
}
din.close();
}
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 11 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | 6c49059084d6c1d4b898f727b94acadd | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.util.*;
import java.io.*;
public class _1604_E {
static final long MOD = 998244353;
public static void main(String[] args) throws IOException {
Reader in = new Reader();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-- > 0) {
int n = in.nextInt();
int[] a = new int[n];
for(int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
int[][] dp = new int[2][100005];
ArrayList<Integer>[] keys = new ArrayList[2];
keys[0] = new ArrayList<Integer>();
keys[1] = new ArrayList<Integer>();
dp[0][a[n - 1]] = 1;
keys[0].add(a[n - 1]);
long res = 0;
int cur = 0;
for(int i = n - 1; i > 0; i--) {
int prev = -1;
for(int key : keys[cur]) {
int ceil = a[i - 1] / key;
if(a[i - 1] % key != 0) ceil++;
int min = a[i - 1] / ceil;
dp[1 - cur][min] += dp[cur][key];
res = modadd(res, modmult(modmult(dp[cur][key], ceil - 1), i));
if(min != prev) {
keys[1 - cur].add(min);
prev = min;
}
dp[cur][key] = 0;
}
dp[1 - cur][a[i - 1]] += 1;
keys[1 - cur].add(a[i - 1]);
keys[cur].clear();
cur = 1 - cur;
}
out.println(res);
}
in.close();
out.close();
}
static long modadd(long a, long b) {
return (a + b + MOD) % MOD;
}
static long modmult(long a, long b) {
return a * b % MOD;
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
} else {
continue;
}
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg) {
return -ret;
}
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) {
buffer[0] = -1;
}
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) {
fillBuffer();
}
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null) {
return;
}
din.close();
}
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 11 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | 3cf08b4eb1ae0e355dcf885437d5a9d3 | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
static int MOD = 998244353;
// After writing solution, quick scan for:
// array out of bounds
// special cases e.g. n=1?
// npe, particularly in maps
//
// Big numbers arithmetic bugs:
// int overflow
// sorting, or taking max, after MOD
void solve() throws IOException {
int T = ri();
// We need to initialize these only once...
int AMAX = 100000;
long[] rightCounts = new long[AMAX+1];
long[] currCounts = new long[AMAX+1];
for (int Ti = 0; Ti < T; Ti++) {
int n = ri();
int[] a = ril(n);
// dp[i][x] is the number of subarrays starting at i, where after the procedure, the
// first element becomes x.
//
// Let split := number of splits needed to turn a[i] into x.
// Then the answer is the sum of contributions over all (i, x) pairs of dp[i][x] * split.
// How to compute dp[i][x]?
// It's pretty difficult to compute dp[i][x] using dp[i][x+e] for e in [0..?] since
// it's hard to tell what that ? should be.
// It's easier to "compute the other way". That is, let's loop over dp[i+1][x]. Now,
// with fixed x as the value to the right, we can add that contribution to
// dp[i][y] where y = floor(a[i] / ceil(a[i] / x)). (see below).
Set<Integer> rightNonzero = new HashSet<>();
Set<Integer> currNonzero = new HashSet<>();
rightCounts[a[n-1]]++;
rightNonzero.add(a[n-1]);
long ans = 0;
for (int i = n-2; i >= 0; i--) {
// Add all the contributions of i+1 to i.
for (int x : rightNonzero) { // x is the value to the right
long count = rightCounts[x]; // number of subarrays [i+1..] with the value x.
int y = a[i] / ((a[i] + x - 1) / x); // y is the value i become
int splits = (a[i] + x - 1) / x; // splits is the new number of elems when i become x.
currCounts[y] += count;
currNonzero.add(y);
ans += (i+1) * count * (splits - 1);
// Clear in preparation for swap
rightCounts[x] = 0;
}
currCounts[a[i]]++;
currNonzero.add(a[i]);
// Swap around the memory we're re-using.
long[] temp = rightCounts; // This is empty
rightCounts = currCounts;
currCounts = temp;
Set<Integer> temp2 = rightNonzero;
rightNonzero = currNonzero;
currNonzero = temp2;
currNonzero.clear();
ans %= MOD;
}
// Clear out rightCounts in preparation for next test case
for (int x : rightNonzero) rightCounts[x] = 0;
pw.println(ans);
}
}
// IMPORTANT
// DID YOU CHECK THE COMMON MISTAKES ABOVE?
// Finds the extreme value for subarray a[i..j].
int solve(int[] a, int l, int r) {
if (l >= r) return 0;
int ans = 0;
int upper = a[r];
for (int i = r-1; i >= l; i--) {
if (a[i] <= upper) {
upper = a[i];
continue;
}
// Want a[i] / k <= upper (k sections)
// So we want k >= a[i] / upper
int k = (a[i] + upper - 1) / upper;
ans += k-1;
upper = a[i] / k;
}
return ans;
}
// Template code below
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) throws IOException {
Main m = new Main();
m.solve();
m.close();
}
void close() throws IOException {
pw.flush();
pw.close();
br.close();
}
int ri() throws IOException {
return Integer.parseInt(br.readLine().trim());
}
long rl() throws IOException {
return Long.parseLong(br.readLine().trim());
}
int[] ril(int n) throws IOException {
int[] nums = new int[n];
int c = 0;
for (int i = 0; i < n; i++) {
int sign = 1;
c = br.read();
int x = 0;
if (c == '-') {
sign = -1;
c = br.read();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = br.read();
}
nums[i] = x * sign;
}
while (c != '\n' && c != -1) c = br.read();
return nums;
}
long[] rll(int n) throws IOException {
long[] nums = new long[n];
int c = 0;
for (int i = 0; i < n; i++) {
int sign = 1;
c = br.read();
long x = 0;
if (c == '-') {
sign = -1;
c = br.read();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = br.read();
}
nums[i] = x * sign;
}
while (c != '\n' && c != -1) c = br.read();
return nums;
}
int[] rkil() throws IOException {
int sign = 1;
int c = br.read();
int x = 0;
if (c == '-') {
sign = -1;
c = br.read();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = br.read();
}
return ril(x);
}
long[] rkll() throws IOException {
int sign = 1;
int c = br.read();
int x = 0;
if (c == '-') {
sign = -1;
c = br.read();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = br.read();
}
return rll(x);
}
char[] rs() throws IOException {
return br.readLine().toCharArray();
}
void sort(int[] A) {
Random r = new Random();
for (int i = A.length-1; i > 0; i--) {
int j = r.nextInt(i+1);
int temp = A[i];
A[i] = A[j];
A[j] = temp;
}
Arrays.sort(A);
}
void printDouble(double d) {
pw.printf("%.16f", d);
}
} | Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 11 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output | |
PASSED | aaafa80b40639f180ce3323d3b24df0d | train_110.jsonl | 1635604500 | For an array $$$b$$$ of $$$n$$$ integers, the extreme value of this array is the minimum number of times (possibly, zero) the following operation has to be performed to make $$$b$$$ non-decreasing: Select an index $$$i$$$ such that $$$1 \le i \le |b|$$$, where $$$|b|$$$ is the current length of $$$b$$$. Replace $$$b_i$$$ with two elements $$$x$$$ and $$$y$$$ such that $$$x$$$ and $$$y$$$ both are positive integers and $$$x + y = b_i$$$. This way, the array $$$b$$$ changes and the next operation is performed on this modified array. For example, if $$$b = [2, 4, 3]$$$ and index $$$2$$$ gets selected, then the possible arrays after this operation are $$$[2, \underline{1}, \underline{3}, 3]$$$, $$$[2, \underline{2}, \underline{2}, 3]$$$, or $$$[2, \underline{3}, \underline{1}, 3]$$$. And consequently, for this array, this single operation is enough to make it non-decreasing: $$$[2, 4, 3] \rightarrow [2, \underline{2}, \underline{2}, 3]$$$.It's easy to see that every array of positive integers can be made non-decreasing this way.YouKn0wWho has an array $$$a$$$ of $$$n$$$ integers. Help him find the sum of extreme values of all nonempty subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. If a subarray appears in $$$a$$$ multiple times, its extreme value should be counted the number of times it appears.An array $$$d$$$ is a subarray of an array $$$c$$$ if $$$d$$$ can be obtained from $$$c$$$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;
public class A1603 {
private static final int P = 998244353;
private static final int M = 100000;
private static final int SQ = 800;
private static final int SQ2 = 400;
private static int[][] g = new int[2][SQ + 1];
private static int getB1(int a, int b) {
return a / ceil(a, b);
}
private static int getMoves(int a, int b) {
return ceil(a, b) - 1;
}
private static int ceil(int a, int b) {
return (a + b - 1) / b;
}
private static int getIndex(int num, int last) {
if (last * 1l * last >= num) {
return num / last;
}
return SQ2 + last;
}
private static int count(int[] a, int n) {
Arrays.fill(g[0], 0);
Arrays.fill(g[1], 0);
int ans = 0;
int active = 0;
for (int i = 1; i < n; i++) {
// compute g[]
Arrays.fill(g[active], 0);
for (int j = 1; j <= SQ2; j++) {
int last = a[i] / j;
if (last == 0) {
g[active][j] = 0;
continue;
}
long res = (getMoves(a[i - 1], last) * 1l * i) % P;
res = (res + g[active ^ 1][getIndex(a[i - 1], getB1(a[i - 1], last))]) % P;
g[active][j] = (int) res;
}
for (int j = SQ2 + 1; j <= SQ; j++) {
// last = (j - sq2), if [a[i] / [a[i] / (j - sq2)]] == j - sq2.
int div = a[i] / (j - SQ2);
if (div <= 0 || a[i] / div != j - SQ2) {
g[active][j] = 0;
continue;
}
int last = j - SQ2;
long res = (getMoves(a[i - 1], last) * 1l * i) % P;
res = (res + g[active ^ 1][getIndex(a[i - 1], getB1(a[i - 1], last))]) % P;
g[active][j] = (int) res;
}
ans = (ans + g[active][1]) % P;
active ^= 1;
}
return ans;
}
public static void main(String[] args) throws IOException {
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int numTests = Integer.parseInt(rd.readLine());
for (int t = 0; t < numTests; t++) {
int n = Integer.parseInt(rd.readLine());
StringTokenizer st = new StringTokenizer(rd.readLine());
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = Integer.parseInt(st.nextToken());
}
pw.println(count(a, n));
}
pw.flush();
pw.close();
rd.close();
}
}
| Java | ["4\n3\n5 4 3\n4\n3 2 1 4\n1\n69\n8\n7264 40515 28226 92776 35285 21709 75124 48163"] | 4 seconds | ["5\n9\n0\n117"] | NoteLet $$$f(l, r)$$$ denote the extreme value of $$$[a_l, a_{l+1}, \ldots, a_r]$$$.In the first test case, $$$f(1, 3) = 3$$$, because YouKn0wWho can perform the following operations on the subarray $$$[5, 4, 3]$$$ (the newly inserted elements are underlined):$$$[5, 4, 3] \rightarrow [\underline{3}, \underline{2}, 4, 3] \rightarrow [3, 2, \underline{2}, \underline{2}, 3] \rightarrow [\underline{1}, \underline{2}, 2, 2, 2, 3]$$$; $$$f(1, 2) = 1$$$, because $$$[5, 4] \rightarrow [\underline{2}, \underline{3}, 4]$$$; $$$f(2, 3) = 1$$$, because $$$[4, 3] \rightarrow [\underline{1}, \underline{3}, 3]$$$; $$$f(1, 1) = f(2, 2) = f(3, 3) = 0$$$, because they are already non-decreasing. So the total sum of extreme values of all subarrays of $$$a = 3 + 1 + 1 + 0 + 0 + 0 = 5$$$. | Java 11 | standard input | [
"dp",
"greedy",
"math",
"number theory"
] | f760f108c66f695e1e51dc6470d29ce7 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) — the number of test cases. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^5$$$). It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$. | 2,300 | For each test case, print a single integer — the sum of extreme values of all subarrays of $$$a$$$ modulo $$$998\,244\,353$$$. | standard output |
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