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PASSED | a0937c2aead66909473e61fef83be063 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class Yoo
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
int i,j,c;
long sum;
while(t-->0)
{
c=0;
sum=0;
int n=sc.nextInt();
long a[]=new long[n];
for(i=0;i<n;i++)
a[i]=sc.nextInt();
for(i=0;i<n;)
{
if(a[i]%2==0)
{
c++;
a[i]=a[i]/2;
}
else
i++;
}
Arrays.sort(a);
a[n-1]=(long)Math.pow(2,c)*a[n-1];
for(i=0;i<n;i++)
sum+=a[i];
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 822bbef04e3a2a311576eda58a7ceaa8 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.*;
import java.io.*;
public class codeforceb
{
//ArrayList<Integer> arr2=new ArrayList<Integer>();
//ArrayList<Integer> arr1=new ArrayList<Integer>();
//Map<Integer,Integer> hm=new HashMap<Integer,Integer>();
//Set<Integer> st1=new HashSet<Integer>();
static int inf = Integer.MAX_VALUE ;
static long infL = Long.MAX_VALUE ;
static int mod=1000000007;
static FastReader sc=new FastReader();
static PrintWriter out=new PrintWriter(System.out);
public static void main (String[] args)
{
/*Arrays.sort(time,new Comparator<Pair>(){
public int compare(Pair p1,Pair p2){
if(p1.a==p2.a)
return p1.b-p2.b;
return p1.a-p2.a;
}
});*/
//int r[][]= {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}
// };
int t=sc.nextInt();
while(t-->0)
{
solve();
}
out.close();
}
public static void solve() {
int n=sc.nextInt();
long arr[]=sc.readLong(n);
int c=0;
for(int i=0;i<n;) {
if(arr[i]%2==0) {
arr[i]/=2;
c++;
}
else i++;
}
Arrays.sort(arr);
arr[n-1]*=myPow((long)2,(long)c);
long sum=0;
for(long i:arr) sum+=i;
out.println(sum);
}
public static boolean check(int r1,int c1,int n,int m) {
if(r1<0 || c1<0 || r1>=n || c1>=m) return false;
return true;
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
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 class Pair{
int a;
int b;
Pair(int x, int y)
{
this.a = x;
this.b = y;
}
}
static boolean isPrime(long x)
{
if(x==1)
return false;
if(x<=3)
return true;
if(x%2==0 || x%3==0)
return false;
for(int i=5;i<=Math.sqrt(x);i+=2)
if(x%i==0)
return false;
return true;
}
static long gcd(long a,long b)
{
return (b==0)?a:gcd(b,a%b);
}
static int gcd(int a,int b)
{
return (b==0)?a:gcd(b,a%b);
}
public static long myPow(long x, long n) {
if(x==0)
return 0;
if(n==0)
return 1;
if(n==1)
return x;
long calc=myPow((long)x,n/2);
//System.out.println(calc+" "+n/2);
if(Math.abs(n)%2==0)
return (calc*calc);
else
return (x*calc*calc);
}
public static long lcm(long a , long b){
return a * (b/gcd(a,b));
}
public static int lcm(int a , int b){
return (a * b)/gcd(a,b);
}
public static void swap(int[] arr, int left , int right){
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
public static void swap(char[] arr, int left , int right){
char temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
public static void reverse(int[] arr){
int left = 0;
int right = arr.length-1;
while(left <= right){
swap(arr, left,right);
left++;
right--;
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c94a246452ad4b00d5977b57188c9c13 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.Math.PI;
import static java.lang.Math.min;
import static java.lang.System.arraycopy;
import static java.lang.System.exit;
import static java.util.Arrays.copyOf;
import java.util.LinkedList;
import java.util.List;
import java.util.Iterator;
import java.io.FileReader;
import java.io.FileWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.Comparator;
import java.lang.StringBuilder;
import java.util.Collections;
import java.text.DecimalFormat;
public class Solution {
static BufferedReader in;
static PrintWriter out;
static StringTokenizer tok;
static long mod= (long)1e9+7;
static boolean isPrime[];
private static void solve() throws IOException{
int n = scanInt();
long []arr = new long[n];
for(int i = 0; i<n; ++i)
arr[i] = scanInt();
int p = 0;
for(int i = 0 ; i<n ;++i){
while(arr[i]%2 == 0){
arr[i] /=2;
++p;
}
}
Arrays.sort(arr);
while(--p >-1){
arr[n-1] *=2;
}
long sum = 0;
for(int i = 0 ; i <n; ++i)
sum+= arr[i];
out.println(sum);
}
public static void main(String[] args) {
try {
long startTime = System.currentTimeMillis();
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
//in = new BufferedReader(new FileReader("input.txt"));
//out = new PrintWriter(new FileWriter("output.txt"));
int test=scanInt();
for(int t=1; t<=test; t++){
// out.print("Case #"+t+": ");
solve();
}
long endTime = System.currentTimeMillis();
long totalTime = endTime - startTime;
//out.println(totalTime+" "+System.currentTimeMillis() );
in.close();
out.close();
} catch (Throwable e) {
e.printStackTrace();
exit(1);
}
}
static int scanInt() throws IOException {
return parseInt(scanString());
}
static long scanLong() throws IOException {
return parseLong(scanString());
}
static double scanDouble() throws IOException {
return parseDouble(scanString());
}
static String scanString() throws IOException {
if (tok == null || !tok.hasMoreTokens()) {
tok = new StringTokenizer(in.readLine());
}
return tok.nextToken();
}
static String scanLine() throws IOException {
return in.readLine();
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | ca15d4622cf5b9dcba6d9a5ca93b48a9 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class Main2 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
long[] out = new long[t];
for (int i = 0; i < t; i++) {
long temp =1 ;
int num = in.nextInt();
long[] a = new long[num];
for (int j = 0; j < num; j++) {
a[j] = in.nextInt();
while(a[j]%2==0){
a[j] /=2;
temp *=2;
}
}
Arrays.sort(a);
a[a.length-1] *= temp;
out[i] = sum(a);
}
for(long j : out ){
System.out.println(j+ " ");
}
}
public static long sum(long[] a) {
long sum = 0;
for (long i : a) {
sum += i;
}
return sum;
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 2b16656b9a6937b32ac6ff96b8613839 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
public class A_Divide_and_Multiply{
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int testCases = scan.nextInt();
while(testCases-->0){
int length = scan.nextInt();
long[] data = new long[length];
int k = 0;
for(int i = 0;i<length;i++){
long input = scan.nextLong();
data[i] = input;
while(data[i]%2==0){
k++;
data[i] = data[i]/2;
}
}
int maxIndex = 0;
for(int i =0;i<data.length;i++){
if(data[maxIndex]<data[i]){
maxIndex = i;
}
}
for(int i = 0;i<k;i++){
data[maxIndex] = data[maxIndex]*2;
}
long totalSum =0;
for(int i = 0;i<length;i++){
totalSum += data[i];
}
System.out.println(totalSum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | d0548466a5e4d0a974e91ddec758adcd | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
public class ProblemA {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int testsCount = scanner.nextInt();
for (int i=0; i < testsCount; i++) {
int arraySize = scanner.nextInt();
long[] array = new long[arraySize];
for (int j = 0; j< arraySize; j++) {
array[j] = scanner.nextLong();
}
System.out.println(maxSum(array));
}
}
public static long maxSum(long[] array) {
long[] tmpArray = array.clone();
long maxSum = 0;
long sum = 0;
for (int i=0; i< array.length; i++) {
for (int j=0; j < array.length; j++) {
if (i != j) {
while ((tmpArray[j]&1) == 0) {
tmpArray[j]>>=1;
tmpArray[i]<<=1;
}
sum+=tmpArray[j];
}
}
sum+=tmpArray[i];
if (sum>maxSum) maxSum = sum;
tmpArray = array.clone();
sum = 0;
}
return maxSum;
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | a67fa18064c4be435428a7d3c72c3ba2 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main{
public static void main (String[] args){
Scanner in=new Scanner(System.in);
int t=in.nextInt(),n;
long a[];
long s=0;
while(t--!=0){
n=in.nextInt();a=new long[n];s=1;
for(int i=0;i<n;i++) a[i]=in.nextLong();
for(int i=0;i<n;i++){
while(a[i]%2==0){
a[i]/=2;
s*=2;
}
}
Arrays.sort(a);
a[n-1]*=s;s=0;
for(int i=0;i<n;i++) s+=a[i];
System.out.println(s);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | f0331c5dad85e49abdbe640aca377610 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
void go() {
// add code
int n = Reader.nextInt();
long[] arr = new long[n];
long ans = 0;
long pow = 1;
for(int i = 0; i < n; i++) {
arr[i] = Reader.nextInt();
while(arr[i] % 2 == 0) {
arr[i] /= 2;
pow *= 2;
}
}
Arrays.sort(arr);
for(int i = 0; i < n - 1; i++) {
ans += arr[i];
}
ans += arr[n - 1] * pow;
Writer.println(ans);
}
void solve() {
for(int T = Reader.nextInt(); T > 0; T--) go();
}
void run() throws Exception {
Reader.init(System.in);
Writer.init(System.out);
solve();
Writer.close();
}
public static void main(String[] args) throws Exception {
new B().run();
}
public static class Reader {
public static StringTokenizer st;
public static BufferedReader br;
public static void init(InputStream in) {
br = new BufferedReader(new InputStreamReader(in));
st = new StringTokenizer("");
}
public static String next() {
while(!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new InputMismatchException();
}
}
return st.nextToken();
}
public static int nextInt() {
return Integer.parseInt(next());
}
public static long nextLong() {
return Long.parseLong(next());
}
public static double nextDouble() {
return Double.parseDouble(next());
}
}
public static class Writer {
public static PrintWriter pw;
public static void init(OutputStream os) {
pw = new PrintWriter(new BufferedOutputStream(os));
}
public static void print(String s) {
pw.print(s);
}
public static void print(int x) {
pw.print(x);
}
public static void print(long x) {
pw.print(x);
}
public static void println(String s) {
pw.println(s);
}
public static void println(int x) {
pw.println(x);
}
public static void println(long x) {
pw.println(x);
}
public static void close() {
pw.close();
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 3e8d6550504539d12b5312eaa884da25 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
void go() {
// add code
int n = Reader.nextInt();
long[] arr = new long[n];
long ans = 0;
for(int i = 0; i < n; i++) {
arr[i] = Reader.nextInt();
}
for(int i = 0; i < n; i++) {
long[] copy = Arrays.copyOf(arr, n);
for(int j = 0; j < n; j++) {
if(j == i) {
continue;
}
while(copy[j] % 2 == 0) {
copy[i] *= 2;
copy[j] /= 2;
}
}
long sum = 0;
for(long x : copy) {
sum += x;
}
ans = Math.max(ans, sum);
}
Writer.println(ans);
}
void solve() {
for(int T = Reader.nextInt(); T > 0; T--) go();
}
void run() throws Exception {
Reader.init(System.in);
Writer.init(System.out);
solve();
Writer.close();
}
public static void main(String[] args) throws Exception {
new B().run();
}
public static class Reader {
public static StringTokenizer st;
public static BufferedReader br;
public static void init(InputStream in) {
br = new BufferedReader(new InputStreamReader(in));
st = new StringTokenizer("");
}
public static String next() {
while(!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new InputMismatchException();
}
}
return st.nextToken();
}
public static int nextInt() {
return Integer.parseInt(next());
}
public static long nextLong() {
return Long.parseLong(next());
}
public static double nextDouble() {
return Double.parseDouble(next());
}
}
public static class Writer {
public static PrintWriter pw;
public static void init(OutputStream os) {
pw = new PrintWriter(new BufferedOutputStream(os));
}
public static void print(String s) {
pw.print(s);
}
public static void print(int x) {
pw.print(x);
}
public static void print(long x) {
pw.print(x);
}
public static void println(String s) {
pw.println(s);
}
public static void println(int x) {
pw.println(x);
}
public static void println(long x) {
pw.println(x);
}
public static void close() {
pw.close();
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | ff4303189e7315cc9d6c589b26bfb3b7 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
public class divideandmultiply {
public static void main(String[] args) {
Scanner scn =new Scanner(System.in);
try{
int t = scn.nextInt();
while(t-- > 0){
int n = scn.nextInt();
long[] arr = new long[n];
for(int i = 0;i<n;i++){
arr[i] = scn.nextLong();
}
Arrays.sort(arr);
long k =0;
for(int i = 0;i<arr.length;i++){
while(arr[i]%2 == 0) {
arr[i] = arr[i]/2;
k++;
}
}
long max = Integer.MIN_VALUE;
for(int i = 0;i<n;i++){
if(arr[i] > max){
max = arr[i];
}
}
long m =0;
long c = 0;
for(int i =0;i<arr.length;i++){
if(arr[i] != max) {
m += arr[i];
}
else if(arr[i] == max){
if(c > 0){
m += arr[i];
}
c++;
}
}
System.out.println((long)(m + Math.pow(2,k)*max));
}
}
catch (Exception e){
return;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 6444ecdcd31df4d245acc205e8174b4f | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | //package com.company;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
int n;
int i;
n = sc.nextInt();
long[] a = new long[n];
long sum = 0;
int cnt = 0;
for(i = 0; i < n; i++)
{
a[i] = sc.nextInt();
if(n == 1)
continue;
while (a[i] % 2 == 0)
{
a[i] /= 2;
cnt++;
}
}
if(n == 1)
{
System.out.println(a[0]);
continue;
}
Arrays.sort(a);
while (cnt-- > 0)
a[a.length - 1] *= 2;
for(i = 0; i < n; i++)
sum += a[i];
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 11 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 3c073fdb55ffcf35d826992ee0bb2ad6 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
public class Main {
static class InputReader {
/**
* The default size of the InputReader's buffer is 2<sup>16</sup>.
*/
private static final int DEFAULT_BUFFER_SIZE = 1 << 16;
/**
* The default stream for the InputReader is standard input.
*/
private static final InputStream DEFAULT_STREAM = System.in;
/**
* The maximum number of accurate decimal digits the method {@link #nextDoubleFast() nextDoubleFast()} can read.
* Currently this value is set to 21 because it is the maximum number of digits a double precision float can have at the moment.
*/
private static final int MAX_DECIMAL_PRECISION = 21;
// 'c' is used to refer to the current character in the stream
private int c;
// Variables associated with the byte buffer.
private byte[] buf;
private int bufferSize, bufIndex, numBytesRead;
private InputStream stream;
// End Of File (EOF) character
private static final byte EOF = -1;
// New line character: '\n'
private static final byte NEW_LINE = 10;
// Space character: ' '
private static final byte SPACE = 32;
// Dash character: '-'
private static final byte DASH = 45;
// Dot character: '.'
private static final byte DOT = 46;
// A reusable character buffer when reading string data.
private char[] charBuffer;
// Primitive data type lookup tables used for optimizations
private static byte[] bytes = new byte[58];
private static int[] ints = new int[58];
private static char[] chars = new char[128];
static {
char ch = ' ';
int value = 0;
byte _byte = 0;
for (int i = 48; i < 58; i++) bytes[i] = _byte++;
for (int i = 48; i < 58; i++) ints[i] = value++;
for (int i = 32; i < 128; i++) chars[i] = ch++;
}
// Primitive double lookup table used for optimizations.
private static final double[][] doubles = {
{0.0d, 0.00d, 0.000d, 0.0000d, 0.00000d, 0.000000d, 0.0000000d, 0.00000000d, 0.000000000d, 0.0000000000d, 0.00000000000d, 0.000000000000d, 0.0000000000000d, 0.00000000000000d, 0.000000000000000d, 0.0000000000000000d, 0.00000000000000000d, 0.000000000000000000d, 0.0000000000000000000d, 0.00000000000000000000d, 0.000000000000000000000d},
{0.1d, 0.01d, 0.001d, 0.0001d, 0.00001d, 0.000001d, 0.0000001d, 0.00000001d, 0.000000001d, 0.0000000001d, 0.00000000001d, 0.000000000001d, 0.0000000000001d, 0.00000000000001d, 0.000000000000001d, 0.0000000000000001d, 0.00000000000000001d, 0.000000000000000001d, 0.0000000000000000001d, 0.00000000000000000001d, 0.000000000000000000001d},
{0.2d, 0.02d, 0.002d, 0.0002d, 0.00002d, 0.000002d, 0.0000002d, 0.00000002d, 0.000000002d, 0.0000000002d, 0.00000000002d, 0.000000000002d, 0.0000000000002d, 0.00000000000002d, 0.000000000000002d, 0.0000000000000002d, 0.00000000000000002d, 0.000000000000000002d, 0.0000000000000000002d, 0.00000000000000000002d, 0.000000000000000000002d},
{0.3d, 0.03d, 0.003d, 0.0003d, 0.00003d, 0.000003d, 0.0000003d, 0.00000003d, 0.000000003d, 0.0000000003d, 0.00000000003d, 0.000000000003d, 0.0000000000003d, 0.00000000000003d, 0.000000000000003d, 0.0000000000000003d, 0.00000000000000003d, 0.000000000000000003d, 0.0000000000000000003d, 0.00000000000000000003d, 0.000000000000000000003d},
{0.4d, 0.04d, 0.004d, 0.0004d, 0.00004d, 0.000004d, 0.0000004d, 0.00000004d, 0.000000004d, 0.0000000004d, 0.00000000004d, 0.000000000004d, 0.0000000000004d, 0.00000000000004d, 0.000000000000004d, 0.0000000000000004d, 0.00000000000000004d, 0.000000000000000004d, 0.0000000000000000004d, 0.00000000000000000004d, 0.000000000000000000004d},
{0.5d, 0.05d, 0.005d, 0.0005d, 0.00005d, 0.000005d, 0.0000005d, 0.00000005d, 0.000000005d, 0.0000000005d, 0.00000000005d, 0.000000000005d, 0.0000000000005d, 0.00000000000005d, 0.000000000000005d, 0.0000000000000005d, 0.00000000000000005d, 0.000000000000000005d, 0.0000000000000000005d, 0.00000000000000000005d, 0.000000000000000000005d},
{0.6d, 0.06d, 0.006d, 0.0006d, 0.00006d, 0.000006d, 0.0000006d, 0.00000006d, 0.000000006d, 0.0000000006d, 0.00000000006d, 0.000000000006d, 0.0000000000006d, 0.00000000000006d, 0.000000000000006d, 0.0000000000000006d, 0.00000000000000006d, 0.000000000000000006d, 0.0000000000000000006d, 0.00000000000000000006d, 0.000000000000000000006d},
{0.7d, 0.07d, 0.007d, 0.0007d, 0.00007d, 0.000007d, 0.0000007d, 0.00000007d, 0.000000007d, 0.0000000007d, 0.00000000007d, 0.000000000007d, 0.0000000000007d, 0.00000000000007d, 0.000000000000007d, 0.0000000000000007d, 0.00000000000000007d, 0.000000000000000007d, 0.0000000000000000007d, 0.00000000000000000007d, 0.000000000000000000007d},
{0.8d, 0.08d, 0.008d, 0.0008d, 0.00008d, 0.000008d, 0.0000008d, 0.00000008d, 0.000000008d, 0.0000000008d, 0.00000000008d, 0.000000000008d, 0.0000000000008d, 0.00000000000008d, 0.000000000000008d, 0.0000000000000008d, 0.00000000000000008d, 0.000000000000000008d, 0.0000000000000000008d, 0.00000000000000000008d, 0.000000000000000000008d},
{0.9d, 0.09d, 0.009d, 0.0009d, 0.00009d, 0.000009d, 0.0000009d, 0.00000009d, 0.000000009d, 0.0000000009d, 0.00000000009d, 0.000000000009d, 0.0000000000009d, 0.00000000000009d, 0.000000000000009d, 0.0000000000000009d, 0.00000000000000009d, 0.000000000000000009d, 0.0000000000000000009d, 0.00000000000000000009d, 0.000000000000000000009d}
};
/**
* Create an InputReader that reads from standard input.
*/
public InputReader() {
this(DEFAULT_STREAM, DEFAULT_BUFFER_SIZE);
}
/**
* Create an InputReader that reads from standard input.
*
* @param bufferSize The buffer size for this input reader.
*/
public InputReader(int bufferSize) {
this(DEFAULT_STREAM, bufferSize);
}
/**
* Create an InputReader that reads from standard input.
*
* @param stream Takes an InputStream as a parameter to read from.
*/
public InputReader(InputStream stream) {
this(stream, DEFAULT_BUFFER_SIZE);
}
/**
* Create an InputReader that reads from standard input.
*
* @param stream Takes an {@link java.io.InputStream#InputStream() InputStream} as a parameter to read from.
* @param bufferSize The size of the buffer to use.
*/
public InputReader(InputStream stream, int bufferSize) {
if (stream == null || bufferSize <= 0)
throw new IllegalArgumentException();
buf = new byte[bufferSize];
charBuffer = new char[128];
this.bufferSize = bufferSize;
this.stream = stream;
}
/**
* Reads a single character from the input stream.
*
* @return Returns the byte value of the next character in the buffer and EOF
* at the end of the stream.
* @throws IOException throws exception if there is no more data to read
*/
private byte read() throws IOException {
if (numBytesRead == EOF) throw new IOException();
if (bufIndex >= numBytesRead) {
bufIndex = 0;
numBytesRead = stream.read(buf);
if (numBytesRead == EOF)
return EOF;
}
return buf[bufIndex++];
}
/**
* Read values from the input stream until you reach a character with a
* higher ASCII value than 'token'.
*
* @param token The token is a value which we use to stop reading junk out of
* the stream.
* @return Returns 0 if a value greater than the token was reached or -1 if
* the end of the stream was reached.
* @throws IOException Throws exception at end of stream.
*/
private int readJunk(int token) throws IOException {
if (numBytesRead == EOF) return EOF;
// Seek to the first valid position index
do {
while (bufIndex < numBytesRead) {
if (buf[bufIndex] > token) return 0;
bufIndex++;
}
// reload buffer
numBytesRead = stream.read(buf);
if (numBytesRead == EOF) return EOF;
bufIndex = 0;
} while (true);
}
/**
* Reads a single byte from the input stream.
*
* @return The next byte in the input stream
* @throws IOException Throws exception at end of stream.
*/
public byte nextByte() throws IOException {
return (byte) nextInt();
}
/**
* Reads a 32 bit signed integer from input stream.
*
* @return The next integer value in the stream.
* @throws IOException Throws exception at end of stream.
*/
public int nextInt() throws IOException {
if (readJunk(DASH - 1) == EOF) throw new IOException();
int sgn = 1, res = 0;
c = buf[bufIndex];
if (c == DASH) {
sgn = -1;
bufIndex++;
}
do {
while (bufIndex < numBytesRead) {
if (buf[bufIndex] > SPACE) {
res = (res << 3) + (res << 1);
res += ints[buf[bufIndex++]];
} else {
bufIndex++;
return res * sgn;
}
}
// Reload buffer
numBytesRead = stream.read(buf);
if (numBytesRead == EOF) return res * sgn;
bufIndex = 0;
} while (true);
}
/**
* Reads a 64 bit signed long from input stream.
*
* @return The next long value in the stream.
* @throws IOException Throws exception at end of stream.
*/
public long nextLong() throws IOException {
if (readJunk(DASH - 1) == EOF) throw new IOException();
int sgn = 1;
long res = 0L;
c = buf[bufIndex];
if (c == DASH) {
sgn = -1;
bufIndex++;
}
do {
while (bufIndex < numBytesRead) {
if (buf[bufIndex] > SPACE) {
res = (res << 3) + (res << 1);
res += ints[buf[bufIndex++]];
} else {
bufIndex++;
return res * sgn;
}
}
// Reload buffer
numBytesRead = stream.read(buf);
if (numBytesRead == EOF) return res * sgn;
bufIndex = 0;
} while (true);
}
/**
* Doubles the size of the internal char buffer for strings
*/
private void doubleCharBufferSize() {
char[] newBuffer = new char[charBuffer.length << 1];
for (int i = 0; i < charBuffer.length; i++) newBuffer[i] = charBuffer[i];
charBuffer = newBuffer;
}
/**
* Reads a line from the input stream.
*
* @return Returns a line from the input stream in the form a String not
* including the new line character. Returns <code>null</code> when there are
* no more lines.
* @throws IOException Throws IOException when something terrible happens.
*/
public String nextLine() throws IOException {
try {
c = read();
} catch (IOException e) {
return null;
}
if (c == NEW_LINE) return ""; // Empty line
if (c == EOF) return null; // EOF
int i = 0;
charBuffer[i++] = (char) c;
do {
while (bufIndex < numBytesRead) {
if (buf[bufIndex] != NEW_LINE) {
if (i == charBuffer.length) doubleCharBufferSize();
charBuffer[i++] = (char) buf[bufIndex++];
} else {
bufIndex++;
return new String(charBuffer, 0, i);
}
}
// Reload buffer
numBytesRead = stream.read(buf);
if (numBytesRead == EOF)
return new String(charBuffer, 0, i);
bufIndex = 0;
} while (true);
}
// Reads a string of characters from the input stream.
// The delimiter separating a string of characters is set to be:
// any ASCII value <= 32 meaning any spaces, new lines, EOF, tabs...
public String nextString() throws IOException {
if (numBytesRead == EOF) return null;
if (readJunk(SPACE) == EOF) return null;
for (int i = 0; ; ) {
while (bufIndex < numBytesRead) {
if (buf[bufIndex] > SPACE) {
if (i == charBuffer.length) doubleCharBufferSize();
charBuffer[i++] = (char) buf[bufIndex++];
} else {
bufIndex++;
return new String(charBuffer, 0, i);
}
}
// Reload buffer
numBytesRead = stream.read(buf);
if (numBytesRead == EOF) return new String(charBuffer, 0, i);
bufIndex = 0;
}
}
// Returns an exact value a double value from the input stream.
public double nextDouble() throws IOException {
String doubleVal = nextString();
if (doubleVal == null) throw new IOException();
return Double.valueOf(doubleVal);
}
// Very quickly reads a double value from the input stream (~3x faster than nextDouble()). However,
// this method may provide a slightly less accurate reading than .nextDouble() if there are a lot
// of digits (~16+). In particular, it will only read double values with at most 21 digits after
// the decimal point and the reading my be as inaccurate as ~5*10^-16 from the true value.
public double nextDoubleFast() throws IOException {
c = read();
int sgn = 1;
while (c <= SPACE) c = read(); // while c is either: ' ', '\n', EOF
if (c == DASH) {
sgn = -1;
c = read();
}
double res = 0.0;
// while c is not: ' ', '\n', '.' or -1
while (c > DOT) {
res *= 10.0;
res += ints[c];
c = read();
}
if (c == DOT) {
int i = 0;
c = read();
// while c is digit and we are less than the maximum decimal precision
while (c > SPACE && i < MAX_DECIMAL_PRECISION) {
res += doubles[ints[c]][i++];
c = read();
}
}
return res * sgn;
}
// Read an array of n byte values
public byte[] nextByteArray(int n) throws IOException {
byte[] ar = new byte[n];
for (int i = 0; i < n; i++) ar[i] = nextByte();
return ar;
}
// Read an integer array of size n
public int[] nextIntArray(int n) throws IOException {
int[] ar = new int[n];
for (int i = 0; i < n; i++) ar[i] = nextInt();
return ar;
}
// Read a long array of size n
public long[] nextLongArray(int n) throws IOException {
long[] ar = new long[n];
for (int i = 0; i < n; i++) ar[i] = nextLong();
return ar;
}
// read an of doubles of size n
public double[] nextDoubleArray(int n) throws IOException {
double[] ar = new double[n];
for (int i = 0; i < n; i++) ar[i] = nextDouble();
return ar;
}
// Quickly read an array of doubles
public double[] nextDoubleArrayFast(int n) throws IOException {
double[] ar = new double[n];
for (int i = 0; i < n; i++) ar[i] = nextDoubleFast();
return ar;
}
// Read a string array of size n
public String[] nextStringArray(int n) throws IOException {
String[] ar = new String[n];
for (int i = 0; i < n; i++) {
String str = nextString();
if (str == null) throw new IOException();
ar[i] = str;
}
return ar;
}
// Read a 1-based byte array of size n+1
public byte[] nextByteArray1(int n) throws IOException {
byte[] ar = new byte[n + 1];
for (int i = 1; i <= n; i++) ar[i] = nextByte();
return ar;
}
// Read a 1-based integer array of size n+1
public int[] nextIntArray1(int n) throws IOException {
int[] ar = new int[n + 1];
for (int i = 1; i <= n; i++) ar[i] = nextInt();
return ar;
}
// Read a 1-based long array of size n+1
public long[] nextLongArray1(int n) throws IOException {
long[] ar = new long[n + 1];
for (int i = 1; i <= n; i++) ar[i] = nextLong();
return ar;
}
// Read a 1-based double array of size n+1
public double[] nextDoubleArray1(int n) throws IOException {
double[] ar = new double[n + 1];
for (int i = 1; i <= n; i++) ar[i] = nextDouble();
return ar;
}
// Quickly read a 1-based double array of size n+1
public double[] nextDoubleArrayFast1(int n) throws IOException {
double[] ar = new double[n + 1];
for (int i = 1; i <= n; i++) ar[i] = nextDoubleFast();
return ar;
}
// Read a 1-based string array of size n+1
public String[] nextStringArray1(int n) throws IOException {
String[] ar = new String[n + 1];
for (int i = 1; i <= n; i++) ar[i] = nextString();
return ar;
}
// Read a two dimensional matrix of bytes of size rows x cols
public byte[][] nextByteMatrix(int rows, int cols) throws IOException {
byte[][] matrix = new byte[rows][cols];
for (int i = 0; i < rows; i++)
for (int j = 0; j < cols; j++)
matrix[i][j] = nextByte();
return matrix;
}
// Read a two dimensional matrix of ints of size rows x cols
public int[][] nextIntMatrix(int rows, int cols) throws IOException {
int[][] matrix = new int[rows][cols];
for (int i = 0; i < rows; i++)
for (int j = 0; j < cols; j++)
matrix[i][j] = nextInt();
return matrix;
}
// Read a two dimensional matrix of longs of size rows x cols
public long[][] nextLongMatrix(int rows, int cols) throws IOException {
long[][] matrix = new long[rows][cols];
for (int i = 0; i < rows; i++)
for (int j = 0; j < cols; j++)
matrix[i][j] = nextLong();
return matrix;
}
// Read a two dimensional matrix of doubles of size rows x cols
public double[][] nextDoubleMatrix(int rows, int cols) throws IOException {
double[][] matrix = new double[rows][cols];
for (int i = 0; i < rows; i++)
for (int j = 0; j < cols; j++)
matrix[i][j] = nextDouble();
return matrix;
}
// Quickly read a two dimensional matrix of doubles of size rows x cols
public double[][] nextDoubleMatrixFast(int rows, int cols) throws IOException {
double[][] matrix = new double[rows][cols];
for (int i = 0; i < rows; i++)
for (int j = 0; j < cols; j++)
matrix[i][j] = nextDoubleFast();
return matrix;
}
// Read a two dimensional matrix of Strings of size rows x cols
public String[][] nextStringMatrix(int rows, int cols) throws IOException {
String[][] matrix = new String[rows][cols];
for (int i = 0; i < rows; i++)
for (int j = 0; j < cols; j++)
matrix[i][j] = nextString();
return matrix;
}
// Read a 1-based two dimensional matrix of bytes of size rows x cols
public byte[][] nextByteMatrix1(int rows, int cols) throws IOException {
byte[][] matrix = new byte[rows + 1][cols + 1];
for (int i = 1; i <= rows; i++)
for (int j = 1; j <= cols; j++)
matrix[i][j] = nextByte();
return matrix;
}
// Read a 1-based two dimensional matrix of ints of size rows x cols
public int[][] nextIntMatrix1(int rows, int cols) throws IOException {
int[][] matrix = new int[rows + 1][cols + 1];
for (int i = 1; i <= rows; i++)
for (int j = 1; j <= cols; j++)
matrix[i][j] = nextInt();
return matrix;
}
// Read a 1-based two dimensional matrix of longs of size rows x cols
public long[][] nextLongMatrix1(int rows, int cols) throws IOException {
long[][] matrix = new long[rows + 1][cols + 1];
for (int i = 1; i <= rows; i++)
for (int j = 1; j <= cols; j++)
matrix[i][j] = nextLong();
return matrix;
}
// Read a 1-based two dimensional matrix of doubles of size rows x cols
public double[][] nextDoubleMatrix1(int rows, int cols) throws IOException {
double[][] matrix = new double[rows + 1][cols + 1];
for (int i = 1; i <= rows; i++)
for (int j = 1; j <= cols; j++)
matrix[i][j] = nextDouble();
return matrix;
}
// Quickly read a 1-based two dimensional matrix of doubles of size rows x cols
public double[][] nextDoubleMatrixFast1(int rows, int cols) throws IOException {
double[][] matrix = new double[rows + 1][cols + 1];
for (int i = 1; i <= rows; i++)
for (int j = 1; j <= cols; j++)
matrix[i][j] = nextDoubleFast();
return matrix;
}
// Read a 1-based two dimensional matrix of Strings of size rows x cols
public String[][] nextStringMatrix1(int rows, int cols) throws IOException {
String[][] matrix = new String[rows + 1][cols + 1];
for (int i = 1; i <= rows; i++)
for (int j = 1; j <= cols; j++)
matrix[i][j] = nextString();
return matrix;
}
// Closes the input stream
public void close() throws IOException {
stream.close();
}
}
static InputReader read = new InputReader();
//static PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) throws IOException {
int t = read.nextInt();
while (t-->0){
int n = read.nextInt();
long[] array = new long[n];
long count = 1;
for (int i = 0; i < n; i++) {
array[i] = read.nextInt();
while (array[i] % 2 == 0) {
array[i] /= 2;
count *=2;
}
}
Arrays.sort(array);
array[n - 1] *= count;
long sum =0;
for(long i : array)
sum+=i;
System.out.println(sum);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 17 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 5ae244d8b5433d35bb89b393fb546989 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | // package com.company.Codeforces;
import java.util.*;
import java.lang.*;
import java.io.*;
public class scratch
{
public static void solve(){
int n = sc.nextInt();
int [] arr = sc.nextIntArray(n);
int k = 0;
int max = 0;
long sum = 0;
for(int i=0; i<n; i++){
if( (arr[i]&1) == 0){
while ( (arr[i]&1) == 0){
arr[i] /= 2;
k++;
}
}
max = Math.max(arr[i], max);
}
boolean flag = false;
for(int i=0; i<n; i++){
if(flag || arr[i] != max){
sum += arr[i];
}
else {
flag = true;
sum += max * power(2, k);
}
}
str.append(sum);
}
static PrintWriter out;
static StringBuilder str;
static Scanner sc;
public static void main (String[] args) throws java.lang.Exception
{
sc = new Scanner();
str = new StringBuilder();
out = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-->0){
solve();
str.append('\n');
}
out.println(str);
out.close();
}
private static long power(long a, long b) {
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
private static long power(long a, long b, long m) {
a %= m;
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = res * a % m;
a = a * a % m;
b >>= 1;
}
return res;
}
private static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
private static long gcd(long a, long b) {
return (b == 0) ? a : gcd(b, a % b);
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() {
br = new BufferedReader(new
InputStreamReader(System.in));
/*
try {
br = new BufferedReader(new
InputStreamReader(new FileInputStream(new File("file_i_o\\input.txt"))));
} catch (FileNotFoundException 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[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
Integer[] nextIntegerArray(int n) {
Integer[] array = new Integer[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; i++) {
array[i] = nextLong();
}
return array;
}
String[] nextStringArray() {
return nextLine().split(" ");
}
String[] nextStringArray(int n) {
String[] array = new String[n];
for (int i = 0; i < n; i++) {
array[i] = next();
}
return array;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 17 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 6a1cd74aba961a269d1aaab1043896aa | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.*;
import java.io.*;
public class Main {
static FastReader in = new FastReader();
static final Random random = new Random();
static long mod = 1000000007L;
static HashMap<String, Integer> map = new HashMap<>();
public static void main(String args[]) throws IOException {
int t = in.nextInt();
StringBuilder res = new StringBuilder();
while (t-- > 0) {
int n = in.nextInt();
int[] arr = new int[n];
for (int i = 0; i < arr.length; i++) {
arr[i] = in.nextInt();
}
// keep track of max
int op = 0;
for (int i = 0; i < arr.length; i++) {
int temp = arr[i];
if (temp % 2 == 0) {
while (temp != 1) {
if (temp % 2 != 0) {
break;
}
op++;
temp = temp / 2;
}
}
arr[i] = temp;
}
// This iteration is to find the max numbers and sum the rest.
int sum = 0;
int max = 0;
for (int i = 0; i < arr.length; i++) {
max = Math.max(max, arr[i]);
sum +=arr[i];
}
sum = sum - max;
long ans = max;
while (op-- > 0) {
ans = ans * 2;
}
System.out.println(sum+ ans);
}
print(res);
}
static int max(int a, int b) {
if (a < b)
return b;
return a;
}
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 <E> void print(E res) {
System.out.println(res);
}
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 int abs(int a) {
if (a < 0)
return -1 * a;
return a;
}
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[] readintarray(int n) {
int res[] = new int[n];
for (int i = 0; i < n; i++)
res[i] = nextInt();
return res;
}
long[] readlongarray(int n) {
long res[] = new long[n];
for (int i = 0; i < n; i++)
res[i] = nextLong();
return res;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 17 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 87bc2bd8f1927acaa4e80aea073a4c22 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.*;
import java.io.*;
public class Main {
static FastReader in = new FastReader();
static final Random random = new Random();
static long mod = 1000000007L;
static HashMap<String, Integer> map = new HashMap<>();
public static void main(String args[]) throws IOException {
int t = in.nextInt();
StringBuilder res = new StringBuilder();
while (t-- > 0) {
int n = in.nextInt();
int[] arr = new int[n];
for (int i = 0; i < arr.length; i++) {
arr[i] = in.nextInt();
}
// keep track of max
int op = 0;
for (int i = 0; i < arr.length; i++) {
int temp = arr[i];
if (temp % 2 == 0) {
while (temp != 1) {
if (temp % 2 != 0) {
break;
}
op++;
temp = temp / 2;
}
}
arr[i] = temp;
}
// This iteration is to find the max numbers and sum the rest.
Arrays.sort(arr);
int sum = 0;
int max = arr[arr.length - 1];
for (int i = 0; i < arr.length - 1; i++) {
sum += arr[i];
}
long ans = max;
while (op-- > 0) {
ans = ans * 2;
}
System.out.println(ans+sum);
}
print(res);
}
static int max(int a, int b) {
if (a < b)
return b;
return a;
}
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 <E> void print(E res) {
System.out.println(res);
}
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 int abs(int a) {
if (a < 0)
return -1 * a;
return a;
}
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[] readintarray(int n) {
int res[] = new int[n];
for (int i = 0; i < n; i++)
res[i] = nextInt();
return res;
}
long[] readlongarray(int n) {
long res[] = new long[n];
for (int i = 0; i < n; i++)
res[i] = nextLong();
return res;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 17 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c88f8c000a42ca3da872da5af24f09c0 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
public class another {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int tc = sc.nextInt();
while(tc-- > 0){
int n = sc.nextInt();
long[] array = new long[n];
long count = 1;
for (int i = 0; i < n; i++) {
array[i] = sc.nextInt();
while (array[i] % 2 == 0) {
array[i] /= 2;
count *=2;
}
}
Arrays.sort(array);
array[n - 1] *= count;
long sum =0;
for(long i : array)
sum+=i;
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 17 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 0ed0aca05d4ee813afd53efded5c4345 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
public class Solution{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
int n1=s.nextInt();
while(n1>0)
{
--n1;
int n=s.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++)
{
arr[i]=s.nextInt();
}
Arrays.sort(arr);
int c=0;
int o=-1;
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
while(arr[i]%2==0)
{
++c;
arr[i]/=2;
}
}
if(arr[i]%2!=0&&arr[i]>o)
{
o=arr[i];
}
}
Long sum=Long.valueOf(arr[n-1]*(long)Math.pow(2, c));
for(int i=0;i<n-1;i++)
{
sum+=arr[i];
}
Long sum1=Long.valueOf(o*(long)Math.pow(2, c));
for(int i=0;i<n;i++)
{
sum1+=arr[i];
}
sum1-=o;
System.out.println(Math.max(sum, sum1));
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 17 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c88a99f8a3f4dd551031bcefdaa489a9 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.util.function.*;
import java.io.*;
// you can compare with output.txt and expected out
public class RoundDeltixAutumn2021A {
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();
RoundDeltixAutumn2021A sol = new RoundDeltixAutumn2021A();
sol.run();
}
private void run() {
boolean isDebug = false;
boolean isFileIO = true;
boolean hasMultipleTests = true;
initIO(isFileIO);
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;
long[] a;
void getInput() {
n = in.nextInt();
a = in.nextLongArray(n);
}
void printOutput() {
out.printlnAns(ans);
}
long ans;
void solve(){
// 2^(4*15) = 2^60
long max = 0;
for(int i=0; i<n; i++) {
long[] b = Arrays.copyOf(a, n);
for(int j=0; j<n; j++) {
if(j == i)
continue;
while(b[j] % 2 == 0) {
b[j] /= 2;
b[i] *= 2;
}
}
long sum = 0;
for(int j=0; j<n; j++)
sum += b[j];
max = Math.max(max, sum);
}
ans = max;
}
// 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) {
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 17 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | b112a36f34f2d33834a97fb7ce022e2a | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.lang.*;
import static java.lang.Math.*;
// Sachin_2961 submission //
public class Codeforces {
static void solve() {
int n = fs.nInt();
int[]ar = new int[n];
long prd = 1,max = 0;
long ans = 0;
for(int i=0;i<n;i++){
ar[i] = fs.nInt();
while( ar[i]%2 == 0){
prd *= 2;
ar[i] /= 2;
}
max = max(max,ar[i]);
}
for(int i=0;i<n;i++){
ans += ar[i];
}
ans -= max;
prd *= max;
ans += prd;
out.println(ans);
}
static class Pair{
int f,s;
Pair(int f,int s){
this.f = f;
this.s = s;
}
}
static boolean multipleTestCase = true;
static FastScanner fs;
static PrintWriter out;
public static void main(String[]args){
try{
out = new PrintWriter(System.out);
fs = new FastScanner();
int tc = multipleTestCase?fs.nInt():1;
while (tc-->0)solve();
out.flush();
out.close();
}catch (Exception e){
e.printStackTrace();
}
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String n() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
String Line()
{
String str = "";
try
{
str = br.readLine();
}catch (IOException e)
{
e.printStackTrace();
}
return str;
}
int nInt() {return Integer.parseInt(n()); }
long nLong() {return Long.parseLong(n());}
double nDouble(){return Double.parseDouble(n());}
int[]aI(int n){
int[]ar = new int[n];
for(int i=0;i<n;i++)
ar[i] = nInt();
return ar;
}
}
public static void sort(int[] arr){
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static void sort(long[] arr){
ArrayList<Long> ls = new ArrayList<>();
for(long x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 2f2cf1c04e4c8a5ff470c864f66fee39 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
Scanner in;
PrintWriter out;
Main() {
in = new Scanner(System.in);
out = new PrintWriter(System.out);
}
void solve() throws IOException {
int t = in.nextInt();
while (t > 0) {
int n = in.nextInt();
long[] ns = new long[n];
for (int i = 0; i < n; i++) {
long v = in.nextInt();
ns[i] = v;
}
int k = 0;
for (int i = 0; i < n; i++) {
while (ns[i] % 2 == 0) {
ns[i] /= 2;
k++;
}
}
Arrays.sort(ns);
for (int i = 0; i < k; i++) {
ns[n - 1] *= 2;
}
long ans = 0;
for (int i = 0; i < n; i++) {
ans += ns[i];
}
out.append(String.format("%d\n", ans));
t--;
}
}
void close() throws IOException {
if (in != null) {
in.close();
}
if (out != null) {
out.flush();
out.close();
}
}
public static void main(String[] args) throws Exception {
FileInputStream inStream = null;
PrintStream outStream = null;
boolean isLocal = System.getProperty("os.name").equals("Mac OS X");
if (isLocal) {
inStream = new FileInputStream("1.in");
// outStream = new PrintStream("1.out");
System.setIn(inStream);
// System.setOut(outStream);
}
Main main = new Main();
main.solve();
main.close();
if (isLocal) {
inStream.close();
// outStream.close();
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 00c92ed7b06505b0b68f410f1172f3f8 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
/*
*
* @author Sabirov Jakhongir
*
*/
public class Test {
static class Coin{
long value;
int cnt;
public Coin(long value, int cnt) {
this.value = value;
this.cnt = cnt;
}
@Override public String toString() {
return "Coin{" +
"value=" + value +
", cnt=" + cnt +
'}';
}
}
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int test = cin.nextInt();
while (test -- > 0)
{
int n = cin.nextInt();
int a[] = new int[n];
for(int i = 0 ;i < n;i ++){
a[i] = cin.nextInt();
}
int k = 0;
long summa = 0;
for(int i = 0 ;i < n; i++)
{
while (a[i] % 2 == 0)
{
k++;
a[i] /= 2;
}
}
int mx = 1;
int ind = 0;
for(int i = 0 ;i < n;i ++)
{
if(mx < a[i])
{
ind = i;
mx = a[i];
}
}
long suma = (long) (mx * Math.pow(2,k));
for(int i = 0 ; i < n ;i ++)
{
if(ind != i){
suma += a[i];
}
}
System.out.println(suma);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | ec9d57dc4ee2a3814259adf4bdb0c2cf | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
public class test {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-->0){
int a = in.nextInt();
long[] arr = new long[a];
for (int i = 0; i < a; i++) {
arr[i] = in.nextLong();
}
long temp = 1;
for (int i = 0; i < a; i++) {
while(arr[i] % 2 == 0){
arr[i] /= 2;
temp *= 2;
}
}
Arrays.sort(arr);
arr[arr.length-1] *= temp;
long sum = 0;
for (int i = 0; i < a; i++) {
sum += arr[i];
}
System.out.println(sum);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 5cee5c05f11403c019a95a8658d7b81e | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | //CP- MASS_2701
import java.util.*;
import java.io.*;
public class A_Divide_and_Multiply {
static FastReader in=new FastReader();
static final Random random=new Random();
static long mod=1000000007L;
static HashMap<String,Integer>map=new HashMap<>();
public static void main(String args[]) throws IOException {
int t=in.nextInt();
int cse=1;
loop:
while(t-->0)
{
StringBuilder res=new StringBuilder();
//res.append("Hello"+"\n");
int n=in.nextInt();
long[] a=new long[n];
for(int i=0;i<n;i++)
{
a[i]=in.nextLong();
}
Arrays.sort(a);
long count=0;
for(int i=0;i<n;i++)
{
while(a[i]%2==0)
{
a[i]=a[i]/2;
count++;
}
}
Arrays.sort(a);
a[n-1]=a[n-1]*(long)Math.pow(2,count);
long sum=a[n-1];
for(int i=0;i<n-1;i++)
{
sum+=a[i];
}
println(sum);
}
}
static int max(int a, int b)
{
if(a<b)
return b;
return a;
}
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 < E > void print(E res)
{
System.out.print(res);
}
static < E > void println(E res)
{
System.out.println(res);
}
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 int abs(int a)
{
if(a<0)
return -1*a;
return a;
}
static boolean isPrime(int 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 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 [] readintarray(int n) {
int res [] = new int [n];
for(int i = 0; i<n; i++)res[i] = nextInt();
return res;
}
long [] readlongarray(int n) {
long res [] = new long [n];
for(int i = 0; i<n; i++)res[i] = nextLong();
return res;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 459cbc622996df7dd3b1c317477d5ce3 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class Multiply
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int test=sc.nextInt();
while(test>0)
{
int n=sc.nextInt();
long arr[]=new long[n];
for(int i=0; i<n; i++)
{
arr[i]=sc.nextLong();
}
Arrays.sort(arr);
long brr[]=new long[n];
//long evensum=0;
int count=0;
for(int i=0; i<n; i++)
{
if(arr[i]%2==0)
{
long x=arr[i];
while(x%2==0 && x>0)
{
x=x/2;
count++;
}
brr[i]=x;
}
else
{
brr[i]=arr[i];
}
}
/*if(arr[n-1]%2==0)
{
evensum=0;
for(int i=0; i<n-1; i++)
{
evensum=evensum+brr[i];
}
evensum=evensum+arr[n-1]*(long)Math.pow(2,count);
}
*/
Arrays.sort(brr);
long oddsum=0;
/*int indx=-1;
for(int i=0; i<n; i++)
{
if(arr[i]%2!=0)
{
indx=i;
break;
}
}
*/
for(int i=0; i<n-1; i++)
{
oddsum=oddsum+brr[i];
}
oddsum=oddsum+brr[n-1]*(long)Math.pow(2,count);
System.out.println(oddsum);
test--;
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | f8bb4b43bbc579399528a0af55d5b831 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.math.BigInteger;
import java.util.*;
public class test {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-->0){
int n = in.nextInt();
BigInteger arr[] = new BigInteger[n];
int co = 0;
for (int i = 0; i < n; i++) {
arr[i] = new BigInteger(in.next());
while (arr[i].mod(BigInteger.valueOf(2)).compareTo(BigInteger.valueOf(0))==0){
arr[i] = arr[i].divide(BigInteger.valueOf(2));
co++;
}
}
Arrays.sort(arr);
for (int i = 0; i < co; i++) {
arr[n-1] =arr[n-1].multiply(BigInteger.valueOf(2));
}
BigInteger sum = new BigInteger("0");
for (int i = 0; i < n; i++) {
sum = sum.add(arr[i]);
}
System.out.println(sum);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | f99e5bedd4a32f1df78ad588a8007610 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class DivideAndMultiply {
static long sum(long[] a){
long sum = 0;
for(int i=0; i<a.length ;i++)
sum += a[i];
return sum;
}
public static void main(String[] args){
Scanner sc =new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
long[] a = new long[n];
long p = 1;
for(int i=0; i<n ;i++) {
a[i] = sc.nextInt();
while(a[i]%2 == 0){
a[i] /= 2;
p *= 2;
}
}
Arrays.sort(a);
a[n-1] *= p;
long sum = sum(a);
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | a452b4959076b94ade1463857a2e3ada | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.*;
public class solution {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
int i=0;
long count=0;long res=0;
for(i=0;i<n;i++)
{ while(arr[i]%2==0)
{count++;
arr[i]=arr[i]/2;
}
}
Arrays.sort(arr);
for(int k=0;k<n-1;k++)
{
res=res+arr[k];
}
res+= arr[n-1]*(long)Math.pow(2,count) ;
System.out.println(res);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 80edc3f7d9b0e3239df089a93e3d749a | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class solution
{ 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) throws java.lang.Exception
{
// your code goes here
OutputStream outputStream =System.out;
PrintWriter out =new PrintWriter(outputStream);
FastReader s = new FastReader();
int t = s.nextInt();
while(t-->0){
int n = s.nextInt();
int arr[] = new int[n];
//int max = 0;
// long key = 0;
//long sum = 0;
//long odd = 0;
//int sum1 = 0;
//int key2 = 0;
for(int i = 0;i<n;i++){
arr[i] = s.nextInt();
// sum1+=arr[i];
// if(arr[i]>=key&&arr[i]%2!=0){
// key = arr[i];
// }
// if(arr[i]>=key2&&arr[i]%2==0){
// key2=arr[i];
// }
}
// long ans = 0;
// //int flag = 0;
// // Arrays.sort(arr);
// if(key==0){
// Arrays.sort(arr);
// key = arr[n-1];
// for(int i = 0;i<n-1;i++){
// if(arr[i]%2==0){
// while(arr[i]%2==0){
// key = key*2;
// arr[i] = arr[i]/2;
// }
// sum+=arr[i];
// }
// ans = key + sum;
// }
// }else{
// long temp = key;
// for(int i = 0;i<n;i++){
// if(arr[i]%2==0){
// while(arr[i]%2==0){
// key = key*2;
// arr[i] = arr[i]/2;
// }
// sum+=arr[i];
// }else{
// odd+=arr[i];
// }
// }
// ans = (key + odd + sum) - temp;
// }
int count = 0;
for(int i = 0;i<n;i++){
while(arr[i]%2==0){
arr[i] = arr[i]/2;
count++;
}
}
Arrays.sort(arr);
// out.println(count);
long max = (long)Math.pow(2,count);
long ans = arr[n-1]*max;
for(int i = 0;i<n-1;i++){
ans+=arr[i];
}
out.println(ans);
}
out.close();
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 693f8ffd6276ade5978ef14825397fc8 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
public class A3 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-- > 0){
long ans = 0;
int n = in.nextInt();
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextLong();
}
int pow = 0;
for (int i = 0; i < n; i++) {
while (a[i] % 2 == 0){
a[i] /= 2;
pow++;
}
}
Arrays.sort(a);
a[n - 1] *= (long)Math.pow(2, pow);
for (int i = 0; i < n; i++) {
ans += a[i];
}
System.out.println(ans);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 0e6db414fcfda8b05e07d18b21eafb7d | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.*;
import java.util.stream.Collectors;
import java.io.*;
import java.math.*;
public class A2 {
public static FastScanner sc;
public static PrintWriter pw;
public static StringBuilder sb;
public static int MOD= 1000000007;
public 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());
}
}
public static void printList(List<Long> al) {
System.out.println(al.stream().map(it -> it.toString()).collect(Collectors.joining(" ")));
}
public static void printList(Deque<Long> al) {
System.out.println(al.stream().map(it -> it.toString()).collect(Collectors.joining(" ")));
}
public static void printArr(int[] arr) {
System.out.println(Arrays.toString(arr).trim().replace("[", "").replace("]","").replace(","," "));
}
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 void decreasingOrder(ArrayList<Long> al) {
Comparator<Long> c = Collections.reverseOrder();
Collections.sort(al,c);
}
public static boolean[] sieveOfEratosthenes(int n) {
boolean isPrime[] = new boolean[n+1];
Arrays.fill(isPrime, true);
isPrime[0]=false;
isPrime[1]=false;
for(int i=2;i*i<n;i++) {
for(int j=2*i;j<n;j+=i) {
isPrime[j]=false;
}
}
return isPrime;
}
public static long nCr(long N, long R) {
double result=1;
for(int i=1;i<=R;i++) result=((result*(N-R+i))/i);
return (long) result;
}
public static long fastPow(long a, long b, int n) {
long result=1;
while(b>0) {
if((b&1)==1)
result=(result*a %n)%n;
a=(a%n * a%n)%n;
b>>=1;
}
return result;
}
public static int BinarySearch(long[] arr, long key) {
int low=0;
int high=arr.length-1;
while(low<=high) {
int mid=(low+high)/2;
if(arr[mid]==key)
return mid;
else if(arr[mid]>key)
high=mid-1;
else
low=mid+1;
}
return low; //High=low-1
}
public static void solve(int t) {
int n=sc.nextInt();
int[] arr = new int[n];
for(int i=0;i<n;i++) arr[i]=sc.nextInt();
Arrays.sort(arr);
long ans=-1;
int index=-1;
int temp=-1;
for(int i=n-1;i>=0;i--) {
int temp2=arr[i];
while(temp2%2==0) temp2/=2;
if(temp2>temp) {
temp=temp2;
ans=arr[i];
index=i;
}
}
for(int i=0;i<n;i++) {
if(i==index) continue;
while(arr[i]%2==0) {
arr[i]=arr[i]/2;
ans*=2;
}
}
for(int i=0;i<n;i++) {
if(i==index) continue;
else {
ans+=arr[i];
}
}
System.out.println(ans);
}
public static void main(String[] args) {
sc = new FastScanner();
pw = new PrintWriter(new OutputStreamWriter(System.out));
sb= new StringBuilder("");
int t=sc.nextInt();
for(int i=1;i<=t;i++) solve(i);
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c2ee731f00b6bbb7a39e238e05b6b7ce | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class A {
public static long gcd(long a, long b) {
return a == 0 ? b : gcd(b, a % b);
}
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter o = new PrintWriter(System.out);
int t1 = sc.nextInt();
while (t1-- > 0) {
int odd = 0;
int n = sc.nextInt();
int[] a = sc.readArray(n);
int sum = 0, x = 0;
long ans = 0,ans1 = 0;
for (int k = 0; k < n; k++) {
ArrayList<Integer> dd = new ArrayList<>();
for (int i = 0; i < n; i++) {
dd.add(a[i]);
}
ans = x = 0;
odd = a[k];
dd.remove(Integer.valueOf(odd));
for (int i = 0; i < 4; i++) {
for (int j = 0; j < n - 1; j++) {
if (dd.get(j) % 2 == 0) {
x += 1;
dd.set(j, dd.get(j) / 2);
}
}
}
ans += odd * (Math.pow(2, x));
for (int i = 0; i < n - 1; i++) {
ans += dd.get(i);
}
ans1 = Math.max(ans, ans1);
}
o.println(ans1);
}
o.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
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 {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public boolean ready() throws IOException {
return br.ready();
}
int[] readArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 1a4e5adb6d4d313b5a8b7901a293e432 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.Scanner;
public class Main {
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
long max = 0;
long res = 1;
long sum = 0;
for(int i = 0;i<n;i++){
int x = sc.nextInt();
while(x%2==0) {
x>>=1;
res<<=1;
}
max = Math.max(x,max);
sum+=x;
}
System.out.println(sum - max+max*res);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 15c2cce12812e1c98d252a64094fd61e | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.*;
public class Main{
public static void main(String args[]) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long a[]=new long[n];
long ans1=0;
int k=0;
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
if(a[i]%2==0) {
while(a[i]%2==0) {
a[i]=a[i]/2;
k++;
}
}
}
long max=-1;
int idx=0;
for(int i=0;i<n;i++) {
if(max<a[i]) {
max=a[i];
idx=i;
}
}
while(k!=0) {
a[idx]=a[idx]*2;
k--;
}
long sum=0;
for(long i:a) {
sum+=i;
}
System.out.println(sum);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 0048765d7501f0e99dde7b3e2eb99406 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
public class practice{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
long[] arr=new long[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextLong();
}
long count=1;
for(int i=0;i<n;i++){
while(arr[i]%2==0){
arr[i]=arr[i]/2;
count*=2;
}
}
Arrays.sort(arr);
arr[n-1]=arr[n-1]*count;
long sum=0;
for(int i=0;i<n;i++){
sum+=arr[i];
}
System.out.println(sum);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | bf10cb9d8c991266f8f7cbfcb13a1a83 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
/*
Challenge 1: Pupil 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!
*/
public class Coder {
static StringBuffer str=new StringBuffer();
static int n;
static List<Long> a;
static void solve(){
int tot=0;
for(int i=0;i<n;i++){
while(a.get(i)%2==0){
a.set(i, a.get(i)/2);
tot++;
}
}
Collections.sort(a);
long sum=a.get(n-1) * (long)Math.pow(2, tot);
for(int i=0;i<n-1;i++) sum+=a.get(i);
str.append(sum).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 t = Integer.parseInt(bf.readLine().trim());
while (t-- > 0) {
n=Integer.parseInt(bf.readLine().trim());
String s[]=bf.readLine().trim().split("\\s+");
a=new ArrayList<>();
for(int i=0;i<n;i++) a.add(Long.parseLong(s[i]));
solve();
}
pw.print(str);
pw.flush();
// System.out.print(str);
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 0f29ea527ed24e15afe182511b7edd04 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class C {
public static void main(String[] args) {
new C().run();
}
BufferedReader br;
PrintWriter out;
long mod = (long) (1e9 + 7), inf = (long) (3e18);
class pair {
int F, S;
pair(int f, int s) {
F = f; S = s;
}
}
void solve() {
int t = ni();
while(t-- > 0) {
//TODO:
int n= ni();
long sum=0;
long k=0, max=0;;
for(int i=0; i<n; i++){
long a=nl();
while(a%2==0){
a/=2;
k++;
}
max=Math.max(max, a);
sum+=a;
}
sum=sum-max+(max<<k);
out.println(sum);
}
}
private long pow(long a, long b){
if(b==0){
return 1;
}
long c=pow(a, b/2);
if(b%2==0){
return c*c;
}
else{
return a*c*c;
}
}
// -------- I/O Template -------------
char nc() {
return ns().charAt(0);
}
String nLine() {
try {
return br.readLine();
} catch(IOException e) {
return "-1";
}
}
double nd() {
return Double.parseDouble(ns());
}
long nl() {
return Long.parseLong(ns());
}
int ni() {
return Integer.parseInt(ns());
}
int[] na(int n) {
int a[] = new int[n];
for(int i = 0; i < n; i++) a[i] = ni();
return a;
}
StringTokenizer ip;
String ns() {
if(ip == null || !ip.hasMoreTokens()) {
try {
ip = new StringTokenizer(br.readLine());
if(ip == null || !ip.hasMoreTokens())
ip = new StringTokenizer(br.readLine());
} catch(IOException e) {
throw new InputMismatchException();
}
}
return ip.nextToken();
}
void run() {
try {
if (System.getProperty("ONLINE_JUDGE") == null) {
br = new BufferedReader(new FileReader("/media/ankanchanda/Data/WORKPLACE/DS and CP/Competitive Programming/VSCODE/IO/input.txt"));
out = new PrintWriter("/media/ankanchanda/Data/WORKPLACE/DS and CP/Competitive Programming/VSCODE/IO/output.txt");
} else {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}
} catch (FileNotFoundException e) {
System.out.println(e);
}
solve();
out.flush();
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | db99f1a99ec95735312f91aab7aca151 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class DivideAndMultiply{
public static void main(String[] args) {
FastReader fr = new FastReader();
PrintWriter out = new PrintWriter(System.out);
Scanner sc= new Scanner (System.in);
//Code From Here----
int t =fr.nextInt();
while (t-->0) {
int n=fr.nextInt();
long k=0;
long[] a=new long[n];
long sum=0;
long max=-1;
for (int i = 0; i < a.length; i++) {
a[i]=fr.nextLong();
while(a[i]%2==0)
{
k=k+1;
a[i]=a[i]/2;
}
sum+=a[i];
max=Math.max(max,a[i]);
}
sum-=max;
sum+=max<<k;
out.println(sum);
}
out.flush();
sc.close();
}
//This RadixSort() is for long method
public static long[] radixSort(long[] f){ return radixSort(f, f.length); }
public static long[] radixSort(long[] f, int n)
{
long[] to = new long[n];
{
int[] b = new int[65537];
for(int i = 0;i < n;i++)b[1+(int)(f[i]&0xffff)]++;
for(int i = 1;i <= 65536;i++)b[i]+=b[i-1];
for(int i = 0;i < n;i++)to[b[(int)(f[i]&0xffff)]++] = f[i];
long[] d = f; f = to;to = d;
}
{
int[] b = new int[65537];
for(int i = 0;i < n;i++)b[1+(int)(f[i]>>>16&0xffff)]++;
for(int i = 1;i <= 65536;i++)b[i]+=b[i-1];
for(int i = 0;i < n;i++)to[b[(int)(f[i]>>>16&0xffff)]++] = f[i];
long[] d = f; f = to;to = d;
}
{
int[] b = new int[65537];
for(int i = 0;i < n;i++)b[1+(int)(f[i]>>>32&0xffff)]++;
for(int i = 1;i <= 65536;i++)b[i]+=b[i-1];
for(int i = 0;i < n;i++)to[b[(int)(f[i]>>>32&0xffff)]++] = f[i];
long[] d = f; f = to;to = d;
}
{
int[] b = new int[65537];
for(int i = 0;i < n;i++)b[1+(int)(f[i]>>>48&0xffff)]++;
for(int i = 1;i <= 65536;i++)b[i]+=b[i-1];
for(int i = 0;i < n;i++)to[b[(int)(f[i]>>>48&0xffff)]++] = f[i];
long[] d = f; f = to; to = d;
}
return f;
}
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 lowerBound(long a[], long 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 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) >>> 1L;
if (a[m] <= x) l = m;
else r = m;
}
return l + 1;
}
static int upperBound(long[] arr, int start, int end, long k) {
int N = end;
while (start < end) {
int mid = start + (end - start) / 2;
if (k >= arr[mid]) {
start = mid + 1;
} else {
end = mid;
}
}
if (start < N && arr[start] <= k) {
start++;
}
return start;
}
static long lowerBound(long[] arr, int start, int end, long k) {
int N = end;
while (start < end) {
int mid = start + (end - start) / 2;
if (k <= arr[mid]) {
end = mid;
} else {
start = mid + 1;
}
}
if (start < N && arr[start] < k) {
start++;
}
return start;
}
// For Fast Input ----
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 6adc50c84a5ee8d10012530d154ca48a | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class DivideMultiply {
public static void main(String[] args) throws IOException {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
for (int i = 1; i <= t; i++) {
int n = in.nextInt();
int[] a = new int[n];
int[] numFactors = new int[n];
int[] reducedNum = new int[n];
for (int j = 0; j < n; j++) {
a[j] = in.nextInt();
if (a[j] == 2) {
numFactors[j] = 1;
reducedNum[j] = 1;
}
else if (a[j] == 4) {
numFactors[j] = 2;
reducedNum[j] = 1;
}
else if (a[j] == 6) {
numFactors[j] = 1;
reducedNum[j] = 3;
}
else if (a[j] == 8) {
numFactors[j] = 3;
reducedNum[j] = 1;
}
else if (a[j] == 10) {
numFactors[j] = 1;
reducedNum[j] = 5;
}
else if (a[j] == 12) {
numFactors[j] = 2;
reducedNum[j] = 3;
}
else if (a[j] == 14) {
numFactors[j] = 1;
reducedNum[j] = 7;
}
else if (a[j] == 16) {
numFactors[j] = 4;
reducedNum[j] = 1;
}
}
long sum = 0;
for (int k = 0; k < n; k++) {
long tempSum = 0;
long max = a[k];
for (int z = 0; z < n; z++) {
if (z != k) {
if (a[z] % 2 == 0) {
tempSum += reducedNum[z];
if (numFactors[z] == 1) {
max = max * 2;
}
else if (numFactors[z] == 2) {
max = max * 4;
}
else if (numFactors[z] == 3) {
max = max * 8;
}
else if (numFactors[z] == 4) {
max = max * 16;
}
}
else {
tempSum += a[z];
}
}
}
tempSum += max;
if (tempSum > sum) {
sum = tempSum;
}
}
out.println(sum);
}
out.close();
}
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) {
// noop
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | d44e1381d8fff3e0fe02597884b540d4 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;
public class A {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for (int i = 0; i < T; i++) {
int n = in.nextInt();
long[] arr = new long[n];
// long max = 0;
// int maxInd = 0;
// long secMax = 0;
// int secMaxInd = 0;
for (int j = 0; j < n; j++) {
arr[j] = in.nextLong();
// if(arr[j] > max) {
// secMax = max;
// secMaxInd = maxInd;
// max = arr[j];
// maxInd = j;
// } else if(arr[j] > secMax) {
// secMax = arr[j];
// secMaxInd = j;
// }
}
long count = 0;
for (int j = 0; j < arr.length; j++) {
count = Math.max(count, getSum(arr, arr[j], j));
}
// long count = getSum(Arrays.copyOf(arr, arr.length), max, maxInd);
// long count2 = getSum(arr, secMax, secMaxInd);
System.out.println(count);
}
}
private static long getSum(long[] arr, long max, int maxInd) {
long[] ar = new long[arr.length];
for (int i = 0; i < arr.length; i++) {
if(i == maxInd) {
continue;
}
long val = arr[i];
while(val % 2 == 0) {
val /= 2;
max *= 2;
}
ar[i] = val;
}
long sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += ar[i];
}
return sum + max;
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 8e05cf9ee73d8dc31cbdf0ac0b52817d | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.io.*;
import java.util.*;
public class A {
public static void main(String[] hi) {
FastIO io = new FastIO();
int t = io.nextInt();
while (t --> 0) {
int n = io.nextInt();
long[] array = new long[n];
long[] array2 = new long[n];
long[] array3 = new long[n];
int index = -1;
for (int i=0; i<n; i++) {
array[i] = io.nextLong();
array2[i] = array[i];
}
int sum = 0, sum2 = 0;
for (int i=0; i<n; i++) {
while (array2[i] % 2 != 1) {
array2[i] = array2[i]/2;
array3[i]++;
}
sum += array2[i];
sum2 += array3[i];
}
long answer = 0;
for (int i=0; i<n; i++) {
long temp = array[i];
for (int k=0; k<sum2-array3[i]; k++) {
temp = temp*2;
}
answer = Math.max(temp+(sum-array2[i]), answer);
}
io.println(answer);
}
io.close();
}
static class FastIO extends PrintWriter {
BufferedReader br;
StringTokenizer st;
public FastIO() {
super(System.out);
br = new BufferedReader(new InputStreamReader(System.in));
}
public String next() {
try {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(br.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());
}
public void close() {
try {
br.close();
}
catch (IOException e) {
e.printStackTrace();
}
super.close();
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | a260e03a401b862f86df03e8b5514413 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
import java.util.Map.Entry;
public class Codechef {
static ArrayList<Integer> adj[];
// static long ans = 0;
static long above = 0;
static List<Integer> primes = new ArrayList<>();
static FastReader f = new FastReader();
static int mod = 1000000007;
public static void solve() {
int n = f.nextInt();
int a[] = new int[n];
int count2sPow = 0;
long max = 1;
for(int i =0 ; i < n ; i++) {
a[i] = f.nextInt();
int temp = countOf2(a[i]);
count2sPow += temp;
a[i] = (a[i] / (1<<temp));
}
for(int i =0 ; i < n ; i++) {
max = Math.max(a[i], max);
}
long ans = 1;
for(int i = 1 ; i <= count2sPow ; i++)
ans *=2;
ans *= max;
for(int i = 0 ; i < n ; i++) {
if(max == a[i]) {
max = -1;
}
else
ans += a[i];
}
System.out.println(ans);
}
static int countOf2(long n) {
int i = 0;
while(n%2 == 0) {
i++;
n/=2;
}
return i;
}
static int nPowerOf2(long n) {
return (int) (Math.log(n)/Math.log(2));
}
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 int highestPowerof2(int n)
{
int res = 0;
for(int i = n; i >= 1; i--)
{
// If i is a power of 2
if ((i & (i-1)) == 0)
{
res = i;
break;
}
}
return res;
}
public static void main(String[] args) throws IOException {
// FastReader f = new FastReader();
int t = f.nextInt();
// int t = 1;
for (int i = 0; i < t; i++) {
solve();
}
}
static List<Integer> fibonacci(int n){
List<Integer> list = new ArrayList<>();
list.add(0);
list.add(1);
for(int i = 2 ; i < n ; i++) {
list.add(list.get(i - 1) + list.get(i - 2));
}
return list;
}
static int nCr(int n, int r) {
return fact(n) / (fact(r) * fact(n - r));
}
// Returns factorial of n
static int fact(int n) {
int res = 1;
for (int i = 2; i <= n; i++)
res = res * i;
return res;
}
public static Map<Integer, Integer> primeFactors(int n) {
// Print the number of 2s that divide n
Map<Integer, Integer> map = new HashMap<>();
int count = 0;
while (n % 2 == 0) {
// System.out.print(2 + " ");
count++;
n /= 2;
}
if (count > 0)
map.put(2, count);
// 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
count = 0;
while (n % i == 0) {
// System.out.print(i + " ");
count++;
n /= i;
}
if (count > 0)
map.put(i, count);
}
// This condition is to handle the case when
// n is a prime number greater than 2
if (n > 2)
map.put(n, map.getOrDefault(n, 0) + 1);
return map;
}
static List<Integer> sieveOfEratosthenes(int n) {
List<Integer> list = new ArrayList();
boolean 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;
}
}
// Print all prime numbers
for (int i = 2; i <= n; i++) {
if (prime[i] == true) {
list.add(i);
}
}
return list;
}
public static int mex(List<Integer> a) {
int prev = 0;
int n = a.size();
for (int i = 0; i < n; i++) {
int curr = a.get(i);
if (curr == prev || curr == prev + 1) {
prev = curr;
} else {
return curr;
}
}
return a.get(n - 1) + 1;
}
//
static List<Integer> getDivisors(long n) {
List<Integer> ans = new ArrayList();
for (int i = 1; i * i <= n; i++) {
if (n % i == 0) {
ans.add(i);
ans.add((int) (n / i));
}
}
return ans;
}
// static void allFactors(int n) {
// for (int i = 1; i <= Math.sqrt(n); i++) {
// if (n % i == 0) {
// if (n / i == i)
// factors.add(i); // divisors are equal then add only one of them.
//
// else {
// factors.add(i); // divisors are different thus add i, and n/i because i * n/i
// = n
// factors.add(n / i);
//
// }
//
// }
// }
// }
public static boolean isLucky(int n) {
boolean flag = true;
int nn = n;
while (n > 0) {
if (n % 10 != 4 && n % 10 != 7) {
flag = false;
break;
}
n /= 10;
}
return flag;
}
public static boolean helper(long arr[], int n, long target) {
int i = 0, start = i + 1, end = n - 1;
while (start < end) {
long val = arr[i] + arr[start] + arr[end];
if (val < target) {
start++;
} else if (val > target) {
end--;
} else {
return true;
}
}
return false;
}
static boolean checkPerfectSquare(double number) {
// calculating the square root of the given number
double sqrt = Math.sqrt(number);
// finds the floor value of the square root and comparing it with zero
return ((sqrt - Math.floor(sqrt)) == 0);
}
public static long smallestDivisor(long n) {
// if divisible by 2
if (n % 2 == 0)
return 2;
// iterate from 3 to sqrt(n)
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return i;
}
return n;
}
public 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 (long i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static long factorize(long n) {
long ans = 0;
// count the number of times 2 divides
while (!(n % 2 > 0)) {
// equivalent to n=n/2;
n >>= 1;
ans++;
}
if (ans > 0)
ans = 1;
// if 2 divides it
// if (count > 0) {
// System.out.println("2" + " " + count);
// }
// check for all the possible
// numbers that can divide it
for (long i = 3; i <= (long) Math.sqrt(n); i += 2) {
int count = 0;
while (n % i == 0) {
count++;
n = n / i;
}
// if (count > 0) {
// System.out.println(i + " " + count);
// }
ans += count;
}
// if n at the end is a prime number.
if (n > 2) {
ans++;
}
return ans;
}
static int countSetBits(long n) {
int count = 0;
while (n > 0) {
n &= (n - 1);
count++;
}
return count;
}
public static long gcd(long a, long b) {
if (a == 0)
// returns y is x==0
return b;
// calling function that returns GCD
return gcd(b % a, a);
}
static long lcm(long a, long b) {
// returns the LCM
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | e613f108eed189a62be2e3cf0caa0794 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class DivideAndMultiply {
/*
* Dipraj Howlader
* [email protected]
*/
public static void main(String[] args) {
Scanner iScanner = new Scanner(System.in);
int test = iScanner.nextInt();
while (test > 0) {
test--;
int n = iScanner.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = iScanner.nextInt();
Arrays.sort(arr);
long countK = 0;
long temp = 0;
long highValue = 0;
for(int i=0;i<n;i++){
while(arr[i]%2==0){
countK++;
arr[i]/=2;
}
highValue+=arr[i];
temp = Long.max(temp, arr[i]);
}
long res = temp*(long) Math.pow(2, countK)+highValue-temp;
System.out.println(res);
}
iScanner.close();
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 2af7065001343fcd4792bbbaac2c5066 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
for (int k = 0; k < t; k++) {
int n = scn.nextInt();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = scn.nextLong();
}
long mult = 1;
for (int i = 0; i < n; i++) {
while(arr[i] %2 == 0){
mult *= 2;
arr[i] /= 2;
}
}
Arrays.sort(arr);
arr[n - 1] *= mult;
long ans = 0;
for (long ele : arr) {
ans = ans + ele;
}
System.out.println(ans);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | dd647208ebb0913512f6678fd7aa6331 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;
public class DivideAndMultiply {
public static Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int tc = scanner.nextInt();
while (tc-->0){
int size = scanner.nextInt();
BigInteger[] array = new BigInteger[size];
int mul =0;
for (int i=0;i< array.length;i++){
array[i]= scanner.nextBigInteger();
while (array[i].remainder(BigInteger.valueOf(2)).equals(BigInteger.valueOf(0))){
mul++;
array[i]=array[i].divide(BigInteger.valueOf(2));
}
}
Arrays.sort(array);
for (int i =0;i<mul;i++){
array[array.length-1]=array[array.length-1].multiply(BigInteger.valueOf(2));
}
BigInteger ans = BigInteger.ZERO;
for (int i =0;i< array.length;i++){
ans = ans.add((array[i]));
}
System.out.println(ans);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 2fd97feab71e77de8cf8a07d36038404 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Pratice{
final long mod = 1000000007;
static StringBuilder sb=new StringBuilder();
static boolean ans;
static int xn = (int) (2e5 + 10);
public static void main(String[] args) throws IOException {
Reader reader = new Reader();
int t = reader.nextInt();
while (t-- > 0) {
int n=reader.nextInt();
int arr[]=new int[n];
for (int i = 0; i < n; i++) {
arr[i]=reader.nextInt();
}
long ans=1;
int k=0;
for(int i=0;i<n;i++)
{
if(arr[i]%2==0) {
while (arr[i]%2 == 0) {
k++;
arr[i] = arr[i] / 2;
}
}
}
// System.out.println(k);
Arrays.sort(arr);
for (int i = 1; i <= k; i++) {
ans*=2;
}
ans*=arr[arr.length-1];
for(int i=0;i<n-1;i++)
{
ans=ans+arr[i];
}
System.out.println(ans);
}
}
// check number is prime or not
public static boolean isPrime(int n) {
return BigInteger.valueOf(n).isProbablePrime(1);
}
// return the gcd of two numbers
public static long gcd(long a, long b)
{
BigInteger b1 = BigInteger.valueOf(a);
BigInteger b2 = BigInteger.valueOf(b);
BigInteger gcd = b1.gcd(b2);
return gcd.longValue();
}
// number of digits in the given number
public static int numberOfDigits(int n)
{
int ans= (int) (Math.floor((Math.log10(n)))+1);
return ans;
}
// return most significant bit in the number
public static long mostSignificantNumber(long n)
{
double k=Math.log10(n);
k=k-Math.floor(k);
int ans=(int)Math.pow(10,k);
return ans;
}
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();
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 7de22b6d8a2d8a78f94842306bc9eeef | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
// A -> CodeForcesProblemSet
public final class A {
static FastReader fr = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static final int gigamod = 1000000007;
static int t = 1;
static double epsilon = 0.0000001;
public static void main(String[] args) {
t = fr.nextInt();
OUTER:
for (int tc = 0; tc < t; tc++) {
int n = fr.nextInt();
long[] arr = fr.nextLongArray(n);
reverseSort(arr);
long mul = 1;
for (int i = 0; i < n; i++) {
// We will remove as much 2 as possible from each element.
while (arr[i] % 2 == 0) {
arr[i] /= 2;
mul *= 2;
}
}
// We'll give mul to largest number that is not a multiple of 2.
reverseSort(arr);
arr[0] *= mul;
long sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
out.println(sum);
}
out.close();
}
// Manual implementation of RB-BST for C++ PBDS functionality.
// Modification required according to requirements.
public static final class RedBlackCountSet {
// Colors for Nodes:
private static final boolean RED = true;
private static final boolean BLACK = false;
// Pointer to the root:
private Node root;
// Public constructor:
public RedBlackCountSet() {
root = null;
}
// Node class for storing key value pairs:
private class Node {
long key;
long value;
Node left, right;
boolean color;
long size; // Size of the subtree rooted at that node.
public Node(long key, long value, boolean color) {
this.key = key;
this.value = value;
this.left = this.right = null;
this.color = color;
this.size = value;
}
}
/***** Invariant Maintenance functions: *****/
// When a temporary 4 node is formed:
private Node flipColors(Node node) {
node.left.color = node.right.color = BLACK;
node.color = RED;
return node;
}
// When there's a right leaning red link and it is not a 3 node:
private Node rotateLeft(Node node) {
Node rn = node.right;
node.right = rn.left;
rn.left = node;
rn.color = node.color;
node.color = RED;
return rn;
}
// When there are 2 red links in a row:
private Node rotateRight(Node node) {
Node ln = node.left;
node.left = ln.right;
ln.right = node;
ln.color = node.color;
node.color = RED;
return ln;
}
/***** Invariant Maintenance functions end *****/
// Public functions:
public void put(long key) {
root = put(root, key);
root.color = BLACK; // One of the invariants.
}
private Node put(Node node, long key) {
// Standard insertion procedure:
if (node == null)
return new Node(key, 1, RED);
// Recursing:
int cmp = Long.compare(key, node.key);
if (cmp < 0)
node.left = put(node.left, key);
else if (cmp > 0)
node.right = put(node.right, key);
else
node.value++;
// Invariant maintenance:
if (node.left != null && node.right != null) {
if (node.right.color = RED && node.left.color == BLACK)
node = rotateLeft(node);
if (node.left.color == RED && node.left.left.color == RED)
node = rotateRight(node);
if (node.left.color == RED && node.right.color == RED)
node = flipColors(node);
}
// Property maintenance:
node.size = node.value + size(node.left) + size(node.right);
return node;
}
private long size(Node node) {
if (node == null)
return 0;
else
return node.size;
}
public long rank(long key) {
return rank(root, key);
}
// Modify according to requirements.
private long rank(Node node, long key) {
if (node == null) return 0;
int cmp = Long.compare(key, node.key);
if (cmp < 0)
return rank(node.left, key);
else if (cmp > 0)
return node.value + size(node.left) + rank(node.right, key);
else
return size(node.left);
}
}
static long modDiv(long a, long b) {
return mod(a * power(b, gigamod - 2));
}
static class SegTree {
// Setting up the parameters.
int n;
long[] sumTree;
long[] ogArr;
long[] maxTree;
long[] minTree;
public static final int sumMode = 0;
public static final int maxMode = 1;
public static final int minMode = 2;
// Constructor
public SegTree(long[] arr, int mode) {
if (mode == sumMode) {
Arrays.fill(sumTree = new long[4 * (n = ((ogArr = arr.clone()).length))], -1);
buildSum(0, n - 1, 0);
} else if (mode == maxMode) {
Arrays.fill(maxTree = new long[4 * (n = ((ogArr = arr.clone()).length))], -1);
buildMax(0, n - 1, 0);
} else if (mode == minMode) {
Arrays.fill(minTree = new long[4 * (n = ((ogArr = arr.clone()).length))], Long.MAX_VALUE - 1);
buildMin(0, n - 1, 0);
}
}
/**********Code for sum***********/
/*********************************/
// Build the segment tree node-by-node
private long buildSum(int l, int r, int segIdx) {
return sumTree[segIdx] = (l == r) ? ogArr[l] :
(buildSum(l, l + (r - l) / 2, 2 * segIdx + 1)
+ buildSum((l + (r - l) / 2) + 1, r, 2 * segIdx + 2));
}
// Change a single element
public void changeForSum(int idx, long to) {
changeForSum(idx, to, 0, n - 1, 0);
}
private long changeForSum(int idx, long to, int l, int r, int segIdx) {
return sumTree[segIdx] =
((l == r) ? (ogArr[idx] = to) :
((0 * ((idx <= (l + (r - l) / 2))
? changeForSum(idx, to, l, l + (r - l) / 2, 2 * segIdx + 1)
: changeForSum(idx, to, (l + (r - l) / 2) + 1, r, 2 * segIdx + 2)))
+ sumTree[2 * segIdx + 1] + sumTree[2 * segIdx + 2]));
}
// Return the sum arr[l, r]
public long segSum(int l, int r) {
if (l > r) return 0;
return segSum(l, r, 0, n - 1, 0);
}
private long segSum(int segL, int segR, int l, int r, int segIdx) {
if (segL == l && segR == r)
return sumTree[segIdx];
if (segR <= l + (r - l) / 2)
return segSum(segL, segR, l, l + (r - l) / 2, 2 * segIdx + 1);
if (segL >= l + (r - l) / 2 + 1)
return segSum(segL, segR, l + (r - l) / 2 + 1, r, 2 * segIdx + 2);
return segSum(segL, l + (r - l) / 2, l, l + (r - l) / 2, 2 * segIdx + 1)
+ segSum(l + (r - l) / 2 + 1, segR, l + (r - l) / 2 + 1, r, 2 * segIdx + 2);
}
/********End of code for sum********/
/***********************************/
/*******Start of code for max*******/
/***********************************/
// Build the max tree node-by-node
private long buildMax(int l, int r, int segIdx) {
return maxTree[segIdx] = (l == r) ? ogArr[l] :
Math.max(buildMax(l, (l + (r - l) / 2), 2 * segIdx + 1),
buildMax(l + (r - l) / 2 + 1, r, 2 * segIdx + 2));
}
// Change a single element
public void changeForMax(int idx, long to) {
changeForMax(0, n - 1, idx, to, 0);
}
private long changeForMax(int l, int r, int idx, long to, int segIdx) {
return maxTree[segIdx] = (l == r && l == idx) ? (ogArr[idx] = to) :
Math.max(changeForMax(l, l + (r - l) / 2, idx, to, 2 * segIdx + 1),
changeForMax(l + (r - l) / 2 + 1, r, idx, to, 2 * segIdx + 2));
}
public long segMax(int segL, int segR) {
if (segL > segR) return 0;
return segMax(0, n - 1, segL, segR, 0);
}
private long segMax(int l, int r, int segL, int segR, int segIdx) {
int midL = l + (r - l) / 2;
if (segL == segR && segL == l)
return ogArr[segL];
if (segR <= midL)
return segMax(l, midL, segL, segR, 2 * segIdx + 1);
if (segL >= midL + 1)
return segMax(midL + 1, r, segL, segR, 2 * segIdx + 2);
return Math.max(segMax(l, midL, segL, midL, 2 * segIdx + 1),
segMax(midL + 1, r, midL + 1, segR, 2 * segIdx + 2));
}
/*******End of code for max*********/
/***********************************/
/*******Start of code for min*******/
/***********************************/
private long buildMin(int l, int r, int segIdx) {
return minTree[segIdx] = (l == r) ? ogArr[l] :
Math.min(buildMin(l, (l + (r - l) / 2), 2 * segIdx + 1),
buildMin(l + (r - l) / 2 + 1, r, 2 * segIdx + 2));
}
// Change a single element
public void changeForMin(int idx, long to) {
changeForMin(0, n - 1, idx, to, 0);
}
private long changeForMin(int l, int r, int idx, long to, int segIdx) {
return minTree[segIdx] = (l == r && l == idx) ? (ogArr[idx] = to) :
Math.min(changeForMin(l, l + (r - l) / 2, idx, to, 2 * segIdx + 1),
changeForMin(l + (r - l) / 2 + 1, r, idx, to, 2 * segIdx + 2));
}
public long segMin(int segL, int segR) {
if (segL > segR) return 0;
return segMin(0, n - 1, segL, segR, 0);
}
private long segMin(int l, int r, int segL, int segR, int segIdx) {
int midL = l + (r - l) / 2;
if (segL == segR && segL == l)
return ogArr[segL];
if (segR <= midL)
return segMin(l, midL, segL, segR, 2 * segIdx + 1);
if (segL >= midL + 1)
return segMin(midL + 1, r, segL, segR, 2 * segIdx + 2);
return Math.min(segMin(l, midL, segL, midL, 2 * segIdx + 1),
segMin(midL + 1, r, midL + 1, segR, 2 * segIdx + 2));
}
/*******End of code for min*********/
/***********************************/
public String toString() {
return A.toString(minTree);
}
}
static void coprimeGenerator(int m, int n, ArrayList<Point> coprimes, int limit, int numCoprimes) {
if (m > limit) return;
if (m <= limit && n <= limit)
coprimes.add(new Point(m, n));
if (coprimes.size() > numCoprimes) return;
coprimeGenerator(2 * m - n, m, coprimes, limit, numCoprimes);
coprimeGenerator(2 * m + n, m, coprimes, limit, numCoprimes);
coprimeGenerator(m + 2 * n, n, coprimes, limit, numCoprimes);
}
// Returns the length of the longest prefix that is also
// a suffix (no overlap).
static int longestPrefixSuffix(String s) {
int n = s.length();
int[] arr = new int[n];
arr[0] = 0;
int len = 0;
for (int i = 1; i < n;) {
if (s.charAt(len) == s.charAt(i)) {
len++;
arr[i++] = len;
} else {
if (len != 0) {
len = arr[len - 1];
} else {
arr[i] = 0;
i++;
}
}
}
return arr[n - 1];
}
static long hash(long i) {
return (i * 2654435761L % gigamod);
}
static long hash2(long key) {
long h = Long.hashCode(key);
h ^= (h >>> 20) ^ (h >>> 12) ^ (h >>> 7) ^ (h >>> 4);
return h & (gigamod-1);
}
static long power(long x, long y)
{
// int p = 998244353;
int p = gigamod;
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
static long power2(long x, long y)
{
// int p = 998244353;
// int p = gigamod;
long res = 1; // Initialize result
// x = x /*% p*/; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) /*% p*/;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) /*% p*/;
}
return res;
}
// Maps elements in a 2D matrix serially to elements in
// a 1D array.
static int mapTo1D(int row, int col, int n, int m) {
return row * m + col;
}
// Inverse of what the one above does.
static int[] mapTo2D(int idx, int n, int m) {
int[] rnc = new int[2];
rnc[0] = idx / m;
rnc[1] = idx % m;
return rnc;
}
// Checks if s has subsequence t.
static boolean hasSubsequence(String s, String t) {
char[] schars = s.toCharArray();
char[] tchars = t.toCharArray();
int slen = schars.length, tlen = tchars.length;
int tctr = 0;
if (slen < tlen) return false;
for (int i = 0; i < slen || i < tlen; i++) {
if (tctr == tlen) break;
if (schars[i] == tchars[tctr]) {
tctr++;
}
}
if (tctr == tlen) return true;
return false;
}
// Returns the binary string of length at least bits.
static String toBinaryString(long num, int bits) {
StringBuilder sb = new StringBuilder(Long.toBinaryString(num));
sb.reverse();
for (int i = sb.length(); i < bits; i++)
sb.append('0');
return sb.reverse().toString();
}
@SuppressWarnings("serial")
static class CountMap<T> extends TreeMap<T, Integer>{
CountMap() {
}
CountMap(T[] arr) {
this.putCM(arr);
}
public Integer putCM(T key) {
if (super.containsKey(key)) {
return super.put(key, super.get(key) + 1);
} else {
return super.put(key, 1);
}
}
public Integer removeCM(T key) {
Integer count = super.get(key);
if (count == null) return -1;
if (count == 1)
return super.remove(key);
else
return super.put(key, super.get(key) - 1);
}
public Integer getCM(T key) {
Integer count = super.get(key);
if (count == null)
return 0;
return count;
}
public void putCM(T[] arr) {
for (T l : arr)
this.putCM(l);
}
}
static class Point implements Comparable<Point> {
long x;
long y;
int id;
Point() {
x = y = id = 0;
}
Point(Point p) {
this.x = p.x;
this.y = p.y;
this.id = p.id;
}
/*Point(double a, double b) {
x = a;
y = b;
}*/
Point(long a, long b, int id) {
this.x = a;
this.y = b;
this.id = id;
}
Point(long a, long b) {
this.x = a;
this.y = b;
}
@Override
public int compareTo(Point o) {
if (this.x > o.x)
return 1;
if (this.x < o.x)
return -1;
if (this.y > o.y)
return 1;
if (this.y < o.y)
return -1;
return 0;
}
public boolean equals(Point that) {
return this.compareTo(that) == 0;
}
}
static class PointComparator implements Comparator<Point> {
@Override
public int compare(Point o1, Point o2) {
// Comparision will be done on the basis of the start
// of the segment and then on the length of the segment.
if (o1.id > o2.id)
return -1;
if (o1.id < o2.id)
return 1;
return 0;
}
}
static double distToOrigin(Point p1) {
return Math.sqrt(Math.pow(p1.y, 2) + Math.pow(p1.x, 2));
}
static double distance(Point p1, Point p2) {
double y2 = p2.y, y1 = p1.y;
double x2 = p2.x, x1 = p1.x;
double distance = Math.sqrt(Math.pow(y2-y1, 2) + Math.pow(x2-x1, 2));
return distance;
}
// Returns the largest power of k that fits into n.
static int largestFittingPower(long n, long k) {
int lo = 0, hi = logk(Long.MAX_VALUE, 3);
int largestPower = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
long val = (long) Math.pow(k, mid);
if (val <= n) {
largestPower = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return largestPower;
}
// Returns map of factor and its power in the number.
static HashMap<Long, Integer> primeFactorization(long num) {
HashMap<Long, Integer> map = new HashMap<>();
while (num % 2 == 0) {
num /= 2;
Integer pwrCnt = map.get(2L);
map.put(2L, pwrCnt != null ? pwrCnt + 1 : 1);
}
for (long i = 3; i * i <= num; i += 2) {
while (num % i == 0) {
num /= i;
Integer pwrCnt = map.get(i);
map.put(i, pwrCnt != null ? pwrCnt + 1 : 1);
}
}
// If the number is prime, we have to add it to the
// map.
if (num != 1)
map.put(num, 1);
return map;
}
// Returns map of factor and its power in the number.
static HashMap<Integer, Integer> primeFactorization(int num) {
HashMap<Integer, Integer> map = new HashMap<>();
while (num % 2 == 0) {
num /= 2;
Integer pwrCnt = map.get(2);
map.put(2, pwrCnt != null ? pwrCnt + 1 : 1);
}
for (int i = 3; i * i <= num; i += 2) {
while (num % i == 0) {
num /= i;
Integer pwrCnt = map.get(i);
map.put(i, pwrCnt != null ? pwrCnt + 1 : 1);
}
}
// If the number is prime, we have to add it to the
// map.
if (num != 1)
map.put(num, 1);
return map;
}
static HashSet<Long> divisors(long num) {
HashSet<Long> divisors = new HashSet<Long>();
divisors.add(1L);
divisors.add(num);
for (long i = 2; i * i <= num; i++) {
if (num % i == 0) {
divisors.add(num/i);
divisors.add(i);
}
}
return divisors;
}
// Returns the index of the first element
// larger than or equal to val.
static int bsearch(int[] arr, int val, int lo, int hi) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
idx = mid;
hi = mid - 1;
} else
lo = mid + 1;
}
return idx;
}
static int bsearch(long[] arr, long val, int lo, int hi) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] >= val) {
idx = mid;
hi = mid - 1;
} else
lo = mid + 1;
}
return idx;
}
// Returns the index of the last element
// smaller than or equal to val.
static int bsearch(long[] arr, long val, int lo, int hi, boolean sMode) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] > val) {
hi = mid - 1;
} else {
idx = mid;
lo = mid + 1;
}
}
return idx;
}
static int bsearch(int[] arr, long val, int lo, int hi, boolean sMode) {
int idx = -1;
while (lo <= hi) {
int mid = lo + (hi - lo)/2;
if (arr[mid] > val) {
hi = mid - 1;
} else {
idx = mid;
lo = mid + 1;
}
}
return idx;
}
static long factorial(long n) {
if (n <= 1)
return 1;
long factorial = 1;
for (int i = 1; i <= n; i++)
factorial = mod(factorial * i);
return factorial;
}
static long factorialInDivision(long a, long b) {
if (a == b)
return 1;
if (b < a) {
long temp = a;
a = b;
b = temp;
}
long factorial = 1;
for (long i = a + 1; i <= b; i++)
factorial = mod(factorial * i);
return factorial;
}
static BigInteger factorialInDivision(BigInteger a, BigInteger b) {
if (a.equals(b))
return BigInteger.ONE;
return a.multiply(factorialInDivision(a.subtract(BigInteger.ONE), b));
}
static long nCr(long n, long r) {
long p = gigamod;
// Base case
if (r == 0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
long fac[] = new long[(int)n + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = fac[i - 1] * i % p;
return (fac[(int)n] * modInverse(fac[(int)r], p) % p
* modInverse(fac[(int)n - (int)r], p) % p) % p;
}
static long modInverse(long n, long p) {
return power(n, p - 2, p);
}
static long power(long x, long y, long p) {
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if ((y & 1)==1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
static long nPr(long n, long r) {
return factorialInDivision(n, n - r);
}
static int log2(long n) {
return (int)(Math.log(n) / Math.log(2));
}
static double log2(long n, boolean doubleMode) {
return (Math.log(n) / Math.log(2));
}
static int logk(long n, long k) {
return (int)(Math.log(n) / Math.log(k));
}
// Sieve of Eratosthenes:
static boolean[] primeGenerator(int upto) {
boolean[] isPrime = new boolean[upto + 1];
Arrays.fill(isPrime, true);
isPrime[1] = isPrime[0] = false;
for (long i = 2; i * i < upto + 1; i++)
if (isPrime[(int) i])
// Mark all the multiples greater than or equal
// to the square of i to be false.
for (long j = i; j * i < upto + 1; j++)
isPrime[(int) j * (int) i] = false;
return isPrime;
}
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 int gcd(int[] arr) {
int n = arr.length;
int gcd = arr[0];
for (int i = 1; i < n; i++) {
gcd = gcd(gcd, arr[i]);
}
return gcd;
}
static long gcd(long[] arr) {
int n = arr.length;
long gcd = arr[0];
for (int i = 1; i < n; i++) {
gcd = gcd(gcd, arr[i]);
}
return gcd;
}
static long lcm(int[] arr) {
long lcm = arr[0];
int n = arr.length;
for (int i = 1; i < n; i++) {
lcm = (lcm * arr[i]) / gcd(lcm, arr[i]);
}
return lcm;
}
static long lcm(long[] arr) {
long lcm = arr[0];
int n = arr.length;
for (int i = 1; i < n; i++) {
lcm = (lcm * arr[i]) / gcd(lcm, arr[i]);
}
return lcm;
}
static long lcm(int a, int b) {
return (a * b)/gcd(a, b);
}
static long lcm(long a, long b) {
return (a * b)/gcd(a, b);
}
static boolean less(int a, int b) {
return a < b ? true : false;
}
static boolean isSorted(int[] a) {
for (int i = 1; i < a.length; i++) {
if (less(a[i], a[i - 1]))
return false;
}
return true;
}
static boolean isSorted(long[] a) {
for (int i = 1; i < a.length; i++) {
if (a[i] < a[i - 1])
return false;
}
return true;
}
static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(long[] a, int i, int j) {
long temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(double[] a, int i, int j) {
double temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void sort(int[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(char[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(long[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void sort(double[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
}
static void reverseSort(int[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(char[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(long[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void reverseSort(double[] arr) {
shuffleArray(arr, 0, arr.length - 1);
Arrays.sort(arr);
int n = arr.length;
for (int i = 0; i < n/2; i++)
swap(arr, i, n - 1 - i);
}
static void shuffleArray(long[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
long tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(int[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
int tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
static void shuffleArray(double[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
double tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
private static void shuffleArray(char[] arr, int startPos, int endPos) {
Random rnd = new Random();
for (int i = startPos; i < endPos; ++i) {
char tmp = arr[i];
int randomPos = i + rnd.nextInt(endPos - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
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 (long i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static String toString(int[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(boolean[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(long[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + " ");
return sb.toString();
}
static String toString(char[] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++)
sb.append(dp[i] + "");
return sb.toString();
}
static String toString(int[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(long[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(double[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static String toString(char[][] dp) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < dp.length; i++) {
for (int j = 0; j < dp[i].length; j++) {
sb.append(dp[i][j] + " ");
}
sb.append('\n');
}
return sb.toString();
}
static char toChar(int i) {
return (char) (i + 48);
}
static long mod(long a, long m) {
return (a%m + m) % m;
}
static long mod(long num) {
return (num % gigamod + gigamod) % gigamod;
}
// Uses weighted quick-union with path compression.
static class UnionFind {
private int[] parent; // parent[i] = parent of i
private int[] size; // size[i] = number of sites in tree rooted at i
// Note: not necessarily correct if i is not a root node
private int count; // number of components
public UnionFind(int n) {
count = n;
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
// Number of connected components.
public int count() {
return count;
}
// Find the root of p.
public int find(int p) {
int root = p;
while (root != parent[root])
root = parent[root];
while (p != root) {
int newp = parent[p];
parent[p] = root;
p = newp;
}
return root;
}
public boolean connected(int p, int q) {
return find(p) == find(q);
}
public int numConnectedTo(int node) {
return size[find(node)];
}
// Weighted union.
public void union(int p, int q) {
int rootP = find(p);
int rootQ = find(q);
if (rootP == rootQ) return;
// make smaller root point to larger one
if (size[rootP] < size[rootQ]) {
parent[rootP] = rootQ;
size[rootQ] += size[rootP];
}
else {
parent[rootQ] = rootP;
size[rootP] += size[rootQ];
}
count--;
}
public static int[] connectedComponents(UnionFind uf) {
// We can do this in nlogn.
int n = uf.size.length;
int[] compoColors = new int[n];
for (int i = 0; i < n; i++)
compoColors[i] = uf.find(i);
HashMap<Integer, Integer> oldToNew = new HashMap<>();
int newCtr = 0;
for (int i = 0; i < n; i++) {
int thisOldColor = compoColors[i];
Integer thisNewColor = oldToNew.get(thisOldColor);
if (thisNewColor == null)
thisNewColor = newCtr++;
oldToNew.put(thisOldColor, thisNewColor);
compoColors[i] = thisNewColor;
}
return compoColors;
}
}
static class UGraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
public UGraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
adj[to].add(from);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int degree(int v) {
return adj[v].size();
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static void dfsMark(int current, boolean[] marked, UGraph g) {
if (marked[current]) return;
marked[current] = true;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, marked, g);
}
public static void dfsMark(int current, int from, long[] distTo, boolean[] marked, UGraph g, ArrayList<Integer> endPoints) {
if (marked[current]) return;
marked[current] = true;
if (from != -1)
distTo[current] = distTo[from] + 1;
HashSet<Integer> adj = g.adj(current);
int alreadyMarkedCtr = 0;
for (int adjc : adj) {
if (marked[adjc]) alreadyMarkedCtr++;
dfsMark(adjc, current, distTo, marked, g, endPoints);
}
if (alreadyMarkedCtr == adj.size())
endPoints.add(current);
}
public static void bfsOrder(int current, UGraph g) {
}
public static void dfsMark(int current, int[] colorIds, int color, UGraph g) {
if (colorIds[current] != -1) return;
colorIds[current] = color;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, colorIds, color, g);
}
public static int[] connectedComponents(UGraph g) {
int n = g.V();
int[] componentId = new int[n];
Arrays.fill(componentId, -1);
int colorCtr = 0;
for (int i = 0; i < n; i++) {
if (componentId[i] != -1) continue;
dfsMark(i, componentId, colorCtr, g);
colorCtr++;
}
return componentId;
}
public static boolean hasCycle(UGraph ug) {
int n = ug.V();
boolean[] marked = new boolean[n];
boolean[] hasCycleFirst = new boolean[1];
for (int i = 0; i < n; i++) {
if (marked[i]) continue;
hcDfsMark(i, ug, marked, hasCycleFirst, -1);
}
return hasCycleFirst[0];
}
// Helper for hasCycle.
private static void hcDfsMark(int current, UGraph ug, boolean[] marked, boolean[] hasCycleFirst, int parent) {
if (marked[current]) return;
if (hasCycleFirst[0]) return;
marked[current] = true;
HashSet<Integer> adjc = ug.adj(current);
for (int adj : adjc) {
if (marked[adj] && adj != parent && parent != -1) {
hasCycleFirst[0] = true;
return;
}
hcDfsMark(adj, ug, marked, hasCycleFirst, current);
}
}
}
static class Digraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
public Digraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public Digraph reversed() {
Digraph dg = new Digraph(V());
for (int i = 0; i < V(); i++)
for (int adjVert : adj(i)) dg.addEdge(adjVert, i);
return dg;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static void dfsMark(int source, boolean[] marked, Digraph g) {
if (marked[source]) return;
marked[source] = true;
Iterable<Integer> adj = g.adj(source);
for (int adjc : adj)
dfsMark(adjc, marked, g);
}
public static void bfsOrder(int source, Digraph g) {
}
public static int[] KosarajuSharirSCC(Digraph dg) {
int[] id = new int[dg.V()];
Digraph reversed = dg.reversed();
// Gotta perform topological sort on this one to get the stack.
Stack<Integer> revStack = Digraph.topologicalSort(reversed);
// Initializing id and idCtr.
id = new int[dg.V()];
int idCtr = -1;
// Creating a 'marked' array.
boolean[] marked = new boolean[dg.V()];
while (!revStack.isEmpty()) {
int vertex = revStack.pop();
if (!marked[vertex])
sccDFS(dg, vertex, marked, ++idCtr, id);
}
return id;
}
private static void sccDFS(Digraph dg, int source, boolean[] marked, int idCtr, int[] id) {
marked[source] = true;
id[source] = idCtr;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex]) sccDFS(dg, adjVertex, marked, idCtr, id);
}
public static Stack<Integer> topologicalSort(Digraph dg) {
// dg has to be a directed acyclic graph.
// We'll have to run dfs on the digraph and push the deepest nodes on stack first.
// We'll need a Stack<Integer> and a int[] marked.
Stack<Integer> topologicalStack = new Stack<Integer>();
boolean[] marked = new boolean[dg.V()];
// Calling dfs
for (int i = 0; i < dg.V(); i++)
if (!marked[i])
runDfs(dg, topologicalStack, marked, i);
return topologicalStack;
}
static void runDfs(Digraph dg, Stack<Integer> topologicalStack, boolean[] marked, int source) {
marked[source] = true;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex])
runDfs(dg, topologicalStack, marked, adjVertex);
topologicalStack.add(source);
}
public static boolean hasCycle(Digraph dg) {
int n = dg.V();
boolean[] marked = new boolean[n];
boolean[] hasCycleFirst = new boolean[1];
for (int i = 0; i < n; i++) {
if (marked[i]) continue;
hcDfsMark(i, dg, marked, hasCycleFirst, -1);
}
return hasCycleFirst[0];
}
// Helper for hasCycle.
private static void hcDfsMark(int current, Digraph dg, boolean[] marked, boolean[] hasCycleFirst, int parent) {
if (marked[current]) return;
if (hasCycleFirst[0]) return;
marked[current] = true;
HashSet<Integer> adjc = dg.adj(current);
for (int adj : adjc) {
if (marked[adj] && adj != parent && parent != -1) {
hasCycleFirst[0] = true;
return;
}
hcDfsMark(adj, dg, marked, hasCycleFirst, current);
}
}
}
static class Edge {
int from;
int to;
long weight;
public Edge(int from, int to, long weight) {
this.from = from;
this.to = to;
this.weight = weight;
}
}
static class WeightedUGraph {
// Adjacency list.
private HashSet<Edge>[] adj;
private static final String NEWLINE = "\n";
private int E;
public WeightedUGraph(int V) {
adj = (HashSet<Edge>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Edge>();
}
public void addEdge(int from, int to, int weight) {
if (adj[from].contains(new Edge(from, to, weight))) return;
E++;
adj[from].add(new Edge(from, to, weight));
adj[to].add(new Edge(to, from, weight));
}
public HashSet<Edge> adj(int from) {
return adj[from];
}
public int degree(int v) {
return adj[v].size();
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (Edge w : adj[v]) {
s.append(w.toString() + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static void dfsMark(int current, boolean[] marked, UGraph g) {
if (marked[current]) return;
marked[current] = true;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, marked, g);
}
public static void dfsMark(int current, int from, long[] distTo, boolean[] marked, UGraph g, ArrayList<Integer> endPoints) {
if (marked[current]) return;
marked[current] = true;
if (from != -1)
distTo[current] = distTo[from] + 1;
HashSet<Integer> adj = g.adj(current);
int alreadyMarkedCtr = 0;
for (int adjc : adj) {
if (marked[adjc]) alreadyMarkedCtr++;
dfsMark(adjc, current, distTo, marked, g, endPoints);
}
if (alreadyMarkedCtr == adj.size())
endPoints.add(current);
}
public static void bfsOrder(int current, UGraph g) {
}
public static void dfsMark(int current, int[] colorIds, int color, UGraph g) {
if (colorIds[current] != -1) return;
colorIds[current] = color;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, colorIds, color, g);
}
public static int[] connectedComponents(UGraph g) {
int n = g.V();
int[] componentId = new int[n];
Arrays.fill(componentId, -1);
int colorCtr = 0;
for (int i = 0; i < n; i++) {
if (componentId[i] != -1) continue;
dfsMark(i, componentId, colorCtr, g);
colorCtr++;
}
return componentId;
}
public static boolean hasCycle(UGraph ug) {
int n = ug.V();
boolean[] marked = new boolean[n];
boolean[] hasCycleFirst = new boolean[1];
for (int i = 0; i < n; i++) {
if (marked[i]) continue;
hcDfsMark(i, ug, marked, hasCycleFirst, -1);
}
return hasCycleFirst[0];
}
// Helper for hasCycle.
private static void hcDfsMark(int current, UGraph ug, boolean[] marked, boolean[] hasCycleFirst, int parent) {
if (marked[current]) return;
if (hasCycleFirst[0]) return;
marked[current] = true;
HashSet<Integer> adjc = ug.adj(current);
for (int adj : adjc) {
if (marked[adj] && adj != parent && parent != -1) {
hasCycleFirst[0] = true;
return;
}
hcDfsMark(adj, ug, marked, hasCycleFirst, current);
}
}
}
static class FastReader {
private BufferedReader bfr;
private StringTokenizer st;
public FastReader() {
bfr = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
if (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(bfr.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());
}
char nextChar() {
return next().toCharArray()[0];
}
String nextString() {
return next();
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
double[] nextDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < arr.length; i++)
arr[i] = nextDouble();
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
int[][] nextIntGrid(int n, int m) {
int[][] grid = new int[n][m];
for (int i = 0; i < n; i++) {
char[] line = fr.next().toCharArray();
for (int j = 0; j < m; j++)
grid[i][j] = line[j] - 48;
}
return grid;
}
}
}
// NOTES:
// ASCII VALUE OF 'A': 65
// ASCII VALUE OF 'a': 97
// Range of long: 9 * 10^18
// ASCII VALUE OF '0': 48
// Primes upto 'n' can be given by (n / (logn)). | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 016a523487c90f320be5bfa5aaae74a4 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.math.*;
import java.util.stream.IntStream;
public class Main {
static class Pair implements Comparable<Pair> {
int index;
long value;
Pair(int index, long value) {
this.index = index;
this.value = value;
}
@Override
public int compareTo(Pair o) {
return Long.compare(this.value, o.value);
}
}
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;
}
}
static int ncr(int n, int r) {
if (r == 0 || n == r) {
return 1;
}
return ncr(n - 1, r - 1) + ncr(n - 1, r);
}
static class Edge {
int v;
int l;
Edge(int v, int l) {
this.v = v;
this.l = l;
}
}
static int find(int parent[], int x) {
if (parent[x] == x) {
return parent[x];
}
parent[x] = find(parent, parent[x]);
return parent[x];
}
static long sum(long n) {
return (long) n * (n + 1) / 2;
}
static long upperbound(long arr[], long key) {
int answer = Integer.MAX_VALUE;
int start = 0;
int end = arr.length - 1;
while (start <= end) {
int mid = (start + end) >> 1;
if (arr[mid] == key) {
return arr[mid];
} else if (arr[mid] > key) {
answer = mid;
end = mid - 1;
} else {
start = mid + 1;
}
}
if (answer != Integer.MAX_VALUE)
return arr[answer];
else {
return answer;
}
}
static long lowerbound(long arr[], long x) {
int start = 0;
int end = arr.length - 1;
int answer = -1;
while (start <= end) {
int mid = (start + end) >> 1;
if (arr[mid] == x) {
return arr[mid];
} else if (arr[mid] < x) {
answer = mid;
start = mid + 1;
} else {
end = mid - 1;
}
}
if (answer != -1)
return arr[answer];
else {
return answer;
}
}
//This elements pair class
static class Pair1 {
int first;
int second;
int third;
Pair1(int first, int second, int third) {
this.first = first;
this.second = second;
this.third = third;
}
}
static boolean check(int value, int a[], int b[]) {
int value1 = 0;
int count = 0;
for (int i = 0; i < a.length; ++i) {
int rich = a[i];
int poor = b[i];
if (rich >= (value - 1) - value1 && poor >= value1) {
++count;
++value1;
}
}
if (count >= value) {
return true;
} else {
return false;
}
}
static void DFS(List<List<Integer>> L, int parent, List<Integer> first, List<Integer> second, boolean check, int dp[]) {
dp[parent] = 1;
if (check == false) {
first.add(parent);
} else {
second.add(parent);
}
for (int connected : L.get(parent)) {
if (dp[connected] != 1) {
DFS(L, connected, first, second, !check, dp);
}
}
}
static void Bfs(long dp[],List<List<Integer>>L,int parent){
Queue<Integer>Q=new LinkedList<>();
Q.add(parent);
boolean bl[]=new boolean[dp.length];
while(Q.size()!=0){
int remove=Q.remove();
for(int connected:L.get(remove)){
if(bl[connected]==false){
bl[remove]=true;
dp[connected]=dp[remove]+1;
Q.add(connected);
}
}
}
}
static void multisourceBFS(long dp[],int friends[],List<List<Integer>>L){
Queue<Integer>Q=new LinkedList<>();
Arrays.fill(dp,Integer.MAX_VALUE);
for(int i=0;i<friends.length;++i){
dp[friends[i]]=0;
Q.add(friends[i]);
}
boolean bl[]=new boolean[dp.length];
while(Q.size()!=0){
int remove=Q.poll();
for(int connected:L.get(remove)){
if(bl[connected]==false){
bl[connected]=true;
dp[connected]=Math.min(dp[connected],dp[remove]+1);
Q.add(connected);
}
}
}
}
public static void main(String[] args) {
FastReader scan=new FastReader();
int t=scan.nextInt();
while(t-->0){
int n=scan.nextInt();
int arr[]=new int[n];
int count=0;
long max=0;
for(int i=0;i<n;++i){
arr[i]= scan.nextInt();
int temp=arr[i];
while(arr[i]%2==0){
++count;
arr[i]/=2;
}
max=Math.max(max,arr[i]);
}
long sum=IntStream.of(arr).sum();
sum-=max;
for(int i=1;i<=count;++i){
max=(long)max*2;
}
System.out.println(sum+max);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 473f1d7dab173b2a3e0a54209eb17d68 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long[]fac=new long[200100];
static long[] two= new long[200100] ;
static long mod=(long)1e9+7;
static int n,k;
static int x=0;
static int[][]perm;
static int[]pe,aa,a,b,c;
static int[]prime=new int[(int)1e5+100];
static TreeSet<Long>p=new TreeSet<Long>();
static long[][]memo;
public static void main(String[] args) throws IOException{
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long[]a=sc.nextArrlong(n);
int c=0;
for (int i = 0; i < a.length; i++) {
while(a[i]%2==0) {
a[i]/=2;
c++;
}
}
Arrays.sort(a);
a[n-1]*=1l<<c;
long sum=0;
for (int i = 0; i < a.length; i++) {
sum+=a[i];
}
pw.println(sum);
}
pw.close();
}
public static int divNum(int n) {
int total = 1;
while(n>1) {
int p=prime[n];
int count =1;
while(n%p==0) {
n/=p;
count++;
}
total*=count;
}
return total;
}
public static void sieve() {
for (int i = 2; i < prime.length; i++) {
if(prime[i]==0) {
p.add((long)i);
for (int j = i; j < prime.length; j+=i) {
if(prime[j]==0)prime[j]=i;
}
}
}
}
public static long[] Extended(long p, long q) {
if (q == 0)
return new long[] { p, 1, 0 };
long[] vals = Extended(q, p % q);
long d = vals[0];
long a = vals[2];
long b = vals[1] - (p / q) * vals[2];
return new long[] { d, a, b };
}
public static int LIS(int[] a) {
int n = a.length;
int[] ser = new int[n];
Arrays.fill(ser, Integer.MAX_VALUE);
int cur = -1;
for (int i = 0; i < n; i++) {
int low = 0;
int high = n - 1;
int mid = (low + high) / 2;
while (low <= high) {
if (ser[mid] < a[i]) {
low = mid + 1;
} else {
high = mid - 1;
}
mid = (low + high) / 2;
}
cur = Math.max(cur, high + 1);
ser[high + 1] = Math.min(ser[high + 1], a[i]);
}
return cur + 1;
}
public static void permutation(int idx,int v) {
if(v==(1<<n)-1) {
perm[x++]=pe.clone();
return ;
}
for (int i = 0; i < n; i++) {
if((v&1<<i)==0) {
pe[idx]=aa[i];
permutation(idx+1, v|1<<i);
}
}
return ;
}
public static void sort(int[]a) {
mergesort(a, 0, a.length-1);
}
public static void sortIdx(long[]a,long[]idx) {
mergesortidx(a, idx, 0, a.length-1);
}
public static long C(int a,int b) {
long x=fac[a];
long y=fac[a-b]*fac[b];
return x*pow(y,mod-2)%mod;
}
public static long pow(long a,long b) {
long ans=1;a%=mod;
for(long i=b;i>0;i/=2) {
if((i&1)!=0)
ans=ans*a%mod;
a=a*a%mod;
}
return ans;
}
public static void pre(){
fac[0]=1;
fac[1]=1;
fac[2]=1;
for (int i = 3; i < fac.length; i++) {
fac[i]=((fac[i-1]*2*i)/2)%mod;
}
}
public static long eval(String s) {
long p=1;
long res=0;
for (int i = 0; i < s.length(); i++) {
res+=p*(s.charAt(s.length()-1-i)=='1'?1:0);
p*=2;
}
return res;
}
public static String binary(long x) {
String s="";
while(x!=0) {
s=(x%2)+s;
x/=2;
}
return s;
}
public static boolean allSame(String s) {
char x=s.charAt(0);
for (int i = 0; i < s.length(); i++) {
if(s.charAt(i)!=x)return false;
}
return true;
}
public static boolean isPalindrom(String s) {
int l=0;
int r=s.length()-1;
while(l<r) {
if(s.charAt(r--)!=s.charAt(l++))return false;
}
return true;
}
public static boolean isSubString(String s,String t) {
int ls=s.length();
int lt=t.length();
boolean res=false;
for (int i = 0; i <=lt-ls; i++) {
if(t.substring(i, i+ls).equals(s)) {
res=true;
break;
}
}
return res;
}
public static boolean isSorted(long[]a) {
for (int i = 0; i < a.length-1; i++) {
if(a[i]>a[i+1])return false;
}
return true;
}
public static long evaln(String x,int n) {
long res=0;
for (int i = 0; i < x.length(); i++) {
res+=Long.parseLong(x.charAt(x.length()-1-i)+"")*Math.pow(n, i);
}
return res;
}
static void mergesort(int[] arr,int b,int e) {
if(b<e) {
int m=b+(e-b)/2;
mergesort(arr,b,m);
mergesort(arr,m+1,e);
merge(arr,b,m,e);
}
return;
}
static void merge(int[] arr,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
int[] l=new int[len1];
int[] r=new int[len2];
for(int i=0;i<len1;i++)l[i]=arr[b+i];
for(int i=0;i<len2;i++)r[i]=arr[m+1+i];
int i=0,j=0,k=b;
while(i<len1 && j<len2) {
if(l[i]<r[j])arr[k++]=l[i++];
else arr[k++]=r[j++];
}
while(i<len1)arr[k++]=l[i++];
while(j<len2)arr[k++]=r[j++];
return;
}
static void mergesortidx(long[] arr,long[]idx,int b,int e) {
if(b<e) {
int m=b+(e-b)/2;
mergesortidx(arr,idx,b,m);
mergesortidx(arr,idx,m+1,e);
mergeidx(arr,idx,b,m,e);
}
return;
}
static void mergeidx(long[] arr,long[]idx,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
long[] l=new long[len1];
long[] lidx=new long[len1];
long[] r=new long[len2];
long[] ridx=new long[len2];
for(int i=0;i<len1;i++) {
l[i]=arr[b+i];
lidx[i]=idx[b+i];
}
for(int i=0;i<len2;i++) {
r[i]=arr[m+1+i];
ridx[i]=idx[m+1+i];
}
int i=0,j=0,k=b;
while(i<len1 && j<len2) {
if(l[i]<=r[j]) {
arr[k++]=l[i++];
idx[k-1]=lidx[i-1];
}
else {
arr[k++]=r[j++];
idx[k-1]=ridx[j-1];
}
}
while(i<len1) {
idx[k]=lidx[i];
arr[k++]=l[i++];
}
while(j<len2) {
idx[k]=ridx[j];
arr[k++]=r[j++];
}
return;
}
static long mergen(int[] arr,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
int[] l=new int[len1];
int[] r=new int[len2];
for(int i=0;i<len1;i++)l[i]=arr[b+i];
for(int i=0;i<len2;i++)r[i]=arr[m+1+i];
int i=0,j=0,k=b;
long c=0;
while(i<len1 && j<len2) {
if(l[i]<r[j])arr[k++]=l[i++];
else {
arr[k++]=r[j++];
c=c+(long)(len1-i);
}
}
while(i<len1)arr[k++]=l[i++];
while(j<len2)arr[k++]=r[j++];
return c;
}
static long mergesortn(int[] arr,int b,int e) {
long c=0;
if(b<e) {
int m=b+(e-b)/2;
c=c+(long)mergesortn(arr,b,m);
c=c+(long)mergesortn(arr,m+1,e);
c=c+(long)mergen(arr,b,m,e);
}
return c;
}
public static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static long sumd(long x) {
long sum=0;
while(x!=0) {
sum+=x%10;
x=x/10;
}
return sum;
}
public static ArrayList<Integer> findDivisors(int n){
ArrayList<Integer>res=new ArrayList<Integer>();
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
// If divisors are equal, print only one
if (n/i == i)
res.add(i);
else {
res.add(i);
res.add(n/i);
}
}
}
return res;
}
public static void sort2darray(Integer[][]a){
Arrays.sort(a,Comparator.<Integer[]>comparingInt(x -> x[0]).thenComparingInt(x -> x[1]));
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(String file) throws FileNotFoundException {
br = new BufferedReader(new FileReader(file));
}
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 boolean ready() throws IOException {
return br.ready();
}
public int[] nextArrint(int size) throws IOException {
int[] a=new int[size];
for (int i = 0; i < a.length; i++) {
a[i]=sc.nextInt();
}
return a;
}
public long[] nextArrlong(int size) throws IOException {
long[] a=new long[size];
for (int i = 0; i < a.length; i++) {
a[i]=sc.nextLong();
}
return a;
}
public int[][] next2dArrint(int rows,int columns) throws IOException{
int[][]a=new int[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
a[i][j]=sc.nextInt();
}
}
return a;
}
public long[][] next2dArrlong(int rows,int columns) throws IOException{
long[][]a=new long[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
a[i][j]=sc.nextLong();
}
}
return a;
}
}
static Scanner sc=new Scanner(System.in);
static PrintWriter pw=new PrintWriter(System.out);
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 3ba8c53eca099d15880a559c77f28b82 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class CodeForces {
// For fast input output
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
try {
br = new BufferedReader(new FileReader("input.txt"));
PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
System.setOut(out);
} catch (Exception e) {
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;
}
}
// end of fast i/o code
public static void main(String[] args) {
FastReader in = new FastReader();
int t = in.nextInt();
for (int i = 1; i <= t; i++) {
int n=in.nextInt();
int arr[]=new int[n];
for(int j = 0; j <n ; j++)
arr[j]=in.nextInt();
helper(n,arr);
System.out.println();
}
}
public static void helper(int n,int arr[])
{
if(n==1)
System.out.print(arr[0]);
else
{
Arrays.sort(arr);
long val=1;
for(int i=0;i<n;i++)
{
if((arr[i]&1)>0)
continue;
int v=arr[i],c=2;
if((v & (v-1))==0)
{
arr[i]=1;
c=v;
}
else if(arr[i]==12)
{
c=4;
arr[i]=arr[i]/4;
}
else
arr[i]=arr[i]/2;
val=val*c;
}
Arrays.sort(arr);
val=val*arr[n-1];
for(int i=0;i<n-1;i++)
val+=arr[i];
System.out.print(val);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 15cb65773ebc3aaf059ce210625050de | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class a_b {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while(t-->0) {
int n = scan.nextInt();
long[] ar = new long[n];
long[] ans = new long[n];
for (int i = 0; i < n; i++) {
ar[i] = scan.nextInt();
}
Arrays.sort(ar);
if (ar.length == 1) {
System.out.println(ar[0]);
}
else {
int l = ar.length;
long res = 0;
int cnt=0;
for (int i = 0; i < l; i++) {
while (ar[i] % 2 == 0) {
ar[i] /= 2;
cnt++;
}
}
for(int i=0;i<l;i++){
ans[i]=ar[i];
}
Arrays.sort(ans);
ans[l-1]*=Math.pow(2,cnt);
for (long e : ans) {
res += e;
}
System.out.println(res);
}
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 9c692000aa2ce202db28d0ac6ebe61fa | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.math.BigDecimal;
import java.math.*;
public class Main{
public static void main(String[] args) {
TaskA solver = new TaskA();
// boolean[]prime=seive(3*100001);
int t = in.nextInt();
for (int i = 1; i <= t ; i++) {
solver.solve(i, in, out);
}
// solver.solve(1, in, out);
out.flush();
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n= in.nextInt();
int []arr=new int[n];
int max=Integer.MIN_VALUE;
int maxEve=Integer.MIN_VALUE;
int count=0;
long sum=0;
for(int i=0;i<n;i++) {
arr[i]=in.nextInt();
}
Arrays.sort(arr);
for(int i=0;i<n;i++) {
if(arr[i]%2==1) {
sum+=arr[i];
max=Math.max(max, arr[i]);
}
else {
while(arr[i]%2!=1) {
arr[i]/=2;count++;
}
maxEve=Math.max(maxEve, arr[i]);
sum+=arr[i];
}
}
long sum1=sum;long sum2=sum;
sum1-=max;
sum1+=max*(long)Math.pow(2, count);
sum2-=maxEve;
sum2+=maxEve*(long)Math.pow(2, count);
println(Math.max(sum1, sum2));
}
}
static long pow(long b, long e) {
long ans = 1;
while (e > 0) {
if (e % 2 == 1)
ans = ans * b % mod;
e >>= 1;
b = b * b % mod;
}
return ans;
}
static int[]dp; //global variable
static int minStep(int n) {
if(n<0) {
return Integer.MAX_VALUE; //this is never possible should be handled bcz n-1,n-5 can be negative
}
if(n==0) {return 0;}
if(dp[n]!=0) {
return dp[n]; //if solution we have already calculated
}
else {
int x=1+Math.min(minStep(n-1),Math.min(minStep(n-3),minStep(n-5)));
dp[n]=x; //update the array after calculation
return x;
}
}
static void sortF(Pair arr[])
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{ if(p1.first==p2.first) {
return p1.second-p2.second;
}
return p1.first - p2.first;
}
});
}
static int[] shift (int[]inp,int i,int j){
int[]arr=new int[j-i+1];
int n=j-i+1;
for(int k=i;k<=j;k++) {
arr[(k-i+1)%n]=inp[k];
}
for(int k=0;k<n;k++ ) {
inp[k+i]=arr[k];
}
return inp;
}
static int sumDigits(long n) {
int total=0;
while(n!=0) {
total+=n%10;
n=n/10;
}
return total;
}
static long[] fac;
static long mod = (long) 1000000007;
static void initFac(long n) {
fac = new long[(int)n + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++) {
fac[i] = (fac[i - 1] * i) % mod;
}
}
static long nck(int n, int k) {
if (n < k)
return 0;
long den = inv((int) (fac[k] * fac[n - k] % mod));
return fac[n] * den % mod;
}
static int[]MakeArr(int n){
int[]arr=new int[n];
for(int i=0;i<n;i++) {
arr[i]=in.nextInt();
}
return arr;
}
static int[]arr(){
int n= in.nextInt();
int[]arr=new int[n];
for(int i=0;i<n;i++) {
arr[i]=in.nextInt();
}
return arr;
}
static long inv(long x) {
return pow(x, mod - 2);
}
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);
}
public static int bfsSize(int source,LinkedList<Integer>[]a,boolean[]visited) {
Queue<Integer>q=new LinkedList<>();
q.add(source);
visited[source]=true;
int distance=0;
while(!q.isEmpty()) {
int curr=q.poll();
distance++;
for(int neighbour:a[curr]) {
if(!visited[neighbour]) {
visited[neighbour]=true;
q.add(neighbour);
}
}
}
return distance;
}
public static Set<Integer>factors(int n){
Set<Integer>ans=new HashSet<>();
ans.add(1);
for(int i=2;i*i<=n;i++) {
if(n%i==0) {
ans.add(i);
ans.add(n/i);
}
}
return ans;
}
public static int bfsSp(int source,int destination,ArrayList<Integer>[]a) {
boolean[]visited=new boolean[a.length];
int[]parent=new int[a.length];
Queue<Integer>q=new LinkedList<>();
int distance=0;
q.add(source);
visited[source]=true;
parent[source]=-1;
while(!q.isEmpty()) {
int curr=q.poll();
if(curr==destination) {
break;
}
for(int neighbour:a[curr]) {
if(neighbour!=-1&&!visited[neighbour]) {
visited[neighbour]=true;
q.add(neighbour);
parent[neighbour]=curr;
}
}
}
int cur=destination;
while(parent[cur]!=-1) {
distance++;
cur=parent[cur];
}
return distance;
}
static int bs(int size,int[]arr) {
int x = -1;
for (int b = size/2; b >= 1; b /= 2) {
while (!ok(arr));
}
int k = x+1;
return k;
}
static boolean ok(int[]x) {
return false;
}
public static int solve1(ArrayList<Integer> A) {
long[]arr =new long[A.size()+1];
int n=A.size();
for(int i=1;i<=A.size();i++) {
arr[i]=((i%2)*((n-i+1)%2))%2;
arr[i]%=2;
}
int ans=0;
for(int i=0;i<A.size();i++) {
if(arr[i+1]==1) {
ans^=A.get(i);
}
}
return ans;
}
public static String printBinary(long a) {
String str="";
for(int i=31;i>=0;i--) {
if((a&(1<<i))!=0) {
str+=1;
}
if((a&(1<<i))==0 && !str.isEmpty()) {
str+=0;
}
}
return str;
}
public static String reverse(long a) {
long rev=0;
String str="";
int x=(int)(Math.log(a)/Math.log(2))+1;
for(int i=0;i<32;i++) {
rev<<=1;
if((a&(1<<i))!=0) {
rev|=1;
str+=1;
}
else {
str+=0;
}
}
return str;
}
////////////////////////////////////////////////////////
static void sortS(Pair arr[])
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return p1.second - p2.second;
}
});
}
static class Pair implements Comparable<Pair>
{
int first ;int second ;
public Pair(int x, int y)
{
this.first = x ;this.second = y ;
}
@Override
public boolean equals(Object obj)
{
if(obj == this)return true ;
if(obj == null)return false ;
if(this.getClass() != obj.getClass()) return false ;
Pair other = (Pair)(obj) ;
if(this.first != other.first)return false ;
if(this.second != other.second)return false ;
return true ;
}
@Override
public int hashCode()
{
return this.first^this.second ;
}
@Override
public String toString() {
String ans = "" ;ans += this.first ; ans += " "; ans += this.second ;
return ans ;
}
@Override
public int compareTo(Main.Pair o) {
{ if(this.first==o.first) {
return this.second-o.second;
}
return this.first - o.first;
}
}
}
//////////////////////////////////////////////////////////////////////////
static int nD(long num) {
String s=String.valueOf(num);
return s.length();
}
static int CommonDigits(int x,int y) {
String s=String.valueOf(x);
String s2=String.valueOf(y);
return 0;
}
static int lcs(String str1, String str2, int m, int n)
{
int L[][] = new int[m + 1][n + 1];
int i, j;
// Following steps build L[m+1][n+1] in
// bottom up fashion. Note that L[i][j]
// contains length of LCS of str1[0..i-1]
// and str2[0..j-1]
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
if (i == 0 || j == 0)
L[i][j] = 0;
else if (str1.charAt(i - 1)
== str2.charAt(j - 1))
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = Math.max(L[i - 1][j],
L[i][j - 1]);
}
}
// L[m][n] contains length of LCS
// for X[0..n-1] and Y[0..m-1]
return L[m][n];
}
/////////////////////////////////
boolean IsPowerOfTwo(int x)
{
return (x != 0) && ((x & (x - 1)) == 0);
}
////////////////////////////////
static long power(long a,long b,long m ) {
long ans=1;
while(b>0) {
if(b%2==1) {
ans=((ans%m)*(a%m))%m; b--;
}
else {
a=(a*a)%m;b/=2;
}
}
return ans%m;
}
///////////////////////////////
public static boolean repeatedSubString(String string) {
return ((string + string).indexOf(string, 1) != string.length());
}
static int search(char[]c,int start,int end,char x) {
for(int i=start;i<end;i++) {
if(c[i]==x) {return i;}
}
return -2;
}
////////////////////////////////
static int gcd(int a, int b) {
while (b != 0) {
int t = b;
b = a % b;
a = t;
}
return a;
}
static long fac(long a)
{
if(a== 0L || a==1L)return 1L ;
return a*fac(a-1L) ;
}
static ArrayList al() {
ArrayList<Integer>a =new ArrayList<>();
return a;
}
static HashSet h() {
return new HashSet<Integer>();
}
static void debug(int[][]a) {
int n= a.length;
for(int i=1;i<=n;i++) {
for(int j=1;j<=n;j++) {
out.print(a[i][j]+" ");
}
out.print("\n");
}
}
static void debug(int[]a) {
out.println(Arrays.toString(a));
}
static void debug(ArrayList<Integer>a) {
out.println(a.toString());
}
static boolean[]seive(int n){
boolean[]b=new boolean[n+1];
for (int i = 2; i <= n; i++)
b[i] = true;
for(int i=2;i*i<=n;i++) {
if(b[i]) {
for(int j=i*i;j<=n;j+=i) {
b[j]=false;
}
}
}
return b;
}
static int longestIncreasingSubseq(int[]arr) {
int[]sizes=new int[arr.length];
Arrays.fill(sizes, 1);
int max=1;
for(int i=1;i<arr.length;i++) {
for(int j=0;j<i;j++) {
if(arr[j]<arr[i]) {
sizes[i]=Math.max(sizes[i],sizes[j]+1);
max=Math.max(max, sizes[i]);
}
}
}
return max;
}
public static ArrayList primeFactors(long n)
{
ArrayList<Long>h= new ArrayList<>();
// Print the number of 2s that divide n
if(n%2 ==0) {h.add(2L);}
// n must be odd at this point. So we can
// skip one element (Note i = i +2)
for (long i = 3; i <= Math.sqrt(n); i+= 2)
{
if(n%i==0) {h.add(i);}
}
if (n > 2)
h.add(n);
return h;
}
static boolean Divisors(long n){
if(n%2==1) {
return true;
}
for (long i=2; i<=Math.sqrt(n); i++){
if (n%i==0 && i%2==1){
return true;
}
}
return false;
}
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(outputStream);
public static void superSet(int[]a,ArrayList<String>al,int i,String s) {
if(i==a.length) {
al.add(s);
return;
}
superSet(a,al,i+1,s);
superSet(a,al,i+1,s+a[i]+" ");
}
public static long[] makeArr() {
long size=in.nextInt();
long []arr=new long[(int)size];
for(int i=0;i<size;i++) {
arr[i]=in.nextInt();
}
return arr;
}
public static long[] arr(int n) {
long []arr=new long[n+1];
for(int i=1;i<n+1;i++) {
arr[i]=in.nextLong();
}
return arr;
}
public static void sort(long arr[], int l, int r)
{
if (l < r)
{
// Find the middle point
int m = (l+r)/2;
// Sort first and second halves
sort(arr, l, m);
sort(arr , m+1, r);
// Merge the sorted halves
merge(arr, l, m, r);
}
}
static void println(long x) {
out.println(x);
}
static void print(long c) {
out.print(c);
}
static void print(int c) {
out.print(c);
}
static void println(int x) {
out.println(x);
}
static void print(String s) {
out.print(s);
}
static void println(String s) {
out.println(s);
}
public static void merge(long arr[], int l, int m, int r)
{
// Find sizes of two subarrays to be merged
int n1 = m - l + 1;
int n2 = r - m;
/* Create temp arrays */
long L[] = new long[n1];
long R[] = new long[n2];
//Copy data to temp arrays
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
public static void reverse(int[] array)
{
int n = array.length;
for (int i = 0; i < n / 2; i++) {
int temp = array[i];
array[i] = array[n - i - 1];
array[n - i - 1] = temp;
}
}
public static long nCr(int n,int k)
{
long ans=1L;
k=k>n-k?n-k:k;
for( int j=1;j<=k;j++,n--)
{
if(n%j==0)
{
ans*=n/j;
}else
if(ans%j==0)
{
ans=ans/j*n;
}else
{
ans=(ans*n)/j;
}
}
return ans;
}
static int searchMax(int index,long[]inp) {
while(index+1<inp.length &&inp[index+1]>inp[index]) {
index+=1;
}
return index;
}
static int searchMin(int index,long[]inp) {
while(index+1<inp.length &&inp[index+1]<inp[index]) {
index+=1;
}
return index;
}
public class ListNode {
int val;
ListNode next;
ListNode() {}
ListNode(int val) { this.val = val; }
ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}
static class Pairr implements Comparable<Pairr>{
private int index;
private int cumsum;
private ArrayList<Integer>indices;
public Pairr(int index,int cumsum) {
this.index=index;
this.cumsum=cumsum;
indices=new ArrayList<Integer>();
}
public int compareTo(Pairr other) {
return Integer.compare(cumsum, other.cumsum);
}
}
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 3afe2725c5a35f47a74eabb5286e188f | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class A {
public static void main(String[] args) throws IOException {
FastReader in = new FastReader(System.in);
PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
int t = in.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = in.nextInt();
long[] a = in.readArrayL(n);
if (n == 1) {
pw.println(a[0]);
continue;
}
long res = 1;
for (int i = 0; i < n; i++) {
if (a[i] % 2 == 0) {
while (a[i] % 2 == 0) {
res *= 2;
a[i] /= 2;
}
}
}
Arrays.sort(a);
a[n - 1] *= res;
for (int i = 1; i < n; i++) a[i] = a[i] + a[i - 1];
pw.println(a[n - 1]);
}
pw.close();
}
static int gcd(int a, int b) {
if (b == 0) return a;
else return gcd(b, a % b);
}
static boolean isPalindrom(String txt) {
return txt.equals(new StringBuilder(txt).reverse().toString());
}
public static boolean isSorted(int[] arr) {
for (int i = 1; i < arr.length; i++)
if (arr[i] < arr[i - 1]) return false;
return true;
}
static long binaryExponentiation(long x, long n) {
if (n == 0) return 1;
else if (n % 2 == 0) return binaryExponentiation(x * x, n / 2);
else return x * binaryExponentiation(x * x, (n - 1) / 2);
}
public static void Sort(int[] a) {
ArrayList<Integer> lst = new ArrayList<>();
for (int i : a) lst.add(i);
Collections.sort(lst);
for (int i = 0; i < lst.size(); i++) a[i] = lst.get(i);
}
static void debug(Object... obj) {
System.err.println(Arrays.deepToString(obj));
}
static class Pair implements Comparable<Pair> {
int first, second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public int compareTo(Pair ob) {
return first - ob.first;
}
}
static class FastReader {
InputStream is;
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
public FastReader(InputStream is) {
this.is = is;
}
public int readByte() {
if (lenbuf == -1) throw new InputMismatchException();
if (ptrbuf >= lenbuf) {
ptrbuf = 0;
try {
lenbuf = is.read(inbuf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (lenbuf <= 0) return -1;
}
return inbuf[ptrbuf++];
}
public boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b)) ;
return b;
}
public String next() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public String nextLine() {
int c = skip();
StringBuilder sb = new StringBuilder();
while (!isEndOfLine(c)) {
sb.appendCodePoint(c);
c = readByte();
}
return sb.toString();
}
public int nextInt() {
int num = 0, b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = (num << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public long nextLong() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = (num << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public double nextDouble() {
return Double.parseDouble(next());
}
public char[] next(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while (p < n && !(isSpaceChar(b))) {
buf[p++] = (char) b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
public char readChar() {
return (char) skip();
}
public long[] readArrayL(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = nextLong();
return arr;
}
public int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = nextInt();
return arr;
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 34c6a44cae8a58ea6bc906046e050d4d | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class A_Divide_and_Multiply {
public 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 {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int T = reader.nextInt();
while(T!=0){
T--;
int N = reader.nextInt();
int[] ip = new int[N];
for(int i=0; i<N; i++){
ip[i] = reader.nextInt();
}
int[] factorsOf2 = new int[]{0,0,2,0,4,0,2,0,8,0,2,0,4,0,2,0,16};
long powersOf2 = 1L;
for(int i=0; i<N; i++){
int temp =factorsOf2[(int)ip[i]];
if(temp!=0){
powersOf2*=temp;
ip[i] = ip[i]/temp;
}
}
Arrays.sort(ip);
long finalVal = powersOf2*(long)ip[N-1];
long sum = 0L;
for(int i=0; i<N-1; i++){
sum+=ip[i];
}
sum+=finalVal;
System.out.println(sum);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | a08a72157d37151cefed2dfa5a1a5db6 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class A {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
Long[] a = new Long[n];
for (int i = 0; i < n; i++) a[i] = sc.nextLong();
long exp = 0;
for (int i = 0; i < n; i++)
while (a[i] % 2 == 0) {
a[i] /= 2;
exp++;
}
Arrays.sort(a);
long sum = (long) ((Math.pow(2, exp)) * a[n-1]);
for (int i = 0; i < n-1; i++) sum += a[i];
out.println(sum);
}
out.close();
}
static TreeMap<Integer, Integer> dist;
static void add(int x) {
dist.put(x, dist.getOrDefault(x, 0) + 1);
}
static void remove(int x) {
dist.put(x, dist.get(x)-1);
if (dist.get(x) == 0) dist.remove(x);
}
static final Random random = new Random();
static void shuffleSort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int r = random.nextInt(n), temp = a[r];
a[r] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s){br = new BufferedReader(new InputStreamReader(s));}
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 {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if(x.charAt(0) == '-') {
neg = true;
start++;
}
for(int i = start; i < x.length(); i++)
if(x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
}
else {
sb.append(x.charAt(i));
if(dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg?-1:1);
}
public boolean ready() throws IOException {return br.ready();}
int[] readArray(int n) throws IOException {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i] = nextInt();
return a;
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | f312eaca11ef2b20971872690928c627 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner scn=new Scanner(System.in);
int t=scn.nextInt();
while(t-->0){
int n=scn.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=scn.nextInt();
}
int pow=0;
for(int i=0;i<n;i++){
while(arr[i]%2==0){
pow++;
arr[i]=arr[i]/2;
}
}
Arrays.sort(arr);
long ans=arr[n-1]*(long)Math.pow(2,pow);
for(int i=0;i<n-1;i++){
ans+=arr[i];
}
System.out.println(ans);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 3c5fb4c947542b3db49ba347524a82ea | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class DivideAndMultiply{
static long mod = 1000000007L;
static MyScanner sc = new MyScanner();
static void solve() {
int n = sc.nextInt();
int arr[] = new int[n];
int max = Integer.MIN_VALUE;
int max2 = Integer.MIN_VALUE;
for(int i = 0;i<n;i++){
arr[i] = sc.nextInt();
}
int count = 0;
for(int i = 0;i<n;i++){
while(arr[i]%2==0){
arr[i]/=2;
count++;
}
}
for(int i = 0;i<n;i++){
max = Math.max(arr[i],max);
}
long ans = 0;
boolean flag = true;
for(int i = 0;i<n;i++){
if(arr[i]==max && flag){
flag = false;
continue;
}
ans += arr[i];
}
long temp = (long)Math.pow(2,count);
temp = temp*max;
ans += temp;
out.println(ans);
}
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 || n==2) return false;
if(n==3) 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;
}
static class Pair implements Comparable<Pair>{
int val;
int freq;
Pair(int v,int f){
val = v;
freq = f;
}
public int compareTo(Pair p){
return this.freq - p.freq;
}
}
public static void main(String[] args) {
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
while(t-- >0){
// solve();
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();
}
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | a26fa08de55f2c1bc635b960a7dfc09f | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
public class Main {
private static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
int t = Integer.parseInt(in.nextLine());
for (int i = 0; i < t; i++) {
int n = Integer.parseInt(in.nextLine());
String array = in.nextLine();
String[] parts = array.split(" ");
int[] arrayInt = new int[parts.length];
int pairsCounter = 0;
int notPair = 0;
int indexMaxOdd = 0;
for (int j = 0; j < parts.length; j++) {
arrayInt[j] = Integer.parseInt(parts[j]);
while(arrayInt[j] %2 == 0) {
pairsCounter++;
arrayInt[j] /= 2;
}
}
notPair = arrayInt[0];
for (int j = 0; j < arrayInt.length; j++) {
if(notPair < arrayInt[j]) {
notPair = arrayInt[j];
indexMaxOdd = j;
}
}
long maximumSum = (long)(notPair * Math.pow(2, pairsCounter));
for (int j = 0; j < arrayInt.length; j++) {
if(j != indexMaxOdd) {
maximumSum += arrayInt[j];
}
}
System.out.println(maximumSum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | ed1a7e6ab0824bd84ac68340e39c3e84 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
public class Maximizar_suma{
public static void main(String args[]) {
Scanner reader=new Scanner(System.in);
int nTimes=0;
nTimes=reader.nextInt();
for(int h=0;h<nTimes;h++){
int t=reader.nextInt();
long[] numArray=new long[t];
int carryCounter=0;
int maxOddPosition=0;
for(int i=0;i<t;i++)
{
numArray[i]=reader.nextInt();
while(numArray[i] % 2==0)
{
numArray[i]/=2;
carryCounter++;
}
if(numArray[maxOddPosition]<numArray[i])
{
maxOddPosition=i;
}
}
numArray[maxOddPosition]*=Math.pow(2,carryCounter);
long total=0;
for(int i=0;i<t;i++)
{
total+=numArray[i];
}
System.out.println(total);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 490c04ce25118be793eb361deb94b814 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
import java.util.ArrayList;
public class Main {
public static Scanner reader=new Scanner(System.in);
public static void main (String[]args) {
ArrayList<Long> numbers=new ArrayList<Long>();
int counter=0;
int t=0;
int casesNum=0;
casesNum=reader.nextInt();
reader.nextLine();
long max=0;
int maxPos=-1;
long sum=0;
for (int y =0; y<casesNum ;y++) {
numbers.clear();
t=reader.nextInt();
counter=0;
maxPos=0;
max=0;
for (int i=0; i<t ;i++) {
numbers.add(reader.nextLong());
}
for (int i=0;i<numbers.size();i++) {
if (numbers.get(i)%2==0) {
while (numbers.get(i)%2==0){
numbers.set(i,numbers.get(i)/2);
counter++;
}
}
}
for (int i=0;i<numbers.size();i++) {
if(numbers.get(i)%2!=0) {
if (numbers.get(i)>max) {
max=numbers.get(i);
maxPos=i;
}
}
}
for (int i=0; i<counter;i++) {
numbers.set(maxPos, numbers.get(maxPos)*2);
}
sum=0;
for (int i=0; i<numbers.size();i++) {
sum+=numbers.get(i);
}
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 586b555f33a0763a2a539e76113509db | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | //
// Source code recreated from a .class file by IntelliJ IDEA
// (powered by Fernflower decompiler)
//
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
public class CFA {
static CFA.Reader sc = new CFA.Reader();
static BufferedWriter bw;
static long count = 0;
public CFA() {
}
static boolean flag = false;
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
process:
for (int p = 0; p < t; p++) {
int n=sc.nextInt();
long[] ar=new long[n];
long maxod=-1L;
long maxindex=-1L;
int count=0;
for(int i=0;i<n;i++){
ar[i]=sc.nextLong();
if(ar[i]%2!=0){
if(ar[i]>maxod){
maxod=ar[i];
maxindex=i;
}
}
while (ar[i]%2==0){
ar[i]/=2;
count++;
}
}
Arrays.sort(ar);
ar[n-1]= (long) (ar[n-1]*Math.pow(2,count));
/*if(maxindex!=-1){
for(int i=0;i<n;i++){
if(ar[i]%2==0&&i!=maxindex){
while(ar[i]%2==0){
ar[i]/=2;
ar[(int) maxindex]*=2;
}
}
}
}else{
Arrays.sort(ar);
//System.out.println(Arrays.toString(ar));
for(int i=0;i<n-1;i++){
while(ar[i]%2==0){
ar[i]/=2;
ar[n-1]=ar[n-1]*2;
}
}
//System.out.println(ar[n-1]);
}
//*/
long sum=0;
for(long val:ar){
sum+=val;
}
System.out.println(sum);
}
bw.flush();
bw.close();
}
public static class Check{
int cost,ram;
public Check(int cost,int ram){
this.cost=cost;
this.ram=ram;
}
}
static void dfs(int i, int j, int n, char[][] ar, boolean[][] reached) {
// reached[i][j] = true;
//System.out.println(i +" "+j);
if (i >= 2 || j >= n) {
return;
}
if (i == 1 && j == (n - 1)) {
flag = true;
return;
}
if (i + 1 < 2) {
if (ar[i + 1][j] == '0' && !reached[i + 1][j]) {
reached[i][j] = true;
dfs(i + 1, j, n, ar, reached);
reached[i][j] = false;
}
}
if (j + 1 < n) {
if (ar[i][j + 1] == '0' && !reached[i][j + 1]) {
reached[i][j] = true;
dfs(i, j + 1, n, ar, reached);
reached[i][j] = false;
}
}
if (i + 1 < 2 && j + 1 < n) {
if (ar[i + 1][j + 1] == '0' && !reached[i + 1][j + 1]) {
reached[i][j] = true;
dfs(i + 1, j + 1, n, ar, reached);
reached[i][j] = false;
}
}
if (i - 1 >= 0) {
if (ar[i - 1][j] == '0' && !reached[i - 1][j]) {
reached[i][j] = true;
dfs(i - 1, j, n, ar, reached);
reached[i][j] = false;
}
}
/* if (j + 1 < n) {
if (ar[i][j + 1] == '0' && !reached[i][j + 1]) {
reached[i][j] = true;
dfs(i, j + 1, n, ar, reached);
reached[i][j] = false;
}
}*/
if (i - 1 >= 0 && j + 1 < n) {
if (ar[i - 1][j + 1] == '0' && !reached[i - 1][j + 1]) {
reached[i][j] = true;
dfs(i - 1, j + 1, n, ar, reached);
reached[i][j] = false;
}
}
}
public static class Timings {
int arr, dept;
public Timings(int arr, int dept) {
this.arr = arr;
this.dept = dept;
}
}
static void merge(long arr[], int l, int m, int r) {
int s1 = m - l + 1;
int s2 = r - (m + 1) + 1;
long[] ar1 = new long[s1];
long[] ar2 = new long[s2];
for (int i = 0; i < s1; i++) {
ar1[i] = arr[l + i];
}
for (int i = 0; i < s2; i++) {
ar2[i] = arr[m + 1 + i];
}
int ind = l;
int i = 0, j = 0;
while (i < s1 && j < s2) {
if (ar1[i] < ar2[j]) {
arr[ind] = ar1[i];
i++;
} else if (ar1[i] == ar2[j]) {
arr[ind] = ar1[i];
i++;
} else {
arr[ind] = ar2[j];
j++;
count += (m + 1) - (l + i);
}
ind++;
}
while (i < s1) {
arr[ind] = ar1[i];
i++;
ind++;
//count++;
}
while (j < s2) {
arr[ind] = ar2[j];
j++;
ind++;
}
//System.out.println(count);
}
static void mergeSort(long arr[], int l, int r) {
//code here
if (r > l) {
int mid = l + (r - l) / 2;
mergeSort(arr, l, mid);
mergeSort(arr, mid + 1, r);
merge(arr, l, mid, r);
}
}
public static int checker(int a) {
if (a % 2 == 0) {
return 1;
} else {
byte b;
do {
b = 2;
} while (a + b != (a ^ b));
return b;
}
}
public static boolean checkPrime(int n) {
if (n != 0 && n != 1) {
for (int j = 2; j * j <= n; ++j) {
if (n % j == 0) {
return false;
}
}
return true;
} else {
return false;
}
}
public static void dfs1(List<List<Integer>> g, boolean[] visited, Stack<Integer> stack, int num) {
visited[num] = true;
Iterator var4 = ((List) g.get(num)).iterator();
while (var4.hasNext()) {
Integer integer = (Integer) var4.next();
if (!visited[integer]) {
dfs1(g, visited, stack, integer);
}
}
stack.push(num);
}
public static void dfs2(List<List<Integer>> g, boolean[] visited, List<Integer> list, int num, int c, int[] raj) {
visited[num] = true;
Iterator var6 = ((List) g.get(num)).iterator();
while (var6.hasNext()) {
Integer integer = (Integer) var6.next();
if (!visited[integer]) {
dfs2(g, visited, list, integer, c, raj);
}
}
raj[num] = c;
list.add(num);
}
public static int inputInt() throws IOException {
return sc.nextInt();
}
public static long inputLong() throws IOException {
return sc.nextLong();
}
public static double inputDouble() throws IOException {
return sc.nextDouble();
}
public static String inputString() throws IOException {
return sc.readLine();
}
public static void print(String a) throws IOException {
bw.write(a);
}
public static void printSp(String a) throws IOException {
bw.write(a + " ");
}
public static void println(String a) throws IOException {
bw.write(a + "\n");
}
public static long getAns(int[] ar, int c, long[][] dp, int i, int sign) {
if (i < 0) {
return 1L;
} else if (c <= 0) {
return 1L;
} else {
dp[i][c] = Math.max(dp[i][c], Math.max((long) ar[i] * getAns(ar, c - 1, dp, i - 1, sign), getAns(ar, c, dp, i - 1, 1)));
return dp[i][c];
}
}
public static long power(long a, long b, long c) {
long ans;
for (ans = 1L; b != 0L; b /= 2L) {
if (b % 2L == 1L) {
ans *= a;
ans %= c;
}
a *= a;
a %= c;
}
return ans;
}
public static long power1(long a, long b, long c) {
long ans;
for (ans = 1L; b != 0L; b /= 2L) {
if (b % 2L == 1L) {
ans = multiply(ans, a, c);
}
a = multiply(a, a, c);
}
return ans;
}
public static long multiply(long a, long b, long c) {
long res = 0L;
for (a %= c; b > 0L; b /= 2L) {
if (b % 2L == 1L) {
res = (res + a) % c;
}
a = (a + a) % c;
}
return res % c;
}
public static long totient(long n) {
long result = n;
for (long i = 2L; i * i <= n; ++i) {
if (n % i == 0L) {
while (n % i == 0L) {
n /= i;
}
result -= result / i;
}
}
if (n > 1L) {
result -= result / n;
}
return result;
}
public static long gcd(long a, long b) {
return b == 0L ? a : gcd(b, a % b);
}
public static boolean[] primes(int n) {
boolean[] p = new boolean[n + 1];
p[0] = false;
p[1] = false;
int i;
for (i = 2; i <= n; ++i) {
p[i] = true;
}
for (i = 2; i * i <= n; ++i) {
if (p[i]) {
for (int j = i * i; j <= n; j += i) {
p[j] = false;
}
}
}
return p;
}
public String LargestEven(String S) {
int[] count = new int[10];
int num;
for (num = 0; num < S.length(); ++num) {
++count[S.charAt(num) - 48];
}
num = -1;
for (int i = 0; i <= 8; i += 2) {
if (count[i] > 0) {
num = i;
break;
}
}
StringBuilder ans = new StringBuilder();
for (int i = 9; i >= 0; --i) {
int j;
if (i != num) {
for (j = 1; j <= count[i]; ++j) {
ans.append(i);
}
} else {
for (j = 1; j <= count[num] - 1; ++j) {
ans.append(num);
}
}
}
if (num != -1 && count[num] > 0) {
ans.append(num);
}
return ans.toString();
}
static {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public static class Score {
int l;
String s;
public Score(int l, String s) {
this.l = l;
this.s = s;
}
}
public static class Ath {
int r1;
int r2;
int r3;
int r4;
int r5;
int index;
public Ath(int r1, int r2, int r3, int r4, int r5, int index) {
this.r1 = r1;
this.r2 = r2;
this.r3 = r3;
this.r4 = r4;
this.r5 = r5;
this.index = index;
}
}
static class Reader {
private final int BUFFER_SIZE = 65536;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer;
private int bytesRead;
public Reader() {
this.din = new DataInputStream(System.in);
this.buffer = new byte[65536];
this.bufferPointer = this.bytesRead = 0;
}
public Reader(String file_name) throws IOException {
this.din = new DataInputStream(new FileInputStream(file_name));
this.buffer = new byte[65536];
this.bufferPointer = this.bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64];
int cnt;
byte c;
for (cnt = 0; (c = this.read()) != -1 && c != 10; buf[cnt++] = (byte) c) {
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c;
for (c = this.read(); c <= 32; c = this.read()) {
}
boolean neg = c == 45;
if (neg) {
c = this.read();
}
do {
ret = ret * 10 + c - 48;
} while ((c = this.read()) >= 48 && c <= 57);
return neg ? -ret : ret;
}
public long nextLong() throws IOException {
long ret = 0L;
byte c;
for (c = this.read(); c <= 32; c = this.read()) {
}
boolean neg = c == 45;
if (neg) {
c = this.read();
}
do {
ret = ret * 10L + (long) c - 48L;
} while ((c = this.read()) >= 48 && c <= 57);
return neg ? -ret : ret;
}
public double nextDouble() throws IOException {
double ret = 0.0D;
double div = 1.0D;
byte c;
for (c = this.read(); c <= 32; c = this.read()) {
}
boolean neg = c == 45;
if (neg) {
c = this.read();
}
do {
ret = ret * 10.0D + (double) c - 48.0D;
} while ((c = this.read()) >= 48 && c <= 57);
if (c == 46) {
while ((c = this.read()) >= 48 && c <= 57) {
ret += (double) (c - 48) / (div *= 10.0D);
}
}
return neg ? -ret : ret;
}
private void fillBuffer() throws IOException {
this.bytesRead = this.din.read(this.buffer, this.bufferPointer = 0, 65536);
if (this.bytesRead == -1) {
this.buffer[0] = -1;
}
}
private byte read() throws IOException {
if (this.bufferPointer == this.bytesRead) {
this.fillBuffer();
}
return this.buffer[this.bufferPointer++];
}
public void close() throws IOException {
if (this.din != null) {
this.din.close();
}
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 0ce28e825a0a08942706766e74d68597 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 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 n = io.nextInt();
long[] data = new long[n];
for (int i = 0; i < n; i++) {
data[i] = io.nextInt();
}
long output = 1;
for (int i = 0; i < n; i++) {
while (data[i] % 2 == 0) {
data[i] /= 2;
output *= 2;
}
}
Arrays.sort(data);
output *= data[n - 1];
for (int i = 0; i < n - 1; i++) {
output += data[i];
}
io.println(output);
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>(a.length);
for (int 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;
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
//--------------------------------------------------------
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c84dbb1d800e6682999799fb903351f1 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;
public class DivideAndMultiply {
static long sum(long[] a){
long sum = 0;
for(int i=0; i<a.length ;i++)
sum += a[i];
return sum;
}
public static void main(String[] args) throws IOException {
Scanner sc =new Scanner(System.in);
int t = sc.nextInt();
PrintWriter pw = new PrintWriter(System.out);
while(t-->0){
int n = sc.nextInt();
long[] a = new long[n];
long p = 1;
for(int i=0; i<n ;i++) {
a[i] = sc.nextInt();
while(a[i]%2 == 0){
a[i] /= 2;
p *= 2;
}
}
Arrays.sort(a);
a[n-1] *= p;
long sum = sum(a);
pw.println(sum);
}
pw.flush();
}
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 1f97941b02b84f6b188ab3e62de2e61a | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class DivideAndMultiply {
static long sum(long[] a){
long sum = 0;
for(int i=0; i<a.length ;i++)
sum += a[i];
return sum;
}
public static void main(String[] args){
Scanner sc =new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
long[] a = new long[n];
long p = 1;
for(int i=0; i<n ;i++) {
a[i] = sc.nextInt();
while(a[i]%2 == 0){
a[i] /= 2;
p *= 2;
}
}
Arrays.sort(a);
a[n-1] *= p;
long sum = sum(a);
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 56c6476642aaf4c8da551f9e9d7572f3 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.math.*;
import java.math.BigInteger;
public final class A
{
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans=new StringBuilder();
static FastReader in=new FastReader();
static ArrayList<Integer> g[];
static long mod=(long) 998244353,INF=Long.MAX_VALUE;
static boolean set[];
static int par[],col[],D[];
static long A[];
public static void main(String args[])throws IOException
{
/*
* star,rope,TPST
* BS,LST,MS,MQ
*/
int T=i();
while(T-->0)
{
int N=i();
long A[]=inputLong(N);
sort(A);
long max=0;
for(int i=0; i<N; i++)
{
long sum=0,pow=1L;
for(int j=0; j<N; j++)
{
if(i==j)continue;
long a=A[j];
while(A[j]%2==0)
{
A[j]/=2;
pow<<=1;
}
sum+=A[j];
A[j]=a;
}
sum+=(pow*A[i]);
max=Math.max(max, sum);
}
ans.append(max+"\n");
}
out.println(ans);
out.close();
}
static boolean f(long X[],long V[],double m)
{
double l=0,r=1e18;
for(int i=0; i<X.length; i++)
{
double d=((double)(V[i]))*m;
double x=(double)X[i];
l=Math.max(l, x-d);
r=Math.min(r, x+d);
// System.out.println(l+" "+r);
}
return l<=r;
}
static int query(long a,TreeNode root)
{
long p=1L<<30;
TreeNode temp=root;
while(p!=0)
{
System.out.println(a+" "+p);
if((a&p)!=0)
{
temp=temp.right;
}
else temp=temp.left;
p>>=1;
}
return temp.index;
}
static void delete(long a,TreeNode root)
{
long p=1L<<30;
TreeNode temp=root;
while(p!=0)
{
if((a&p)!=0)
{
temp.right.cnt--;
temp=temp.right;
}
else
{
temp.left.cnt--;
temp=temp.left;
}
p>>=1;
}
}
static void insert(long a,TreeNode root,int i)
{
long p=1L<<30;
TreeNode temp=root;
while(p!=0)
{
if((a&p)!=0)
{
temp.right.cnt++;
temp=temp.right;
}
else
{
temp.left.cnt++;
temp=temp.left;
}
p>>=1;
}
temp.index=i;
}
static TreeNode create(int p)
{
TreeNode root=new TreeNode(0);
if(p!=0)
{
root.left=create(p-1);
root.right=create(p-1);
}
return root;
}
static boolean f(long A[],long m,int N)
{
long B[]=new long[N];
for(int i=0; i<N; i++)
{
B[i]=A[i];
}
for(int i=N-1; i>=0; i--)
{
if(B[i]<m)return false;
if(i>=2)
{
long extra=Math.min(B[i]-m, A[i]);
long x=extra/3L;
B[i-2]+=2L*x;
B[i-1]+=x;
}
}
return true;
}
static int f(int l,int r,long A[],long x)
{
while(r-l>1)
{
int m=(l+r)/2;
if(A[m]>=x)l=m;
else r=m;
}
return r;
}
static boolean f(long m,long H,long A[],int N)
{
long s=m;
for(int i=0; i<N-1;i++)
{
s+=Math.min(m, A[i+1]-A[i]);
}
return s>=H;
}
static boolean f(int i,int j,long last,boolean win[])
{
if(i>j)return false;
if(A[i]<=last && A[j]<=last)return false;
long a=A[i],b=A[j];
//System.out.println(a+" "+b);
if(a==b)
{
return win[i] | win[j];
}
else if(a>b)
{
boolean f=false;
if(b>last)f=!f(i,j-1,b,win);
return win[i] | f;
}
else
{
boolean f=false;
if(a>last)f=!f(i+1,j,a,win);
return win[j] | f;
}
}
static long ask(long l,long r)
{
System.out.println("? "+l+" "+r);
return l();
}
static long f(long N,long M)
{
long s=0;
if(N%3==0)
{
N/=3;
s=N*M;
}
else
{
long b=N%3;
N/=3;
N++;
s=N*M;
N--;
long a=N*M;
if(M%3==0)
{
M/=3;
a+=(b*M);
}
else
{
M/=3;
M++;
a+=(b*M);
}
s=Math.min(s, a);
}
return s;
}
static int ask(StringBuilder sb,int a)
{
System.out.println(sb+""+a);
return i();
}
static void swap(char X[],int i,int j)
{
char x=X[i];
X[i]=X[j];
X[j]=x;
}
static int min(int a,int b,int c)
{
return Math.min(Math.min(a, b), c);
}
static long and(int i,int j)
{
System.out.println("and "+i+" "+j);
return l();
}
static long or(int i,int j)
{
System.out.println("or "+i+" "+j);
return l();
}
static int len=0,number=0;
static void f(char X[],int i,int num,int l)
{
if(i==X.length)
{
if(num==0)return;
//update our num
if(isPrime(num))return;
if(l<len)
{
len=l;
number=num;
}
return;
}
int a=X[i]-'0';
f(X,i+1,num*10+a,l+1);
f(X,i+1,num,l);
}
static boolean is_Sorted(int A[])
{
int N=A.length;
for(int i=1; i<=N; i++)if(A[i-1]!=i)return false;
return true;
}
static boolean f(StringBuilder sb,String Y,String order)
{
StringBuilder res=new StringBuilder(sb.toString());
HashSet<Character> set=new HashSet<>();
for(char ch:order.toCharArray())
{
set.add(ch);
for(int i=0; i<sb.length(); i++)
{
char x=sb.charAt(i);
if(set.contains(x))continue;
res.append(x);
}
}
String str=res.toString();
return str.equals(Y);
}
static boolean all_Zero(int f[])
{
for(int a:f)if(a!=0)return false;
return true;
}
static long form(int a,int l)
{
long x=0;
while(l-->0)
{
x*=10;
x+=a;
}
return x;
}
static int count(String X)
{
HashSet<Integer> set=new HashSet<>();
for(char x:X.toCharArray())set.add(x-'0');
return set.size();
}
static int f(long K)
{
long l=0,r=K;
while(r-l>1)
{
long m=(l+r)/2;
if(m*m<K)l=m;
else r=m;
}
return (int)l;
}
// static void build(int v,int tl,int tr,long A[])
// {
// if(tl==tr)
// {
// seg[v]=A[tl];
// }
// else
// {
// int tm=(tl+tr)/2;
// build(v*2,tl,tm,A);
// build(v*2+1,tm+1,tr,A);
// seg[v]=Math.min(seg[v*2], seg[v*2+1]);
// }
// }
static int [] sub(int A[],int B[])
{
int N=A.length;
int f[]=new int[N];
for(int i=N-1; i>=0; i--)
{
if(B[i]<A[i])
{
B[i]+=26;
B[i-1]-=1;
}
f[i]=B[i]-A[i];
}
for(int i=0; i<N; i++)
{
if(f[i]%2!=0)f[i+1]+=26;
f[i]/=2;
}
return f;
}
static int[] f(int N)
{
char X[]=in.next().toCharArray();
int A[]=new int[N];
for(int i=0; i<N; i++)A[i]=X[i]-'a';
return A;
}
static int max(int a ,int b,int c,int d)
{
a=Math.max(a, b);
c=Math.max(c,d);
return Math.max(a, c);
}
static int min(int a ,int b,int c,int d)
{
a=Math.min(a, b);
c=Math.min(c,d);
return Math.min(a, c);
}
static HashMap<Integer,Integer> Hash(int A[])
{
HashMap<Integer,Integer> mp=new HashMap<>();
for(int a:A)
{
int f=mp.getOrDefault(a,0)+1;
mp.put(a, f);
}
return mp;
}
static long mul(long a, long b)
{
return ( a %mod * 1L * b%mod )%mod;
}
static void swap(int A[],int a,int b)
{
int t=A[a];
A[a]=A[b];
A[b]=t;
}
static int find(int a)
{
if(par[a]<0)return a;
return par[a]=find(par[a]);
}
static void union(int a,int b)
{
a=find(a);
b=find(b);
if(a!=b)
{
par[a]+=par[b];
par[b]=a;
}
}
static boolean isSorted(int A[])
{
for(int i=1; i<A.length; i++)
{
if(A[i]<A[i-1])return false;
}
return true;
}
static boolean isDivisible(StringBuilder X,int i,long num)
{
long r=0;
for(; i<X.length(); i++)
{
r=r*10+(X.charAt(i)-'0');
r=r%num;
}
return r==0;
}
static int lower_Bound(int A[],int low,int high, int x)
{
if (low > high)
if (x >= A[high])
return A[high];
int mid = (low + high) / 2;
if (A[mid] == x)
return A[mid];
if (mid > 0 && A[mid - 1] <= x && x < A[mid])
return A[mid - 1];
if (x < A[mid])
return lower_Bound( A, low, mid - 1, x);
return lower_Bound(A, mid + 1, high, x);
}
static String f(String A)
{
String X="";
for(int i=A.length()-1; i>=0; i--)
{
int c=A.charAt(i)-'0';
X+=(c+1)%2;
}
return X;
}
static void sort(long[] a) //check for long
{
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 String swap(String X,int i,int j)
{
char ch[]=X.toCharArray();
char a=ch[i];
ch[i]=ch[j];
ch[j]=a;
return new String(ch);
}
static int sD(long n)
{
if (n % 2 == 0 )
return 2;
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0 )
return i;
}
return (int)n;
}
static void setGraph(int N)
{
par=new int[N+1];
D=new int[N+1];
g=new ArrayList[N+1];
set=new boolean[N+1];
col=new int[N+1];
for(int i=0; i<=N; i++)
{
col[i]=-1;
g[i]=new ArrayList<>();
D[i]=Integer.MAX_VALUE;
}
}
static long pow(long a,long b)
{
//long mod=1000000007;
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 toggleBits(long x)//one's complement || Toggle bits
{
int n=(int)(Math.floor(Math.log(x)/Math.log(2)))+1;
return ((1<<n)-1)^x;
}
static int countBits(long a)
{
return (int)(Math.log(a)/Math.log(2)+1);
}
static long fact(long N)
{
long n=2;
if(N<=1)return 1;
else
{
for(int i=3; i<=N; i++)n=(n*i)%mod;
}
return n;
}
static int kadane(int A[])
{
int lsum=A[0],gsum=A[0];
for(int i=1; i<A.length; i++)
{
lsum=Math.max(lsum+A[i],A[i]);
gsum=Math.max(gsum,lsum);
}
return gsum;
}
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 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;
}
static void print(char A[])
{
for(char c:A)System.out.print(c+" ");
System.out.println();
}
static void print(boolean A[])
{
for(boolean c:A)System.out.print(c+" ");
System.out.println();
}
static void print(int A[])
{
for(int a:A)System.out.print(a+" ");
System.out.println();
}
static void print(long A[])
{
for(long i:A)System.out.print(i+ " ");
System.out.println();
}
static void print(boolean A[][])
{
for(boolean a[]:A)print(a);
}
static void print(long A[][])
{
for(long a[]:A)print(a);
}
static void print(int A[][])
{
for(int a[]:A)print(a);
}
static void print(ArrayList<Integer> A)
{
for(int a:A)System.out.print(a+" ");
System.out.println();
}
static int i()
{
return in.nextInt();
}
static long l()
{
return in.nextLong();
}
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 long GCD(long a,long b)
{
if(b==0)
{
return a;
}
else return GCD(b,a%b );
}
}
class pair implements Comparable<pair>
{
long a,l,r;
boolean left;
int index;
pair(long a,long l,int i)
{
this.a=a;
index=i;
this.l=l;
}
public int compareTo(pair x)
{
if(this.a==x.a)
{
if(this.l>x.l)return -1;
else if(x.l>this.l)return 1;
return 0;
}
if(this.a>x.a)return 1;
return -1;
}
}
class TreeNode
{
int cnt,index;
TreeNode left,right;
TreeNode(int c)
{
cnt=c;
index=-1;
}
TreeNode(int c,int index)
{
cnt=c;
this.index=index;
}
}
//Code For FastReader
//Code For FastReader
//Code For FastReader
//Code For FastReader
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | b003d3d02f8a4282ffc4717c2b0bfd8a | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | /*
stream Butter!
eggyHide eggyVengeance
I need U
xiao rerun when
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
import java.math.*;
public class x1609A
{
public static void main(String hi[]) 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());
int N = Integer.parseInt(st.nextToken());
int[] arr = readArr(N, infile, st);
Arrays.sort(arr);
long res = 0L;
for(int a=0; a < N; a++)
{
int cnt = 0;
long temp = 0L;
for(int b=0; b < N; b++)
if(a != b)
{
int val = arr[b];
while(val%2 == 0)
{
val /= 2;
cnt++;
}
temp += val;
}
temp += arr[a]*(1L<<cnt);
res = max(res, temp);
}
sb.append(res+"\n");
}
System.out.print(sb);
}
public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
{
int[] arr = new int[N];
st = new StringTokenizer(infile.readLine());
for(int i=0; i < N; i++)
arr[i] = Integer.parseInt(st.nextToken());
return arr;
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | cf9010d7d00a69186cfb086b0eafc497 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class pre1{
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 obj = new FastReader();
int tc = obj.nextInt();
while(tc--!=0){
int n = obj.nextInt();
long arr[] = new long[n],count2 = 0;
for(int i=0;i<n;i++){
long l = obj.nextLong();
while((l&1l)==0){
l>>=1;
count2++;
}
arr[i] = l;
}
// System.out.println(count2);
// for(int i=0;i<n;i++) System.out.print(arr[i]+" ");
// System.out.println();
long ans = Long.MIN_VALUE;
for(int i=0;i<n;i++){
long sum = 0;
for(int j=0;j<n;j++){
if(i==j) sum += arr[j]*((2l<<(count2-1)==0)?1:2l<<(count2-1));
else sum += arr[j];
}
ans = Math.max(ans,sum);
}
System.out.println(ans);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 6d87c775f2e78cc342c123803fdf9e99 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;
public class Practice2 {
static int[] sort(int[] arr) {
int n=arr.length;
ArrayList<Integer> al=new ArrayList<>();
for(int i=0;i<n;i++) {
al.add(arr[i]);
}
Collections.sort(al);
for(int i=0;i<n;i++) {
arr[i]=al.get(i);
}
return arr;
}
public static void main (String[] args) {
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
// out.print();
//out.println();
FastReader sc=new FastReader();
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++) {
arr[i]=sc.nextInt();
}
long val=1;
for(int i=0;i<n;i++) {
while(arr[i]%2==0) {
arr[i]/=2;
val *=2;
}
}
int pos=-1;
int max=Integer.MIN_VALUE;
for(int i=0;i<n;i++) {
if(arr[i]>max) {
pos=i;
max=arr[i];
}
}
if(pos==-1) {
pos=0;
}
//out.println("pos : "+pos+" val : "+val);
long sum=0;
for(int i=0;i<n;i++) {
//out.println("arr[i] : "+arr[i]);
if(i==pos) {
sum +=arr[i]*val;
}else {
sum +=arr[i];
}
}
out.println(sum);
}
out.close();
}
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 0516cbca1b949e2fc17dd1682a0aa923 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
public class DivideAndMultiply {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.valueOf(br.readLine().split(" ")[0]);
PrintWriter pr = new PrintWriter(System.out);
while(t>0){
t--;
String[] ints = br.readLine().split(" ");
int n = Integer.valueOf(ints[0]);
ints = br.readLine().split(" ");
List<Long> a = new ArrayList<>();
for(int i = 0;i<n;i++){
a.add(Long.valueOf(ints[i]));
}
Long pow2 = 1L;
for(int i = 0;i<n;i++){
Long x = a.get(i);
while(x%2 == 0){
pow2*=2;
x = x/2;
}
a.set(i,x);
}
a.sort(Comparator.naturalOrder());
Long sum = 0L;
for(int i = 0;i<n;i++){
if (i!=n-1){
sum+=a.get(i);
}
else{
sum+=a.get(i)*pow2;
}
}
pr.println(sum);
}
pr.close();
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 4fb0e0d6e023ed885b933a64022aa93c | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.StringReader;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Date;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Hashtable;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Random;
import java.util.Scanner;
import java.util.Stack;
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;
}
}
static int gcd(int no, int sum) {
if (sum == 0)
return no;
return gcd(sum, no % sum);
}
static int lcm(int a , int b){
return (a / gcd(a , b))*b;
}
static void countSort(int[] arr)
{
int max = Arrays.stream(arr).max().getAsInt();
int min = Arrays.stream(arr).min().getAsInt();
int range = max - min + 1;
int count[] = new int[range];
int output[] = new int[arr.length];
for (int i = 0; i < arr.length; i++) {
count[arr[i] - min]++;
}
for (int i = 1; i < count.length; i++) {
count[i] += count[i - 1];
}
for (int i = arr.length - 1; i >= 0; i--) {
output[count[arr[i] - min] - 1] = arr[i];
count[arr[i] - min]--;
}
for (int i = 0; i < arr.length; i++) {
arr[i] = output[i];
}
}
static void RandomShuffleSort(int [] a , int start , int end){
Random random = new Random();
for(int i = start; i<end; i++){
int j = random.nextInt(end);
int temp = a[j];
a[j] = a[i];
a[i] = temp;
}
Arrays.sort(a , start , end);
}
public static void main(String[] args) throws IOException , ParseException{
BufferedWriter out = new BufferedWriter(
new OutputStreamWriter(System.out));
FastReader sc = new FastReader();
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
int count = 0;
int maxNumber = Integer.MIN_VALUE;
int index = 0;
for(int i = 0; i<n; i++){
a[i] = sc.nextInt();
while(a[i]%2 == 0){
a[i]/=2;
count++;
}
if(a[i] > maxNumber){
maxNumber = a[i];
index = i;
}
}
long sum = 0;
for(int i = 0; i<n; i++){
if(i == index){
long sum1 = a[i];
for(int j = 0; j<count; j++){
sum1*=2;
}
sum+=sum1;
}
else sum+=a[i];
}
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 39f7773882ad8670517694e21deef71b | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
import java.lang.Math;
public class DivideandMultiply
{
public static void main(String args[])
{
Scanner s = new Scanner(System.in);
int n = s.nextInt();
for(int i = 0; i<n; i++)
{
int max = 0, k=0;
long sum = 0;
int arraylength = s.nextInt();
int a[] = new int[arraylength];
for(int j = 0; j<arraylength; j++)
{
a[j] = s.nextInt();
while(a[j]%2==0)
{
a[j] = a[j]/2;
k++;
}
if(a[j]>a[max])
max = j;
}
for(int j = 0; j<arraylength; j++)
if(j!=max)
sum = sum + a[j];
System.out.println(sum + (long)Math.pow(2,k)*a[max]);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 10909fe334bd94f8e515f64fa599e9d5 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class Divide_Multiply {
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());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
double nextDouble() {
return Double.parseDouble(next());
}
}
public static void main(String[] args) {
FastReader in = new FastReader();
int t = in.nextInt();
StringBuilder sb = new StringBuilder();
while(t-- > 0) {
int n = in.nextInt();
int [] arr = new int [n];
int max = Integer.MIN_VALUE;
int ind = -1;
int pow = 0;
for (int i = 0; i < n; i++) {
arr[i] = in.nextInt();
ArrayList<Integer> na = fun(arr[i]);
pow += na.get(1);
arr[i] = na.get(0);
}
for (int i = 0; i < n; i++) {
if (max < arr[i]) {
max = arr[i];
ind = i;
}
}
long sum = 0;
for (int i = 0; i < n; i++) {
if (i != ind) {
sum += arr[i];
}
}
sum += (Math.pow(2,pow) * arr[ind]);
sb.append(sum+"\n");
}
System.out.println(sb);
}
static ArrayList<Integer> fun(int n) {
int cnt = 0;
while(n > 1 && n % 2 == 0) {
n/=2;
cnt++;
}
ArrayList<Integer> val = new ArrayList<>();
val.add(n);
val.add(cnt);
return val;
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 07e7f0a3fce5a5ae6928ebee065cbdd2 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.BigInteger;
//code by tishrah_
public class _practise {
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[] ia(int n)
{
int a[]=new int[n];
for(int i=0;i<n;i++)a[i]=nextInt();
return a;
}
int[][] ia(int n , int m)
{
int a[][]=new int[n][m];
for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextInt();
return a;
}
long[][] la(int n , int m)
{
long a[][]=new long[n][m];
for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextLong();
return a;
}
char[][] ca(int n , int m)
{
char a[][]=new char[n][m];
for(int i=0;i<n;i++)
{
String x =next();
for(int j=0;j<m ;j++) a[i][j]=x.charAt(j);
}
return a;
}
long[] la(int n)
{
long a[]=new long[n];
for(int i=0;i<n;i++)a[i]=nextLong();
return a;
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static void sort(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 sort(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);}
public static long sum(long a[])
{long sum=0; for(long i : a) sum+=i; return(sum);}
public static long count(long a[] , long x)
{long c=0; for(long i : a) if(i==x) c++; return(c);}
public static int sum(int a[])
{ int sum=0; for(int i : a) sum+=i; return(sum);}
public static int count(int a[] ,int x)
{int c=0; for(int i : a) if(i==x) c++; return(c);}
public static int count(String s ,char ch)
{int c=0; char x[] = s.toCharArray(); for(char i : x) if(ch==i) c++; return(c);}
public static boolean prime(int n)
{for(int i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
public static boolean prime(long n)
{for(long i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
public static int gcd(int n1, int n2)
{ if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
public static long gcd(long n1, long n2)
{ if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
public static int[] freq(int a[], int n)
{ int f[]=new int[n+1]; for(int i:a) f[i]++; return f;}
public static int[] pos(int a[], int n)
{ int f[]=new int[n+1]; for(int i=0; i<n ;i++) f[a[i]]=i; return f;}
public static int[] rev(int a[])
{
for(int i=0 ; i<(a.length+1)/2;i++)
{
int temp=a[i];
a[i]=a[a.length-1-i];
a[a.length-1-i]=temp;
}
return a;
}
public static long mindig(long n)
{
long ans=9; while(n>0) { if(n%10<ans) ans=n%10; n/=10; }
return ans;
}
public static long maxdig(long n)
{
long ans=0; while(n>0) { if(n%10>ans) ans=n%10; n/=10; }
return ans;
}
public static boolean palin(String s)
{
StringBuilder sb = new StringBuilder();
sb.append(s);
String str=String.valueOf(sb.reverse());
if(s.equals(str))
return true;
else return false;
}
public static int[] f1(int a[])
{
int n = a.length/2;
int x[] = new int[2*n];
for(int i=n ; i<2*n ; i++) x[i-n]=a[i];
for(int i=0 ; i<n ; i++) x[i+n]=a[i];
return x;
}
public static int[] f2(int a[])
{
int n = a.length/2;
int x[] = a.clone();
for(int i=0 ; i<2*n ; i+=2)
{
int temp=x[i];
x[i]=x[i+1];
x[i+1]=temp;
}
return x;
}
public static void main(String args[])
{
FastReader in=new FastReader();
PrintWriter so = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
_practise ob = new _practise();
int T = in.nextInt();
//int T = 1;
tc : while(T-->0)
{
int n = in.nextInt();
long a[] = in.la(n);
sort(a);
long ans=0;
for(int i=0 ; i<n ; i++)
{
long f[] = a.clone();
for(int j=0 ; j<n ; j++)
{
if(j!=i)
{
while(f[j]%2==0)
{
f[j]/=2;
f[i]*=2;
}
}
}
ans=Math.max(sum(f), ans);
}
so.println(ans);
}
so.flush();
/*String s = in.next();
* Arrays.stream(f).min().getAsInt()
* BigInteger f = new BigInteger("1"); 1 ke jagah koi bhi value ho skta jo aap
* initial value banan chahte
int a[] = new int[n];
Stack<Integer> stack = new Stack<Integer>();
Deque<Integer> q = new LinkedList<>(); or Deque<Integer> q = new ArrayDeque<Integer>();
PriorityQueue<Long> pq = new PriorityQueue<Long>();
ArrayList<Integer> al = new ArrayList<Integer>();
StringBuilder sb = new StringBuilder();
HashSet<Integer> st = new LinkedHashSet<Integer>();
Set<Integer> s = new HashSet<Integer>();
Map<Long,Integer> hm = new HashMap<Long, Integer>(); //<key,value>
for(Map.Entry<Integer, Integer> i :hm.entrySet())
HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
so.println("HELLO");
Arrays.sort(a,Comparator.comparingDouble(o->o[0]))
Arrays.sort(a, (aa, bb) -> Integer.compare(aa[1], bb[1]));
Set<String> ts = new TreeSet<>();*/
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | ae0b265e421d1fbdc119b462cef2d796 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;
/*
* Author : Imtiaz Adar
*/
public class x1609A {
public static void main(String[] args) {
FastScanner scan = new FastScanner();
int T = scan.nextInt();
while(T-- > 0) {
int temp = 0;
int num = scan.nextInt();
long[] arr = new long[num];
for(int i = 0; i < arr.length; i++) {
arr[i] = scan.nextLong();
while(arr[i] % 2 == 0) {
arr[i] /= 2;
temp++;
}
}
long maxi = arr[0];
for(Long it : arr)
maxi = Math.max(maxi, it);
long total = 0;
for(Long tot : arr) total += tot;
long finalres = (long)(maxi * Math.pow(2, temp)) + total - maxi;
System.out.println(finalres);
}
}
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());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readIntArray(int size) {
int[] x = new int[size];
for(int i = 0; i < x.length; i++) {
x[i] = nextInt();
}
return x;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c2ef231a48bead8ac9595a3a5a867a06 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.*;
public class Pentagram {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int tc = sc.nextInt();
while(tc-->0){
int n = sc.nextInt();
Long[] arr = new Long[n]; for(int i = 0; i<n; i++)arr[i] = sc.nextLong();
long multi = 1;
for(int i = 0; i<n; i++){
while(arr[i] % 2 == 0){
multi *= 2;
arr[i]/=2;
}
}
Arrays.sort(arr);
arr[n - 1]*=multi;
long ans = 0;
for(long x: arr)ans += x;
pw.println(ans);
}
pw.flush();
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c6184ca05b2eccf1af48ce76a9c0200c | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
public class DivideAndMultiply {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
long a[] = new long[n];
long sum = 0;
int c = 0;
int max = 0;
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
while (a[i] % 2 == 0) {
a[i] /= 2;
c++;
}
if (max < a[i])
max = (int) a[i];
sum += a[i];
}
System.out.println(sum - max + ((long) max << c));
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 5350585a9a8573e1d708a354f45f5c85 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class deltix_21_a {
public static void main(String args[]){
FScanner in = new FScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-->0) {
int n=in.nextInt();
int a[]=in.readArray(n);
long sum=0;
int c=0;
for(int i=0;i<a.length;i++)
{
if(a[i]%2==0)
{
while(a[i]%2==0)
{
c++;
a[i]=a[i]/2;
}
}
}
//out.println(c);
Arrays.sort(a);
long max=a[a.length-1];
for(int i=0;i<n-1;i++)
{
sum+=a[i];
}
while(c!=0)
{
max=max*2;
c--;
}
out.println(sum+max);
}
out.close();
}
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 73ad26d4e2a58daa09b3b811a665e6a8 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.awt.Container;
import java.awt.image.SampleModel;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.concurrent.CountDownLatch;
import javax.naming.TimeLimitExceededException;
import java.io.PrintStream;
import java.security.KeyStore.Entry;
public class Solution {
static class Pair<T,V>{
T first;
V second;
public Pair(T first, V second) {
super();
this.first = first;
this.second = second;
}
}
// public static final boolean LOCAL = System.getProperty("ONLINE_JUDGE")==null;
private static FastScanner fs=new FastScanner();
private static Scanner sc=new Scanner(System.in);
static int upperBound(long[] arr,int l,int r,long x) {
int ans=r+1;
while (l <= r) {
// mid = l + (r - l) / 2;
int mid = (l + r) / 2;
if (arr[mid] >= x) {
ans = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return ans;
}
private static int log2(int n) {
if(n<=1) {
return 0;
}
return 1+log2(n/2);
}
public static void main(String[] args) throws Exception {
int tcr=1;
//tcr = sc.nextInt();
tcr=fs.nextInt();
while(tcr-->0) {
solve(tcr);
}
System.gc();
}
private static void solve(int TC) throws Exception{
int n=fs.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++) {
arr[i]=fs.nextLong();
}
long cnt=0;
for(int i=0;i<n;i++) {
while( arr[i]%2==0 ) {
cnt++;
arr[i]/=2;
}
}
Arrays.parallelSort(arr);
arr[n-1]*=Math.pow(2, cnt);
long sum=0;
for(int i=0;i<n;i++) {
sum+=arr[i];
}
System.out.println(sum);
}
static boolean isPalindrome(ArrayList<Integer> list) {
int i=0,j=list.size()-1;
while(i<j) {
if( !list.get(i).equals( list.get(j) ) ) {
return false;
}
i++;
j--;
}
return true;
}
static class FastScanner {
BufferedReader br;
StringTokenizer st ;
FastScanner(){
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer("");
}
FastScanner(String file) {
try {
br = new BufferedReader(new InputStreamReader(new FileInputStream(file)));
st = new StringTokenizer("");
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
System.out.println("file not found");
e.printStackTrace();
}
}
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());
}
String readLine() throws IOException{
return br.readLine();
}
}
public static boolean isPrime(int n) {
for(int i=2;i<n;i++) {
if(n%i==0) {
return false;
}
}
return true;
}
public static long gcd(long a,long b){
if(b == 0l){
return a;
}
return gcd(b,a%b);
}
public static List<Integer> sieve(){
List<Integer> prime = new ArrayList<>();
int arr[] = new int[100001];
Arrays.fill(arr,1);
arr[1] = 0;
arr[2] = 1;
for(int i=2;i<=100000;i++){
if(arr[i] == 1){
prime.add(i);
for(long j = (i*1l*i);j<100001;j+=i){
arr[(int)j] = 0;
}
}
}
return prime;
}
static boolean isPower(long n,long a){
long log = (long)(Math.log(n)/Math.log(a));
long power = (long)Math.pow(a,log);
if(power == n){return true;}
return false;
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 855dc141e5ec027f973f55724dd32589 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
public class _1609A {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
long[] arr;
long[] brr = new long[n];
long sum = 0;
for (int i = 0; i < n; i++)
brr[i] = sc.nextInt();
long max = -1;
for (int idx = 0; idx < n; idx++) {
arr=Arrays.copyOf(brr,n);
sum=0;
for (int i = 0; i < n; i++) {
if (i != idx) {
while (arr[i] % 2 == 0) {
arr[i] /= 2;
arr[idx] *= 2;
}
sum += arr[i];
}
}
max = Math.max(max, sum + arr[idx]);
}
System.out.println(max);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | cfd1e26765ac67ee049620ed16a4fd81 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class DivideandMultiply {
public static void solve(int n,long[] arr) {
Arrays.sort(arr);
long[] arr1=new long[arr.length];
arr1=arr.clone();
long[] arr2=new long[arr.length];
arr2=arr.clone();
int oddm=-1;
for(int i=n-1;i>=0;i--) {
if(arr[i]%2!=0) {
oddm=i;
break;
}
}
// System.out.println(oddm);
for(int i=0;i<n-1;i++) {
while(arr1[i]%2==0) {
// System.out.println("hi");
arr1[n-1]=arr1[n-1]*2;
arr1[i]=arr1[i]/2;
}
}
long sum=0L;
long sumo=0L;
for(int i=0;i<n;i++) {
sum+=arr1[i];
}
if(oddm!=-1) {
for(int i=0;i<n;i++) {
while(arr2[i]%2==0&&i!=oddm) {
// System.out.println(arr2[i]);
arr2[oddm]=arr2[oddm]*2;
arr2[i]=arr2[i]/2;
}
}
for(int i=0;i<n;i++) {
sum+=arr[2];
}
System.out.println(sumo);
}
System.out.println(Math.max(sumo, sum));
}
public static int gcd (int a, int b) {
if (b == 0)
return a;
else
return gcd (b, a % b);
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in=new Scanner(System.in);
int t=in.nextInt();
for (int i = 0; i < t; i++) {
int n=in.nextInt();
long[] arr=new long[n];
for (int j = 0; j < n; j++) {
arr[j]=in.nextLong();
}
// solve(n, arr);
Arrays.sort(arr);
long res = Integer.MIN_VALUE;
for (int i1 = 0; i1 < n; i1++) {
long ans = arr[i1];
long tot = 0;
for (int j = 0; j < n; j++) {
if (i1 != j) {
long num = arr[j];
if (num % 2 == 0) {
while (num % 2 == 0) {
ans = ans * 2;
num = num / 2;
}
}
tot = tot + num;
}
}
ans = ans + tot;
res = Math.max(res, ans);
}
System.out.println(res);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | fd64da48067606def1d21851a77bf9d9 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Scanner;
public class Main
{
public static void main(String[] args) {
Scanner a = new Scanner(System.in);
int[] multiple = new int[17];
int[] div = new int[17];
for(int i=1;i<multiple.length;i++)
{
int count=0;
int num =i;
while(num%2==0)
{
count++;
num=num/2;
}
div[i] = num;
multiple[i] = count;
}
int test = a.nextInt();
while(test-->0)
{
int size = a.nextInt();
int[] arr = new int[size];
for(int i=0;i<size;i++)
{arr[i] = a.nextInt();}
helper(size,arr,multiple,div,0);
}
}
public static void helper(int size,int[] arr,int[] multiple,int[] div,int max)
{
long ans1 = 0;
for(max=0;max<arr.length;max++) {
int power=0;
long ans=0;
for (int i = 0; i < arr.length; i++) {
if (i != max) {
power += multiple[arr[i]];
ans += div[arr[i]];
}
}
double m = arr[max]*Math.pow(2,power);
ans1 = Math.max(ans1,(long)(m+ans));
}
System.out.println(ans1);
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 09cc08aae2d9c6b173dc2f4a85f1d2ab | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class A {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
int t = sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt();
int[] a = new int[n];
for(int i = 0; i < n; i++)
a[i] = sc.nextInt();
int count = 0;
for(int i = 0; i < n; i++) {
while(a[i] % 2 == 0) {
a[i] /= 2;
count++;
}
}
Arrays.sort(a);
BigInteger res = BigInteger.ONE.shiftLeft(count);
res = res.multiply(BigInteger.valueOf(a[n - 1]));
for(int i = 0; i < n - 1; i++) {
res = res.add(BigInteger.valueOf(a[i]));
}
System.out.println(res);
}
}
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;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 2dff9aed529008af11a80f349df1b783 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class A {
public static void main(String args[]){
FScanner in = new FScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-->0) {
int n=in.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++)
arr[i]=in.nextInt();
Arrays.sort(arr);
long sum=0;
long max2=0;int p=-1;
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
if(arr[i]>max2)
{
max2=arr[i];
p=i;
}
}
long tmp=arr[i];
while(tmp%2==0)
{
tmp/=2;
}
if(tmp!=1&&tmp>max2)
{
max2=tmp;
p=i;
}
}
//out.println(p);
if(p==-1)
p=0;
for(int i=0;i<n;i++)
{
if(i==p)
continue;
if(arr[i]%2==0)
{
while(arr[i]%2==0)
{
arr[i]/=2;
arr[p]*=2;
}
}
}
for(int i=0;i<n;i++)
{
//out.println(arr[i]);
sum+=arr[i];
}
out.println(sum);
}
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 773bd8a70448f455e921759a0a94c4da | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | //package com.company;
import java.io.FileInputStream;
import java.io.InputStream;
import java.util.*;
public class Main {
public static class Dsu{
int[] parent;
Set<Integer>[] rest;
Set<Integer>[] restrictions;
Dsu(int n, int[][] restrictions)
{
parent = new int[n];
rest = new Set[n];
for(int i =0; i < n; i ++)
parent[i]=i;
for(int i =0; i < n; i++)
{
rest[i] = new HashSet<>();
rest[i].add(i);
}
this.restrictions = new Set[n];
for(int i =0; i < n; i++)
{
this.restrictions[i] = new HashSet<>();
}
for(int[] restriction : restrictions)
{
this.restrictions[restriction[0]].add(restriction[1]);
this.restrictions[restriction[1]].add(restriction[0]);
}
}
int find(int son)
{
if(parent[son]==son) return son;
return find(parent[son]);
}
boolean union(int p, int q)
{
int pp = find(p);
int pq = find(q);
if(pp==pq) return true;
if(valid(rest[pp],restrictions[pq]))
{
rest[pp].addAll(rest[pq]);
parent[pq] = pp;
return true;
}
return false;
}
boolean valid(Set<Integer> s1, Set<Integer> s2)
{
for(Integer i : s1)
{
if(s2.contains(i)) return false;
//System.out.println(i);
}
return true;
}
}
public static boolean[] friendRequests(int n, int[][] restrictions, int[][] requests) {
boolean[] ans = new boolean[requests.length];
Dsu dsu = new Dsu(n,restrictions);
int i =0;
for(int[] req : requests)
{
ans[i++] = dsu.union(req[0],req[1]);
}
return ans;
}
public static void main(String[] args) {
// write your code here
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while(t-->0)
{
int n = scanner.nextInt();
long[] arr = new long[n];
for(int i =0; i < n;i++)
arr[i] = scanner.nextLong();
int times = 0;
for(int i =0; i < n;i++)
{
while(arr[i]%2==0)
{
times++;
arr[i]/=2;
}
}
int max =0;
for(int i =0; i < n;i++)
{
if(arr[max]<arr[i])
max=i;
}
for(int i =0; i < times;i++)
arr[max]*=2;
long sol = 0;
for(long ele:arr)
sol += ele;
System.out.println(sol);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | cde3d837b506bb04fc740fda0c2dc5c4 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;
public class a729 {
public static void main(String[] args) {
// try {
BufferedWriter out = new BufferedWriter(
new OutputStreamWriter(System.out));
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
PrintWriter pt = new PrintWriter(System.out);
FastReader sc = new FastReader();
// System.out.println(1);
int t = sc.nextInt();
for(int o = 0 ; o<t;o++){
int n = sc.nextInt();
long[] arr = new long[n];
for(int i = 0 ; i<n;i++) {
arr[i] = sc.nextLong();
}
//Arrays.sort(arr);
long mul = 1;
for(int i = 0 ; i<n;i++) {
while(arr[i]%2==0) {
arr[i] = arr[i]/2;
mul = mul * 2;
}
}
Arrays.sort(arr);
arr[n-1] = arr[n-1] * mul;
long sum = 0;
for(int i = 0 ; i<n;i++) {
sum += arr[i];
}
System.out.println(sum);
//int max = -1;
//int od = 0;
//int id = -1;
//for(int i = 0 ; i<n;i++){
// if(arr[i]%2==1) {
// od++;
// max = Math.max(max, arr[i]);
// id = i;
// }
//}
//
//int ev = n - od;
//long mul = 1;
//for(int i = 0 ; i<n;i++) {
// if(arr[i]%2==0) {
// mul = (long)mul*arr[i];
// }
//}
//long ans = 0;
//if(od==0) {
// ans = mul + n-1;
//}else {
// for(int i = 0 ; i<n;i++) {
// if(i == id) {
// ans += (long)arr[i]*mul;
// }else if(arr[i]%2==0) {
// ans++;
// }else {
// ans+=arr[i];
// }
// }
//}
//System.out.println(ans);
}
//}catch(Exception e) {
// return;
//}
}
}
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 | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 0dd168161c4da5cb50220789d800e86a | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStreamReader;
import java.util.*;
import java.util.Collections;
import java.util.StringTokenizer;
public class Solution{
static FastScanner fs = new FastScanner();
static PrintWriter out = new PrintWriter(System.out);
static int[] fun(int n){
int[] a = new int[2];
int c = 0;
while(n%2 == 0){
c++;
n /= 2;
}
a[0] = c;
a[1] = n;
return a;
}
static void output(){
int n = fs.nextInt();
int[] a = fs.readArray(n);
sort(a);
long ans = 0;
int[][] b = new int[n][];
long powSum = 0;
for(int i=0;i<n;i++){
b[i] = fun(a[i]);
powSum += b[i][0];
}
for(int i=0;i<n;i++){
long sum = 0;
for(int j=0;j<n;j++){
sum += b[j][1];
}
sum += (a[i]*Math.pow((long)2,(long)(powSum-b[i][0]))) - b[i][1];
ans = Math.max(ans,sum);
}
out.println(ans);
}
public static void main(String[] args) {
int T = 1;
T=fs.nextInt();
for (int tt=0; tt<T; tt++) {
output();
out.flush();
}
}
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 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());
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 4bb4070bb8f94905291c2fd68dddd5a4 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int numberOfLines = Integer.parseInt(scanner.nextLine());
for (int i = 0; i < numberOfLines; i++) {
int n = scanner.nextInt();
scanner.nextLine();
long[] numbers = new long[n];
for (int j = 0; j < n; j++) {
numbers[j] = scanner.nextInt();
}
int counter = 0;
for (int j = 0; j < n; j++) {
while (numbers[j] % 2 == 0) {
numbers[j] /= 2;
counter++;
}
}
int index = 0;
long max = 0;
for (int j = 0; j < n; j++) {
if (numbers[j] > max) {
max = numbers[j];
index = j;
}
}
numbers[index] *= Math.pow(2, counter);
long sum = 0;
for (int j = 0; j < n; j++) {
sum += numbers[j];
}
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 9f9802d1c1f11f58c159ab6b4a023d65 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.io.*;
public class s1 {
public static FastScanner scan;
public static PrintWriter out;
public static void main(String[] args) throws Exception {
scan=new FastScanner(System.in);
out=new PrintWriter(System.out);
// int T=1;
int T=scan.nextInt();
while(T-->0) {
int n=scan.nextInt();
long[] a=new long[n];
for(int i=0;i<n;i++) a[i]=scan.nextInt();
long ans=0;
for(int i=0;i<n;i++) {
for(int j=0;j<n;j++) {
if(i!=j) {
while(a[j]%2==0) {
a[j]/=2;
a[i]*=2;
}
}
}
long sum=0;
for(int j=0;j<n;j++) sum+=a[j];
ans=Math.max(ans,sum);
}
out.println(ans);
} out.close();
}
}
class Pair {
int a,b;
Pair(int a,int b) {
this.a=a;
this.b=b;
}
}
class FastScanner {
private InputStream stream;
private byte[] buf=new byte[1024];
private int curChar;
private int numChars;
public FastScanner(InputStream stream) { this.stream=stream; }
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++];
}
boolean isSpaceChar(int c) { return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1; }
boolean isEndline(int c) { return c=='\n'||c=='\r'||c==-1; }
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() { return Double.parseDouble(next()); }
String next() {
int c=read();
while(isSpaceChar(c)) c=read();
StringBuilder res=new StringBuilder();
do {
res.appendCodePoint(c);
c=read();
} while(!isSpaceChar(c));
return res.toString();
}
String nextLine() {
int c=read();
while(isEndline(c)) c=read();
StringBuilder res=new StringBuilder();
do {
res.appendCodePoint(c);
c=read();
} while(!isEndline(c));
return res.toString();
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 6abc89e4d4162fb39da8b9578c34ea96 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | //package codeforce.div2.deltix.Autumn2021;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.ArrayList;
import java.util.List;
import java.util.HashMap;
import java.util.Map;
import java.util.Set;
import java.util.HashSet;
import java.util.stream.LongStream;
import static java.lang.System.out;
import static java.util.stream.Collectors.joining;
/**
* @author pribic (Priyank Doshi)
* @see <a href="https://codeforces.com/contest/1609/problem/A" target="_top">https://codeforces.com/contest/1609/problem/A</a>
* @since 28/11/21 8:09 PM
*/
public class A {
static FastScanner sc = new FastScanner(System.in);
public static void main(String[] args) {
try (PrintWriter out = new PrintWriter(System.out)) {
int T = sc.nextInt();
for (int tt = 1; tt <= T; tt++) {
int n = sc.nextInt();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
Arrays.sort(arr);
long max = 0;
for (int j = 0; j < n; j++) {
long[] c = arr.clone();
for (int i = 0; i < n; i++) {
if (i != j) {
while (c[i] % 2 == 0) {
c[i] = c[i] / 2;
c[j] *= 2;
}
}
}
max = Math.max(max, LongStream.of(c).sum());
}
System.out.println(max);
}
}
}
//2 3 4
//1 24 1
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f), 32768);
}
String next() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return null;
st = new StringTokenizer(s);
}
return st.nextToken();
}
boolean hasMoreTokens() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return false;
st = new StringTokenizer(s);
}
return true;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 09065e42fe2e2ca8f8bc265a03c09200 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.beans.DesignMode;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.concurrent.LinkedBlockingDeque;
import java.util.concurrent.CompletableFuture.AsynchronousCompletionTask;
import org.xml.sax.ErrorHandler;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.io.DataInputStream;
public class Solution {
//TEMPLATE -------------------------------------------------------------------------------------
public static boolean Local(){
try{
return System.getenv("LOCAL_SYS")!=null;
}catch(Exception e){
return false;
}
}
public static boolean LOCAL;
static class FastScanner {
BufferedReader br;
StringTokenizer st ;
FastScanner(){
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer("");
}
FastScanner(String file) {
try{
br = new BufferedReader(new InputStreamReader(new FileInputStream(file)));
st = new StringTokenizer("");
}catch(FileNotFoundException e) {
// TODO Auto-generated catch block
System.out.println("file not found");
e.printStackTrace();
}
}
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());
}
double nextDouble(){
return Double.parseDouble(next());
}
String readLine() throws IOException{
return br.readLine();
}
}
static class Pair<T,X> {
T first;
X second;
Pair(T first,X second){
this.first = first;
this.second = second;
}
@Override
public int hashCode(){
return Objects.hash(first,second);
}
@Override
public boolean equals(Object obj){
return obj.hashCode() == this.hashCode();
}
}
static PrintStream debug = null;
static long mod = (long)(Math.pow(10,9) + 7);
//TEMPLATE -------------------------------------------------------------------------------------END//
public static void main(String[] args) throws Exception {
FastScanner s = new FastScanner();
LOCAL = Local();
//PrintWriter pw = new PrintWriter(System.out);
if(LOCAL){
s = new FastScanner("src/input.txt");
PrintStream o = new PrintStream("src/sampleout.txt");
debug = new PrintStream("src/debug.txt");
System.setOut(o);
// pw = new PrintWriter(o);
} long mod = 1000000007;
int tcr = s.nextInt();
StringBuilder sb = new StringBuilder();
for(int tc=0;tc<tcr;tc++){
int n = s.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++){
arr[i] = s.nextInt();
}
long ans = 0;
int cnt = 0;
int max = 0;
for(int i=0;i<n;i++){
while((arr[i] % 2)== 0){
cnt++;
arr[i] = arr[i]/2;
}
if(arr[i] > max){
max = arr[i];
}
}
for(int i=0;i<n;i++){
ans+=arr[i];
}
ans-=max;
ans += (max * 1l * (Math.pow(2,cnt)));
sb.append(ans+"\n");
}
print(sb.toString());
}
public static List<int[]> print_prime_factors(int n){
List<int[]> list = new ArrayList<>();
for(int i=2;i<=(int)(Math.sqrt(n));i++){
if(n % i == 0){
int cnt = 0;
while( (n % i) == 0){
n = n/i;
cnt++;
}
list.add(new int[]{i,cnt});
}
}
if(n!=1){
list.add(new int[]{n,1});
}
return list;
}
public static List<int[]> prime_factors(int n,List<Integer> sieve){
List<int[]> list = new ArrayList<>();
int index = 0;
while(n > 1 && sieve.get(index) <= Math.sqrt(n)){
int curr = sieve.get(index);
int cnt = 0;
while((n % curr) == 0){
n = n/curr;
cnt++;
}
if(cnt >= 1){
list.add(new int[]{curr,cnt});
}
index++;
}
if(n > 1){
list.add(new int[]{n,1});
}
return list;
}
public static boolean inRange(int r1,int r2,int val){
return ((val >= r1) && (val <= r2));
}
static int len(long num){
return Long.toString(num).length();
}
static long mulmod(long a, long b,long mod)
{
long ans = 0l;
while(b > 0){
long curr = (b & 1l);
if(curr == 1l){
ans = ((ans % mod) + a) % mod;
}
a = (a + a) % mod;
b = b >> 1;
}
return ans;
}
public static void dbg(PrintStream ps,Object... o) throws Exception{
if(ps == null){
return;
}
Debug.dbg(ps,o);
}
public static long modpow(long num,long pow,long mod){
long val = num;
long ans = 1l;
while(pow > 0l){
long bit = pow & 1l;
if(bit == 1){
ans = (ans * (val%mod))%mod;
}
val = (val * val) % mod;
pow = pow >> 1;
}
return ans;
}
public static char get(int n){
return (char)('a' + n);
}
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 int[] sort(int arr[]){
List<Integer> list = new ArrayList<>();
for(int n : arr){list.add(n);}
Collections.sort(list);
for(int i=0;i<arr.length;i++){
arr[i] = list.get(i);
}
return arr;
}
// return the (index + 1)
// where index is the pos of just smaller element
// i.e count of elemets strictly less than num
public static int justSmaller(long arr[],long num){
// System.out.println(num+"@");
int st = 0;
int e = arr.length - 1;
int ans = -1;
while(st <= e){
int mid = (st + e)/2;
if(arr[mid] >= num){
e = mid - 1;
}else{
ans = mid;
st = mid + 1;
}
}
return ans + 1;
}
public static int justSmaller(int arr[],int num){
// System.out.println(num+"@");
int st = 0;
int e = arr.length - 1;
int ans = -1;
while(st <= e){
int mid = (st + e)/2;
if(arr[mid] >= num){
e = mid - 1;
}else{
ans = mid;
st = mid + 1;
}
}
return ans + 1;
}
//return (index of just greater element)
//count of elements smaller than or equal to num
public static int justGreater(long arr[],long num){
int st = 0;
int e = arr.length - 1;
int ans = arr.length;
while(st <= e){
int mid = (st + e)/2;
if(arr[mid] <= num){
st = mid + 1;
}else{
ans = mid;
e = mid - 1;
}
}
return ans;
}
public static int justGreater(int arr[],int num){
int st = 0;
int e = arr.length - 1;
int ans = arr.length;
while(st <= e){
int mid = (st + e)/2;
if(arr[mid] <= num){
st = mid + 1;
}else{
ans = mid;
e = mid - 1;
}
}
return ans;
}
public static void println(Object obj){
System.out.println(obj.toString());
}
public static void print(Object obj){
System.out.print(obj.toString());
}
public static int gcd(int a,int b){
if(b == 0){return a;}
return gcd(b,a%b);
}
public static long gcd(long a,long b){
if(b == 0l){
return a;
}
return gcd(b,a%b);
}
public static int find(int parent[],int v){
if(parent[v] == v){
return v;
}
return parent[v] = find(parent, parent[v]);
}
public static List<Integer> sieve(){
List<Integer> prime = new ArrayList<>();
int arr[] = new int[1000001];
Arrays.fill(arr,1);
arr[1] = 0;
arr[2] = 1;
for(int i=2;i<1000001;i++){
if(arr[i] == 1){
prime.add(i);
for(long j = (i*1l*i);j<1000001;j+=i){
arr[(int)j] = 0;
}
}
}
return prime;
}
static boolean isPower(long n,long a){
long log = (long)(Math.log(n)/Math.log(a));
long power = (long)Math.pow(a,log);
if(power == n){return true;}
return false;
}
private static int mergeAndCount(int[] arr, int l,int m, int r)
{
// Left subarray
int[] left = Arrays.copyOfRange(arr, l, m + 1);
// Right subarray
int[] right = Arrays.copyOfRange(arr, m + 1, r + 1);
int i = 0, j = 0, k = l, swaps = 0;
while (i < left.length && j < right.length) {
if (left[i] <= right[j])
arr[k++] = left[i++];
else {
arr[k++] = right[j++];
swaps += (m + 1) - (l + i);
}
}
while (i < left.length)
arr[k++] = left[i++];
while (j < right.length)
arr[k++] = right[j++];
return swaps;
}
// Merge sort function
private static int mergeSortAndCount(int[] arr, int l,int r)
{
// Keeps track of the inversion count at a
// particular node of the recursion tree
int count = 0;
if (l < r) {
int m = (l + r) / 2;
// Total inversion count = left subarray count
// + right subarray count + merge count
// Left subarray count
count += mergeSortAndCount(arr, l, m);
// Right subarray count
count += mergeSortAndCount(arr, m + 1, r);
// Merge count
count += mergeAndCount(arr, l, m, r);
}
return count;
}
static class Debug{
//change to System.getProperty("ONLINE_JUDGE")==null; for CodeForces
public static final boolean LOCAL = System.getProperty("ONLINE_JUDGE")==null;
private static <T> String ts(T t) {
if(t==null) {
return "null";
}
try {
return ts((Iterable) t);
}catch(ClassCastException e) {
if(t instanceof int[]) {
String s = Arrays.toString((int[]) t);
return "{"+s.substring(1, s.length()-1)+"}\n";
}else if(t instanceof long[]) {
String s = Arrays.toString((long[]) t);
return "{"+s.substring(1, s.length()-1)+"}\n";
}else if(t instanceof char[]) {
String s = Arrays.toString((char[]) t);
return "{"+s.substring(1, s.length()-1)+"}\n";
}else if(t instanceof double[]) {
String s = Arrays.toString((double[]) t);
return "{"+s.substring(1, s.length()-1)+"}\n";
}else if(t instanceof boolean[]) {
String s = Arrays.toString((boolean[]) t);
return "{"+s.substring(1, s.length()-1)+"}\n";
}
try {
return ts((Object[]) t);
}catch(ClassCastException e1) {
return t.toString();
}
}
}
private static <T> String ts(T[] arr) {
StringBuilder ret = new StringBuilder();
ret.append("{");
boolean first = true;
for(T t: arr) {
if(!first) {
ret.append(", ");
}
first = false;
ret.append(ts(t));
}
ret.append("}");
return ret.toString();
}
private static <T> String ts(Iterable<T> iter) {
StringBuilder ret = new StringBuilder();
ret.append("{");
boolean first = true;
for(T t: iter) {
if(!first) {
ret.append(", ");
}
first = false;
ret.append(ts(t));
}
ret.append("}\n");
return ret.toString();
}
public static void dbg(PrintStream ps,Object... o) throws Exception {
if(LOCAL) {
System.setErr(ps);
System.err.print("Line #"+Thread.currentThread().getStackTrace()[2].getLineNumber()+": [\n");
for(int i = 0; i<o.length; i++) {
if(i!=0) {
System.err.print(", ");
}
System.err.print(ts(o[i]));
}
System.err.println("]");
}
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 5b0bf687d03d96531bdd49b7714127fe | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 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() {
int n = i();
int[] a = input(n);
long evenMul = 1;
for (int i = 0; i < n; i++) {
while (a[i] % 2 == 0) {
a[i] /= 2;
evenMul *= 2;
}
}
long ans = 0;
Arrays.sort(a);
for (int i = 0; i < n - 1; i++) {
ans += a[i];
}
ans += a[n - 1] * evenMul;
out.println(ans);
}
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 long[] pre(long[] 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 void swap(char A[], int a, int b) {
char 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 void reverse(int[] arr) {
for (int i = 0; i < arr.length / 2; i++) {
int tmp = arr[i];
arr[arr.length - 1 - i] = tmp;
arr[i] = arr[arr.length - 1 - i];
}
}
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;
int 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 val;
public Node(int val) {
this.val = val;
}
}
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(0);
}
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;
}
return new Node(a.val | b.val);
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 402eacfcb3ddf0ab1c7a578ac98ea25e | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
public class A {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
int n = sc.nextInt();
long sum = 0;
int k = 0;
long max = 0;
for (int j = 0; j < n; j++) {
long s = sc.nextInt();
if (s % 2 == 0) {
while (s % 2 == 0) {
k++;
s /= 2;
}
}
max = Math.max(max, s);
sum += s;
}
sum -= max;
for (int j = 0; j < k; j++) {
max *= 2;
}
sum += max;
System.out.println(sum);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | b4605853fd9231abe4212e73f152fea5 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 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.StringTokenizer;
public class A {
static BufferedReader br;
static StringTokenizer st;
static PrintWriter pw;
static String nextToken() {
try {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(br.readLine());
}
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
static int nextInt() {
return Integer.parseInt(nextToken());
}
static long nextLong() {
return Long.parseLong(nextToken());
}
static double nextDouble() {
return Double.parseDouble(nextToken());
}
static String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
throw new IllegalArgumentException();
}
}
static char nextChar() {
try {
return (char) br.read();
} catch (IOException e) {
throw new IllegalArgumentException(e);
}
}
public static void main(String[] args) throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(System.out);
int t = 1;
t = nextInt();
while (t-- > 0) {
solve();
}
pw.close();
}
private static void solve() {
int n= nextInt();
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextLong();
}
int k = 0;
for (int i = 0; i < arr.length; i++) {
while (arr[i]%2==0) {
k++;
arr[i]/=2;
}
}
Arrays.sort(arr);
while (k-->0) {
arr[n-1]*=2;
}
long sum = 0;
for (int i = 0; i < n; i++) {
sum+=arr[i];
}
pw.println(sum);
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 268e7cc3b79a743cfebce531a3528dd0 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st;
static PrintWriter pr = new PrintWriter(System.out);
static String readLine() throws IOException {
return br.readLine();
}
static String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(readLine());
return st.nextToken();
}
static int readInt() throws IOException {
return Integer.parseInt(next());
}
static long readLong() throws IOException {
return Long.parseLong(next());
}
static double readDouble() throws IOException {
return Double.parseDouble(next());
}
static char readChar() throws IOException {
return next().charAt(0);
}
static class Pair implements Comparable<Pair> {
int f, s;
Pair(int f, int s) {
this.f = f; this.s = s;
}
public int compareTo(Pair other) {
if (this.f != other.f) return this.f - other.f;
return this.s - other.s;
}
}
static void solve() throws IOException {
int n = readInt(), max = 0;
long a[] = new long[n + 1], mul = 1;
for (int i = 1; i <= n; ++i) {
a[i] = readInt();
while (a[i]%2 == 0) {
a[i] /= 2;
mul *= 2;
}
if (a[i] > a[max]) max = i;
}
long ans = 0;
for (int i = 1; i <= n; ++i) {
if (i == max) ans += a[i]*mul;
else ans += a[i];
}
pr.println(ans);
}
public static void main(String[] args) throws IOException {
//solve();
for (int t = readInt(); t > 0; --t) solve();
pr.close();
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 9a53dc840e06aa08bc9dc8a8c629b5f6 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | /**
* @author vivek
* programming is thinking, not typing
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class A {
private static void solveTC(int __) {
/* For Google */
// ans.append("Case #").append(__).append(": ");
//code start
int n= scn.nextInt();
long[] arr = scn.nextLongArray(n);
long max = Long.MIN_VALUE;
for (int maxInd = 0; maxInd < n; maxInd++) {
int count = 0;
for (int i = 0; i < n; i++) {
if (i != maxInd) {
while (arr[i] % 2 == 0) {
count++;
arr[i] /= 2;
}
}
}
arr[maxInd] *= 1L << count;
long ans = 0;
for (long ele : arr) {
ans += ele;
}
max = Math.max(ans,max);
}
print(max);
//code end
print("\n");
}
public static void main(String[] args) {
scn = new Scanner();
ans = new StringBuilder();
int t = scn.nextInt();
// int t = 1;
// int limit= ;
// sieve(limit);
/*
try {
System.setOut(new PrintStream(new File("file_i_o\\output.txt")));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
*/
for (int i = 1; i <= t; i++) {
solveTC(i);
}
System.out.print(ans);
}
//Stuff for prime start
/**
* sorting algos
*/
private static void sort(int[] arr) {
ArrayList<Integer> li = new ArrayList<>(arr.length);
for (int ele : arr) li.add(ele);
Collections.sort(li);
for (int i = 0; i < li.size(); i++) {
arr[i] = li.get(i);
}
}
private static void sort(long[] arr) {
ArrayList<Long> li = new ArrayList<>(arr.length);
for (long ele : arr) li.add(ele);
Collections.sort(li);
for (int i = 0; i < li.size(); i++) {
arr[i] = li.get(i);
}
}
private static void sort(float[] arr) {
ArrayList<Float> li = new ArrayList<>(arr.length);
for (float ele : arr) li.add(ele);
Collections.sort(li);
for (int i = 0; i < li.size(); i++) {
arr[i] = li.get(i);
}
}
private static void sort(double[] arr) {
ArrayList<Double> li = new ArrayList<>(arr.length);
for (double ele : arr) li.add(ele);
Collections.sort(li);
for (int i = 0; i < li.size(); i++) {
arr[i] = li.get(i);
}
}
/**
* List containing prime numbers <br>
* <b>i<sup>th</sup></b> position contains <b>i<sup>th</sup></b> prime number <br>
* 0th index is <b>null</b>
*/
private static ArrayList<Integer> listOfPrimes;
/**
* query <b>i<sup>th</sup></b> element to get if its prime of not
*/
private static boolean[] isPrime;
/**
* Performs Sieve of Erathosnesis and initialise isPrime array and listOfPrimes list
*
* @param limit the number till which sieve is to be performed
*/
private static void sieve(int limit) {
listOfPrimes = new ArrayList<>();
listOfPrimes.add(null);
boolean[] array = new boolean[limit + 1];
Arrays.fill(array, true);
array[0] = false;
array[1] = false;
for (int i = 2; i <= limit; i++) {
if (array[i]) {
for (int j = i * i; j <= limit; j += i) {
array[j] = false;
}
}
}
isPrime = array;
for (int i = 0; i <= limit; i++) {
if (array[i]) {
listOfPrimes.add(i);
}
}
}
//stuff for prime end
/**
* Calculates the Least Common Multiple of two numbers
*
* @param a First number
* @param b Second Number
* @return Least Common Multiple of <b>a</b> and <b>b</b>
*/
private static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
/**
* Calculates the Greatest Common Divisor of two numbers
*
* @param a First number
* @param b Second Number
* @return Greatest Common Divisor of <b>a</b> and <b>b</b>
*/
private static long gcd(long a, long b) {
return (b == 0) ? a : gcd(b, a % b);
}
static void print(Object obj) {
ans.append(obj.toString());
}
static Scanner scn;
static StringBuilder ans;
//Fast Scanner
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() {
br = new BufferedReader(new
InputStreamReader(System.in));
/*
try {
br = new BufferedReader(new
InputStreamReader(new FileInputStream(new File("file_i_o\\input.txt"))));
} catch (FileNotFoundException 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[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
Integer[] nextIntegerArray(int n) {
Integer[] array = new Integer[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; i++) {
array[i] = nextLong();
}
return array;
}
String[] nextStringArray() {
return nextLine().split(" ");
}
String[] nextStringArray(int n) {
String[] array = new String[n];
for (int i = 0; i < n; i++) {
array[i] = next();
}
return array;
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | c77ded018ecbfb7af5aaab89357ef289 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int tc = 0; tc < t; ++tc) {
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < a.length; ++i) {
a[i] = sc.nextInt();
}
System.out.println(solve(a));
}
sc.close();
}
static long solve(int[] a) {
int exp2 = 0;
for (int i = 0; i < a.length; ++i) {
while (a[i] % 2 == 0) {
a[i] /= 2;
++exp2;
}
}
long[] sorted =
Arrays.stream(a).boxed().sorted(Comparator.reverseOrder()).mapToLong(x -> x).toArray();
for (int i = 0; i < exp2; ++i) {
sorted[0] *= 2;
}
return Arrays.stream(sorted).sum();
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 868fe21ec54870c866044b05b7e2fe97 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.*;
import static java.lang.System.out;
import static java.lang.Math.*;
public class pre267 {
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;
}
}
static class MultiSet<K> {
TreeMap<K, Integer> map;
MultiSet() {
map = new TreeMap<>();
}
void add(K a) {
map.put(a, map.getOrDefault(a, 0) + 1);
}
boolean contains(K a) {
return map.containsKey(a);
}
void remove(K a) {
map.put(a, map.get(a) - 1);
if (map.get(a) == 0) map.remove(a);
}
int occrence(K a) {
return map.get(a);
}
K floor(K a) {
return map.floorKey(a);
}
K ceil(K a) {
return map.ceilingKey(a);
}
@Override
public String toString() {
ArrayList<K> set = new ArrayList<>();
for (Map.Entry<K, Integer> i : map.entrySet()) {
for (int j = 1; j <= i.getValue(); j++) set.add(i.getKey());
}
return set.toString();
}
}
static class Pair<K, V> {
K value1;
V value2;
Pair(K a, V b) {
value1 = a;
value2 = b;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof Pair)) return false;
Pair<?, ?> p = (Pair<?, ?>) o;
return Objects.equals(this.value1, p.value1) && Objects.equals(this.value2, p.value2);
}
@Override
public int hashCode() {
int result = this.value1.hashCode();
result = 31 * result + this.value2.hashCode();
return result;
}
@Override
public String toString() {
return ("[" + value1 + " <=> " + value2 + "]");
}
}
static ArrayList<Integer> primes;
static void setPrimes(int n) {
boolean p[] = new boolean[n];
Arrays.fill(p, true);
for (int i = 2; i < p.length; i++) {
if (p[i]) {
primes.add(i);
for (int j = i * 2; j < p.length; j += i) p[j] = false;
}
}
}
static int mod = (int) (1e9 + 7);
static int gcd(int a, int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
public static void main(String args[]) {
FastReader obj = new FastReader();
int tc = obj.nextInt();
while(tc--!=0){
int n = obj.nextInt(); long arr[] = new long[n];
for(int i=0;i<n;i++) arr[i] = obj.nextInt();
int two = 0;
for(int i=0;i<n;i++){
while(arr[i]%2==0){
arr[i]/=2;
two++;
}
}
//out.println(Arrays.toString(arr)+" "+two);
Arrays.sort(arr);
arr[n-1] *= (long)(pow(2,two));
for(int i=0;i<n-1;i++) arr[n-1]+=arr[i];
out.println(arr[n-1]);
}
}
}
| Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | 437d6fc6575a2e15bc5e3ab21444b2b2 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | // package faltu;
import java.util.*;
import java.util.Map.Entry;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class Main {
// ***********************MATHS--STARTS************************************************* //
private static ArrayList<Long> get_divisor(long x) {
ArrayList<Long>a=new ArrayList<Long>();
for(long i=1;i*i<=x;i++) {
if(x%i==0) {
a.add((long) i);
if(x/i!=i)a.add(x/i);
}
}
return a;
}
static long[] sieve;
static long[] smallestPrime;
public static void sieve()
{
int n=4000000+1;
sieve=new long[n];
smallestPrime=new long[n];
sieve[0]=1;
sieve[1]=1;
for(int i=2;i<n;i++){
sieve[i]=i;
smallestPrime[i]=i;
}
for(int i=2;i*i<n;i++){
if(sieve[i]==i){
for(int j=i*i;j<n;j+=i){
if(sieve[j]==j)sieve[j]=1;
if(smallestPrime[j]==j||smallestPrime[j]>i)smallestPrime[j]=i;
}
}
}
}
static long nCr(long n,long r,long MOD) {
if(n<r)return 0;
if(r==0)return 1;
return fact[(int) n]*mod_inv(fact[(int) r],MOD)%MOD*mod_inv(fact[(int) (n-r)],MOD)%MOD;
}
static long[]fact;
static void computeFact(long n,long MOD) {
fact=new long[(int)n+1];
fact[0]=1;
for(int i=1;i<=n;i++)fact[i]=(fact[i-1]*i%MOD)%MOD;
}
static long bin_expo(long a,long b,long MOD) {
if(b == 0)return 1;
long ans = bin_expo(a,b/2,MOD);
ans = (ans*ans)%MOD;
if(b % 2!=0){
ans = (ans*a)%MOD;
}
return ans%MOD;
}
static long mod_add(long a, long b, long m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
static long mod_mul(long a, long b, long m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
static long mod_sub(long a, long b, long m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
static long mod_inv(long n,long p) {return bin_expo(n,p-2,p);}
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;}
static long min(long x,long y) {return Math.min(x, y);}static long max(long x,long y) {return Math.max(x, y);}
static int min(int x,int y) {return Math.min(x, y);}static int max(int x,int y) {return Math.max(x, y);}
static ArrayList<String>powof2s;
static void powof2S() {
long i=1;
while(i<(long)2e18) {
powof2s.add(String.valueOf(i));
i*=2;
}
}
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 boolean coprime(int a, long l){return (gcd(a, l) == 1);}
// ****************************MATHS-ENDS*****************************************************
// ***********************BINARY-SEARCH STARTS***********************************************
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 int lowerBound(ArrayList<Integer> ar,int k){
int s=0,e=ar.size();
while (s!=e){
int mid = s+e>>1;
if (ar.get(mid) <k)s=mid+1;
else e=mid;
}
if(s==ar.size())return -1;
return s;
}
public static int upperBound(ArrayList<Integer> ar,int k){
int s=0,e=ar.size();
while (s!=e){
int mid = s+e>>1;
if (ar.get(mid) <=k)s=mid+1;
else e=mid;
}
if(s==ar.size())return -1;
return s;
}
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 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
}
public static int searchindex(long arr[], long t){int index = Arrays.binarySearch(arr, t);return (index < 0) ? -1 : index;}
public static long[] sort(long[] a) {ArrayList<Long> al = new ArrayList<>();for(int i=0;i<a.length;i++) al.add(a[i]);Collections.sort(al);for(int i=0;i<a.length;i++) a[i]=al.get(i);return a;}
public static int[] sort(int[] a) {ArrayList<Integer> al = new ArrayList<>();for(int i=0;i<a.length;i++) al.add(a[i]);Collections.sort(al);for(int i=0;i<a.length;i++) a[i]=al.get(i);return a;}
// *******************************BINARY-SEARCH ENDS***********************************************
// *********************************GRAPHS-STARTS****************************************************
// *******----SEGMENT 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];
}
long query(int[]arr,int[]tree,int start,int end,int treeNode,int qleft,int qright) {
if(start>=qleft&&end<=qright)return tree[treeNode];
if(start>qright||end<qleft)return 0;
int mid=(start+end)/2;
long valLeft=query(arr,tree,start,mid-1,treeNode*2,qleft,qright);
long valRight=query(arr,tree,mid+1,end,treeNode*2+1,qleft,qright);
return valLeft+valRight;
}
// -------------ENDS---------------
//***********************DSU IMPLEMENT START*************************
static int parent[];
static int rank[];
static int[]Size;
static void makeSet(int n){
parent=new int[n];
rank=new int[n];
Size=new int[n];
for(int i=0;i<n;i++){
parent[i]=i;
rank[i]=0;
Size[i]=1;
}
}
static void union(int u,int v){
u=findpar(u);
v=findpar(v);
if(rank[u]<rank[v]) {
parent[u]=v;
Size[v]+=Size[u];
}
else if(rank[v]<rank[u]) {
parent[v]=u;
Size[u]+=Size[v];
}
else{
parent[v]=u;
rank[u]++;
Size[u]+=Size[v];
}
}
private static int findpar(int node){
if(node==parent[node])return node;
return parent[node]=findpar(parent[node]);
}
// *********************DSU IMPLEMENT ENDS*************************
// ****__________PRIMS ALGO______________________****
private static int prim(ArrayList<node>[] adj,int N,int node) {
int key[] = new int[N+1];
int parent[] = new int[N+1];
boolean mstSet[] = new boolean[N+1];
for(int i = 0;i<N;i++) {
key[i] = 100000000;
mstSet[i] = false;
}
PriorityQueue<node> pq = new PriorityQueue<node>(N, new node());
key[node] = 0;
parent[node] = -1;
pq.add(new node( node,key[node]));
for(int i = 0;i<N-1;i++) {
int u = pq.poll().getV();
mstSet[u] = true;
for(node it: adj[u]) {
if(mstSet[it.getV()] == false && it.getW() < key[it.getV()]) {
parent[it.getV()] = u;
key[it.getV()] = (int) it.getW();
pq.add(new node(it.getV(), key[it.getV()]));
}
}
}
int sum=0;
for(int i=1;i<N;i++) {
System.out.println(key[i]);
sum+=key[i];
}
System.out.println(sum);
return sum;
}
// ****____________DIJKSTRAS ALGO___________****
static int[]dist;
static int dijkstra(int u,int n,ArrayList<node>adj[]) {
dist=new int[n];
Arrays.fill(dist,Integer.MAX_VALUE);
dist[u]=0;
PriorityQueue<node>pq=new PriorityQueue<node>(new node());
pq.add(new node(u,0));
while(!pq.isEmpty()) {
node v=pq.poll();
for(node it:adj[v.getV()]) {
if(dist[it.getV()]>it.getW()+dist[v.getV()]) {
dist[it.getV()]=(int) (it.getW()+dist[v.getV()]);
pq.add(new node(it.getV(),dist[it.getV()]));
}
}
}
int sum=0;
for(int i=1;i<n;i++){
System.out.println(dist[i]);
sum+=dist[i];
}
return sum;
}
private static void setGraph(int n,int m){
vis=new boolean[n+1];
indeg=new int[n+1];
// adj=new ArrayList<ArrayList<Integer>>();
// for(int i=0;i<=n;i++)adj.add(new ArrayList<>());
// for(int i=0;i<m;i++){
// int u=s.nextInt(),v=s.nextInt();
// adj.get(u).add(v);
// adj.get(v).add(u);
// }
adj=new ArrayList[n+1];
// backadj=new ArrayList[n+1];
for(int i=0;i<=n;i++){
adj[i]=new ArrayList<Integer>();
// backadj[i]=new ArrayList<Integer>();
}
for(int i=0;i<m;i++){
int u=s.nextInt(),v=s.nextInt();
adj[u].add(v);
adj[v].add(u);
// backadj[v].add(u);
indeg[v]++;
indeg[u]++;
}
// weighted adj
// adj=new ArrayList[n+1];
// for(int i=0;i<=n;i++){
// adj[i]=new ArrayList<node>();
// }
// for(int i=0;i<m;i++){
// int u=s.nextInt(),v=s.nextInt();
// long w=s.nextInt();
// adj[u].add(new node(v,w));
//// adj[v].add(new node(u,w));
// }
}
// *********************************GRAPHS-ENDS****************************************************
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}}; //d-u-r-l
static long MOD=(long) (1e9+7);
static int prebitsum[][];
static boolean[] vis;
static int[]indeg;
// static ArrayList<ArrayList<Integer>>adj;
static ArrayList<Integer> adj[];
static ArrayList<Integer> backadj[];
static FastReader s = new FastReader(System.in);
public static void main(String[] args) throws IOException
{
// sieve();
// computeFact((int)1e7+1,MOD);
// prebitsum=new int[2147483648][31];
// presumbit(prebitsum);
// powof2S();
//
// try {
int tt = s.nextInt();
// int tt=1;
for(int i=1;i<=tt;i++) {
solver();
}
// catch(Exception e) {return;}
}
private static void solver() {
int n=s.nextInt();
int[]a=new int[n];
long max=Integer.MIN_VALUE;
long maxInd=-1;
long power2=1;
for(int i=0;i<n;i++) {
a[i]=s.nextInt();
while(a[i]%2==0) {
power2*=2;
a[i]/=2;
}
if(a[i]>max) {
max=a[i];
maxInd=i;
}
}
// printA(a);
// System.out.println(power2);
long ans=0;
for(int i=0;i<n;i++) {
if(i==maxInd)ans+=a[i]*power2;
else ans+=a[i];
}
System.out.println(ans);
}
/* *********************BITS && TOOLS &&DEBUG STARTS***********************************************/
static boolean issafe(int i, int j, int r,int c,boolean[][]vis){
if (i < 0 || j < 0 || i >= r || j >= c||vis[i][j]==true)return false;
else return true;
}
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 String revStr(String str){String input = str;StringBuilder input1 = new StringBuilder();input1.append(input);input1.reverse();return input1.toString();}
static void printA(long[] x) {for(int i=0;i<x.length;i++)System.out.print(x[i]+" ");System.out.println();}
static void printA(int[] x) {for(int i=0;i<x.length;i++)System.out.print(x[i]+" ");System.out.println();}
static void pc2d(boolean[][] vis) {
int n=vis.length;
int m=vis[0].length;
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
System.out.print(vis[i][j]+" ");
}
System.out.println();
}
}
static void pi2d(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 p1d(int[]a) {
for(int i=0;i<a.length;i++)System.out.print(a[i]+" ");
System.out.println();
}
// *****************BITS && TOOLS &&DEBUG ENDS***********************************************
}
// **************************I/O*************************
class FastReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public FastReader(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());}
public double nextDouble() {return Double.parseDouble(next());}
public String nextLine() {String str = "";try {str = reader.readLine();}catch (IOException e) {e.printStackTrace();}return str;}
}
class dsu{
int n;
static int parent[];
static int rank[];
static int[]Size;
public dsu(int n) {this.n=n;this.parent=new int[n];this.rank=new int[n];this.Size=new int[n];
for(int i=0;i<n;i++){parent[i]=i;rank[i]=0;Size[i]=1;}
}
static int findpar(int node) {if(node==parent[node])return node;return parent[node]=findpar(parent[node]);}
static void union(int u,int v){
u=findpar(u);v=findpar(v);
if(u!=v) {
if(rank[u]<rank[v]) {parent[u]=v;Size[v]+=Size[u];}
else if(rank[v]<rank[u]) {parent[v]=u;Size[u]+=Size[v];}
else{parent[v]=u;rank[u]++;Size[u]+=Size[v];}
}
}
}
class pair{
int x;int y;
long u,v;
public pair(int x,int y){this.x=x;this.y=y;}
public pair(long u,long v) {this.u=u;this.v=v;}
}
class node implements Comparator<node>{
private int v;
private long w;
node(int _v, long _w) { v = _v; w = _w; }
node() {}
int getV() { return v; }
long getW() { return w; }
@Override
public int compare(node node1, node node2) {
if (node1.w < node2.w) return -1;
if (node1.w > node2.w) return 1;
return 0;
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | cba79ff96fe88da28968798eeb0e35d1 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.StringTokenizer;
public class Divide_and_Multiply
{
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();
}
}
public static void main(String Args[]) throws java.lang.Exception
{
try
{
Reader obj = new Reader();
int t = obj.nextInt();
while (t > 0)
{
t--;
int n = obj.nextInt();
long arr[] = new long[n];
int i;
long sum = 0;
for (i=0;i<n;i++)
{
arr[i] = obj.nextLong();
}
Arrays.sort (arr);
for (i=0;i<n-1;i++)
{
if (arr[i] % 2 == 0)
{
int c = 0;
while (arr[i] % 2 == 0)
{
arr[i] /= 2;
c++;
}
sum += arr[i];
arr[n - 1] = (long)Math.pow (2, c) * arr[n - 1];
}
else
{
sum += arr[i];
}
}
sum += arr[n-1];
long s = 0, max = 0;
for (i=0;i<n-1;i++)
{
max = Math.max (max, arr[i]);
s += arr[i];
}
s -= max;
while (arr[n - 1] % 2 == 0)
{
max *= 2;
arr[n-1] /= 2;
}
s += max + arr[n-1];
System.out.println (Math.max(sum, s));
}
}catch(Exception e){
return;
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | cda4d8c8fcd3058992544d688b6dffd6 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class divMul
{
public static void main (String[] args) throws java.lang.Exception
{
try {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int n=sc.nextInt();
int a[]=new int[n];
int ct=0;int sum=0;int max=0;
for (int i=0;i<n;i++) {
a[i]=sc.nextInt();
while(a[i]%2==0)
{
ct++;
a[i]/=2;
}
if(a[i]>max)
max=a[i];
sum+=a[i];
}
long mx=max*(long)(Math.pow(2,ct));
System.out.println((sum-max)+mx);
}
} catch(Exception e) {
return;
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output | |
PASSED | cb8bdf8c75c6eaa8a2798cd732574f90 | train_110.jsonl | 1638110100 | William has array of $$$n$$$ numbers $$$a_1, a_2, \dots, a_n$$$. He can perform the following sequence of operations any number of times: Pick any two items from array $$$a_i$$$ and $$$a_j$$$, where $$$a_i$$$ must be a multiple of $$$2$$$ $$$a_i = \frac{a_i}{2}$$$ $$$a_j = a_j \cdot 2$$$ Help William find the maximal sum of array elements, which he can get by performing the sequence of operations described above. | 256 megabytes |
import java.util.*;
public class solution {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
int i=0;
long count=0;long res=0;
for(i=0;i<n;i++)
{ while(arr[i]%2==0)
{count++;
arr[i]=arr[i]/2;
}
}
Arrays.sort(arr);
for(int k=0;k<n-1;k++)
{
res=res+arr[k];
}
res+= arr[n-1]*(long)Math.pow(2,count) ;
System.out.println(res);
}
}
} | Java | ["5\n3\n6 4 2\n5\n1 2 3 4 5\n1\n10\n3\n2 3 4\n15\n8 8 8 8 8 8 8 8 8 8 8 8 8 8 8"] | 1 second | ["50\n46\n10\n26\n35184372088846"] | NoteIn the first example test case the optimal sequence would be: Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{4}{2} = 2$$$ and $$$a_1 = 6 \cdot 2 = 12$$$, making the array look as: [12, 2, 2]. Pick $$$i = 2$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_2 = \frac{2}{2} = 1$$$ and $$$a_1 = 12 \cdot 2 = 24$$$, making the array look as: [24, 1, 2]. Pick $$$i = 3$$$ and $$$j = 1$$$. After performing a sequence of operations $$$a_3 = \frac{2}{2} = 1$$$ and $$$a_1 = 24 \cdot 2 = 48$$$, making the array look as: [48, 1, 1]. The final answer $$$48 + 1 + 1 = 50$$$.In the third example test case there is no way to change the sum of elements, so the answer is $$$10$$$. | Java 8 | standard input | [
"greedy",
"implementation",
"math",
"number theory"
] | f5de1e9b059bddf8f8dd46c18ce12683 | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows. The first line of each test case contains an integer $$$n$$$ $$$(1 \le n \le 15)$$$, the number of elements in William's array. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i < 16)$$$, the contents of William's array. | 900 | For each test case output the maximal sum of array elements after performing an optimal sequence of operations. | standard output |
Subsets and Splits