exec_outcome
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32
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stringclasses 111
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10
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3.8k
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65.5k
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2.37k
⌀ | difficulty
int64 -1
3.5k
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
PASSED | fab7fd23951252ccc3f41471114e632c | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 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 Codechef
{
public static void main (String[] args) throws IOException
{
BufferedReader f=new BufferedReader(new InputStreamReader(System.in));
PrintWriter out=new PrintWriter(System.out);
int t=Integer.parseInt(f.readLine());
while(t-->0){
StringTokenizer st=new StringTokenizer(f.readLine());
int a=Integer.parseInt(st.nextToken());
int b=Integer.parseInt(st.nextToken());
int c=Integer.parseInt(st.nextToken());
int val=Math.abs(b-c)+c-1;
if(val<a-1){
out.println(2);
}
else if(val>a-1){
out.println(1);
}
else{
out.println(3);
}
}
out.close();
f.close();
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | aca4b1ee79f63eb35b2f3d428eb26af3 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class a {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int ci = 0; ci < t; ci++) {
long a = in.nextLong();
long b = in.nextLong();
long c = in.nextLong();
long ax = Math.abs(a - 1);
long bx = Math.abs(b - c) + Math.abs(c - 1);
System.out.println(ax < bx ? 1 : (ax > bx ? 2 : 3));
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 610225ca3f077d35f01e17d469c53efb | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes |
import java.util.Scanner;
public class SolutionA{
static int twoElevators(int a,int b,int c){
int counta=a-1;
int countb=Math.abs(b-c)+(c-1); //java.lang.Math
if(counta<countb) return 1;
if(counta>countb) return 2;
return 3;
}
public static void main(String args[]){
Scanner input=new Scanner(System.in);
int n=Integer.parseInt(input.nextLine());
for(int i=0;i<n;i++){
System.out.println(twoElevators(input.nextInt(),input.nextInt(),input.nextInt()));
input.nextLine();
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 345accdde12e0969022bc6c3cc2c6753 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class ProbB {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long t = sc.nextLong();
while(t-->0) {
long a = sc.nextLong();
long b = sc.nextLong();
long c = sc.nextLong();
//long x = a-1;
long y = Math.abs(c-b)+c;
if(a==y) {
System.out.println(3);
}
else if(a<y) {
System.out.println(1);
}
else {
System.out.println(2);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 2742998fb497fd744b59e90e371fda75 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | // package codeforce;
import java.util.*;
import java.net.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.io.*;
public class A {
static class Node {
int id1;
int id2;
Node(int v1, int w1){
this.id1= v1;
this.id2=w1;
}
Node(){}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try{
str = br.readLine();
}
catch (IOException e){
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
}
static boolean[] seiveofEratoSthenes(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=j+i) {
isPrime[j]= false;
}
}
return isPrime;
}
static int in = 2;
// Function check whether a number
// is prime or not
public static boolean isPrime(int n)
{
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static class SortingComparator implements Comparator<Node>{
@Override
public int compare(Node p1, Node p2) {
// int n = p1.id1-p2.id1;
// if(n!=0)return n;
return p1.id2-p2.id2;
}
}
static int pp =1;
static long[] dp = new long[500001];
public static void main(String[] args) throws UnknownHostException {
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while(t--!=0) {
long a = sc.nextLong();
long b = sc.nextLong();
long c = sc.nextLong();
long a1 = Math.abs(a-1);
long b1= Math.abs(b-c)+Math.abs(c-1);
if(a1<b1)System.out.println(1);
else if(b1<a1)System.out.println(2);
else System.out.println(3);
}
out.close();
}
static long nextPowerOf2(long N)
{
// if N is a power of two simply return it
if ((N & (N - 1)) == 0)
return N;
return 0x4000000000000000L
>> (Long.numberOfLeadingZeros(N) - 2);
}
public static int[] solve(int[] arr, int[] it) {
HashMap<Integer, Integer> hm = new HashMap<>();
for(int i=0; i<arr.length; i++) {
hm.put(arr[i], hm.getOrDefault(arr[i], 0)+1);
}
for(int i=0; i<arr.length; i++) {
it[i]= hm.get(arr[i]);
}
return it;
}
public static void rec(int l, int e, int[] arr) {
if(l>e || e>=arr.length || l<=0)return;
int mid = (e-l+1)%2==0?(e+l-1)/2:(e+l)/2;
if(arr[mid]==0 ) {
arr[mid]=pp;
pp++;
// if(pp>1)return;
}
if(e-mid-1<=mid-l-1) {rec(l, mid-1, arr);
rec(mid+1, e, arr);
}
else {
rec(mid+1, e, arr);
rec(l, mid-1, arr);
}
}
static double fact(double n)
{
int i=1;
double fact=1;
while(i<=n)
{
fact=fact*i;
i++;
}
return fact;
}
static double combination(int n,int r)
{
double com=fact(n)/(fact(n-r)*fact(r));
return com;
}
static boolean digit(long n) {
long ans = n;
while(n>0) {
long rem = n%10;
if(rem!=0 && ans%rem!=0)return false;
n=n/10;
}
return true;
}
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;
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l, Collections.reverseOrder());
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static boolean isPalindrome(char[] arr, int i, int j) {
while(i<j) {
if(arr[i]!=arr[j])return false;
i++;
j--;
}
return true;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static int max =0;
static void dfs(int i, boolean[] vis , ArrayList<ArrayList<Integer>> adj) {
max = Math.max(max, i);
vis[i]= true;
for(int e: adj.get(i)) {
if(vis[e]==false) {
dfs(e, vis, adj);
}
}
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static double pow(int a, int b) {
long res= 1;
while(b>0) {
if((b&1)!=0) {
res= (res*a);
}
a= (a*a);
b= b>>1;
}
return res;
}
static void permute(String s , String answer, HashSet<String> hs)
{
if (s.length() == 0)
{
hs.add(answer);
return;
}
for(int i = 0 ;i < s.length(); i++)
{
char ch = s.charAt(i);
String left_substr = s.substring(0, i);
String right_substr = s.substring(i + 1);
String rest = left_substr + right_substr;
permute(rest, answer + ch, hs);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 2625a5feb68bb85dae631c0e29fe5742 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;
public class quesA {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int i=0;i<t;i++) {
int a = sc.nextInt();//input.get(0);
int b = sc.nextInt();//input.get(1);
int c = sc.nextInt();//input.get(2);
if (a==1) {
System.out.println("1");
}else if (Math.abs(a-1)<(Math.abs(b-c)+Math.abs(c-1))) {
System.out.println("1");
}else if (Math.abs(a-1)>(Math.abs(b-c)+Math.abs(c-1))) {
System.out.println("2");
}else System.out.println("3");
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | f35bf3eb5dae97e937a9b65ee61ed03a | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes |
import java.util.Scanner;
public class TwoElevators {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0 ) {
int a = (sc.nextInt());
int b = (sc.nextInt());
int c = (sc.nextInt());
int ml = Math.abs(a - 1);
int m = Math.abs(c - b );
m+=Math.abs(c-1);
if (ml < m) System.out.println(1);
else if (m < ml) System.out.println(2);
else System.out.println(3);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 975273517aedea29917e927b27e8e5f8 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | /***** ---> :) Vijender Srivastava (: <--- *****/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static FastReader sc =new FastReader();
static PrintWriter out=new PrintWriter(System.out);
static long mod=(long)32768;
static StringBuilder sb = new StringBuilder();
/* start */
public static void main(String [] args)
{
int testcases = 1;
testcases = i();
// calc();
while(testcases-->0)
{
solve();
}
out.flush();
out.close();
}
static void solve()
{
long a= l(),b = l(),c = l();
long x = a-1,y = Math.abs(c-b)+Math.abs(c-1);
if(x==y)pl(3);
else if(x>y)pl(2);
else pl(1);
}
/* end */
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;
}
}
// print code start
static void p(Object o)
{
out.print(o);
}
static void pl(Object o)
{
out.println(o);
}
static void pl()
{
out.println("");
}
// print code end //
static int i() {
return sc.nextInt();
}
static String s() {
return sc.next();
}
static long l() {
return sc.nextLong();
}
static char[] inputC()
{
String s = sc.nextLine();
return s.toCharArray();
}
static int[] input(int n) {
int A[]=new int[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextInt();
}
return A;
}
static long[] inputL(int n) {
long A[]=new long[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextLong();
}
return A;
}
static long[] putL(long a[]) {
long A[]=new long[a.length];
for(int i=0;i<a.length;i++) {
A[i]=a[i];
}
return A;
}
static String[] inputS(int n) {
String A[]=new String[n];
for(int i=0;i<n;i++) {
A[i]=sc.next();
}
return A;
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static String reverse(String s) {
StringBuffer p=new StringBuffer(s);
p.reverse();
return p.toString();
}
static int min(int a,int b) {
return Math.min(a, b);
}
static int min(int a,int b,int c) {
return Math.min(a, Math.min(b, c));
}
static int min(int a,int b,int c,int d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static int max(int a,int b) {
return Math.max(a, b);
}
static int max(int a,int b,int c) {
return Math.max(a, Math.max(b, c));
}
static int max(int a,int b,int c,int d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long min(long a,long b) {
return Math.min(a, b);
}
static long min(long a,long b,long c) {
return Math.min(a, Math.min(b, c));
}
static long min(long a,long b,long c,long d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static long max(long a,long b) {
return Math.max(a, b);
}
static long max(long a,long b,long c) {
return Math.max(a, Math.max(b, c));
}
static long max(long a,long b,long c,long d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long sum(int A[]) {
long sum=0;
for(int i : A) {
sum+=i;
}
return sum;
}
static long sum(long A[]) {
long sum=0;
for(long i : A) {
sum+=i;
}
return sum;
}
static long mod(long x) {
return ((x%mod + mod)%mod);
}
static long power(long x, long y)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) ;
y = y >> 1;
x = (x * x);
}
return res;
}
static boolean prime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static boolean prime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static long[] sort(long a[]) {
ArrayList<Long> arr = new ArrayList<>();
for(long i : a) {
arr.add(i);
}
Collections.sort(arr);
for(int i = 0; i < arr.size(); i++) {
a[i] = arr.get(i);
}
return a;
}
static int[] sort(int a[])
{
ArrayList<Integer> arr = new ArrayList<>();
for(Integer i : a) {
arr.add(i);
}
Collections.sort(arr);
for(int i = 0; i < arr.size(); i++) {
a[i] = arr.get(i);
}
return a;
}
//pair class
private static class Pair implements Comparable<Pair> {
long first, second;
public Pair(long f, long s) {
first = f;
second = s;
}
@Override
public int compareTo(Pair p) {
if (first < p.first)
return 1;
else if (first > p.first)
return -1;
else {
if (second > p.second)
return 1;
else if (second < p.second)
return -1;
else
return 0;
}
}
}
// segment t start
static long seg[] ;
static void build(long a[], int v, int tl, int tr) {
if (tl == tr) {
seg[v] = a[tl];
} else {
int tm = (tl + tr) / 2;
build(a, v*2, tl, tm);
build(a, v*2+1, tm+1, tr);
seg[v] = Math.min(seg[v*2] , seg[v*2+1]);
}
}
static long query(int v, int tl, int tr, int l, int r) {
if (l > r || tr < tl)
return Integer.MAX_VALUE;
if (l == tl && r == tr) {
return seg[v];
}
int tm = (tl + tr) / 2;
return (query(v*2, tl, tm, l, min(r, tm))+ query(v*2+1, tm+1, tr, max(l, tm+1), r));
}
static void update(int v, int tl, int tr, int pos, long new_val) {
if (tl == tr) {
seg[v] = new_val;
} else {
int tm = (tl + tr) / 2;
if (pos <= tm)
update(v*2, tl, tm, pos, new_val);
else
update(v*2+1, tm+1, tr, pos, new_val);
seg[v] = Math.min(seg[v*2] , seg[v*2+1]);
}
}
// segment t end //
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | d1db25e7093b08b2d25f7568a7a1f04f | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class A {
public static void main(String[] args) {
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = fs.nextInt();
while (t-- > 0) {
int n = fs.nextInt();
int k=fs.nextInt();
int m=fs.nextInt();
// int arr[]=fs.readArray(n);
System.out.println(check(n,k,m));
}
// out.close();
}
static int check(int a,int b,int c){
int x=a-1;
int y=Math.abs(c-b)+c-1;
int z=Math.min(x,y);
if(x==y)
return 3;
if(z==x)
return 1;
return 2;
}
/* HELPER FUNCTION's */
static final Random random = new Random();
static final int mod = 1_000_000_007;
static void ruffleSort(int[] a) {
int n = a.length;//shuffle, then sort
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 long add(long a, long b) {
return (a + b) % mod;
}
static long sub(long a, long b) {
return ((a - b) % mod + mod) % mod;
}
static long mul(long a, long b) {
return (a * b) % mod;
}
/* fast exponentiation */
static long exp(long base, long exp) {
if (exp == 0) return 1;
long half = exp(base, exp / 2);
if (exp % 2 == 0) return mul(half, half);
return mul(half, mul(half, base));
}
/* end of fast exponentiation */
static long[] factorials = new long[2_000_001];
static long[] invFactorials = new long[2_000_001];
static void precompFacts() {
factorials[0] = invFactorials[0] = 1;
for (int i = 1; i < factorials.length; i++) factorials[i] = mul(factorials[i - 1], i);
invFactorials[factorials.length - 1] = exp(factorials[factorials.length - 1], mod - 2);
for (int i = invFactorials.length - 2; i >= 0; i--)
invFactorials[i] = mul(invFactorials[i + 1], i + 1);
}
static long nCk(int n, int k) {
return mul(factorials[n], mul(invFactorials[k], invFactorials[n - k]));
}
/* sort ascending and descending both */
static void sort(int[] a,int x) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
if(x==0) {
Collections.sort(l);
}
if(x==1){
Collections.sort(l,Collections.reverseOrder());
}
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
/* sort String acc. to character */
static String sortString(String s){
char ch[]=s.toCharArray();
Arrays.sort(ch);
String s1=String.valueOf(ch);
return s1;
}
// lcm(a,b) * gcd(a,b) = a * b
public static long _lcm(long a, long b) {
return (a / _gcd(a, b)) * b;
}
// euclidean algorithm time O(max (loga ,logb))
public static long _gcd(long a, long b) {
while (a > 0) {
long x = a;
a = b % a;
b = x;
}
return b;
}
/* Pair Class implementation */
static class Pair<K, V> {
K ff;
V ss;
public Pair(K ff, V ss) {
this.ff = ff;
this.ss = ss;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || this.getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return ff.equals(pair.ff) && ss.equals(pair.ss);
}
@Override
public int hashCode() {
return Objects.hash(ff, ss);
}
@Override
public String toString() {
return ff.toString() + " " + ss.toString();
}
}
/* pair class ends here */
/* fast input output class */
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[] readArrayL(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 7e578a623b69323a100c62a742f5b02b | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class A {
public static void main(String[] args) {
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = fs.nextInt();
while (t-- > 0) {
int n = fs.nextInt();
int k=fs.nextInt();
int m=fs.nextInt();
// int arr[]=fs.readArray(n);
System.out.println(check(n,k,m));
}
// out.close();
}
static int check(int a,int b,int c){
if(c<b){
if(a>b)
return 2;
else if(a==b)
return 3;
else
return 1;
}
if(b>1) {
if ((2 * Math.abs(b - c))+(b-1) > Math.abs(a - 1))
return 1;
else if ((2 * Math.abs(b - c)) +(b-1)< Math.abs(a - 1))
return 2;
else
return 3;
}
if ((2 * Math.abs(b - c)) > Math.abs(a - 1))
return 1;
else if ((2 * Math.abs(b - c))< Math.abs(a - 1))
return 2;
else
return 3;
}
/* HELPER FUNCTION's */
static final Random random = new Random();
static final int mod = 1_000_000_007;
static void ruffleSort(int[] a) {
int n = a.length;//shuffle, then sort
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 long add(long a, long b) {
return (a + b) % mod;
}
static long sub(long a, long b) {
return ((a - b) % mod + mod) % mod;
}
static long mul(long a, long b) {
return (a * b) % mod;
}
/* fast exponentiation */
static long exp(long base, long exp) {
if (exp == 0) return 1;
long half = exp(base, exp / 2);
if (exp % 2 == 0) return mul(half, half);
return mul(half, mul(half, base));
}
/* end of fast exponentiation */
static long[] factorials = new long[2_000_001];
static long[] invFactorials = new long[2_000_001];
static void precompFacts() {
factorials[0] = invFactorials[0] = 1;
for (int i = 1; i < factorials.length; i++) factorials[i] = mul(factorials[i - 1], i);
invFactorials[factorials.length - 1] = exp(factorials[factorials.length - 1], mod - 2);
for (int i = invFactorials.length - 2; i >= 0; i--)
invFactorials[i] = mul(invFactorials[i + 1], i + 1);
}
static long nCk(int n, int k) {
return mul(factorials[n], mul(invFactorials[k], invFactorials[n - k]));
}
/* sort ascending and descending both */
static void sort(int[] a,int x) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
if(x==0) {
Collections.sort(l);
}
if(x==1){
Collections.sort(l,Collections.reverseOrder());
}
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
/* sort String acc. to character */
static String sortString(String s){
char ch[]=s.toCharArray();
Arrays.sort(ch);
String s1=String.valueOf(ch);
return s1;
}
// lcm(a,b) * gcd(a,b) = a * b
public static long _lcm(long a, long b) {
return (a / _gcd(a, b)) * b;
}
// euclidean algorithm time O(max (loga ,logb))
public static long _gcd(long a, long b) {
while (a > 0) {
long x = a;
a = b % a;
b = x;
}
return b;
}
/* Pair Class implementation */
static class Pair<K, V> {
K ff;
V ss;
public Pair(K ff, V ss) {
this.ff = ff;
this.ss = ss;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || this.getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return ff.equals(pair.ff) && ss.equals(pair.ss);
}
@Override
public int hashCode() {
return Objects.hash(ff, ss);
}
@Override
public String toString() {
return ff.toString() + " " + ss.toString();
}
}
/* pair class ends here */
/* fast input output class */
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[] readArrayL(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 4c3a23a35d117df5a6ac2de0a5c3c063 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.util.function.Function;
import java.util.stream.Collectors;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.*;
import java.math.*;
public class A
{
static StringBuilder sb;
static dsu dsu;
static long fact[];
static long mod=(long)(1e9+7);
static ArrayList<Integer>prime;
//static int x[];
static long A[],dp[];
static int n;
static ArrayList<Integer>adj[];
static ArrayList<Long>ll;
static boolean visited[];
static int max;
static int min;
static void solve()
{
int a=i(),b=i(),c=i();
int ta=a-1;
int tb=0;
if(c>=b)
tb=(c-b)+(c-1);
else
tb=(b-c)+(c-1);
if(ta<tb)
{
System.out.println(1);
}
else if(ta>tb)
{
System.out.println(2);
}
else
System.out.println(3);
}
public static void main(String[] args)
{
sb=new StringBuilder();
int t=i();
while(t-->0)
{
solve();
}
System.out.println(sb);
}
//*******************************************Seive OF Erassosthenes n log n*******************************************************
//int prime[]=new int[100000];
//
// Arrays.fill(prime,-1);
//
// for(int i=2;i*i<prime.length;i++)
// {
// if(prime[i]==-1)
// for(int j=i*i;j<prime.length;j+=i)
// {
// prime[j]=1;
// }
//
//
// }
// ArrayList<Integer>l=new ArrayList<>();
// for(int i=2;i<prime.length;i++)
// {
// if(prime[i]==-1)
// {
// l.add(i);
// }
// }
// System.out.println(l.size());
//*******************************************Seive OF Erassosthenes O(n) *******************************************************
//int n=100;
//
// int lp[]=new int[n+1];// Array of minimum prime factor of n numbers
// ArrayList<Integer>pr=new ArrayList<>();// list of prime numbers
//
// for(int i=2;i<n;i++)
// {
// if(lp[i]==0)
// {
// lp[i]=i;
// pr.add(i);
// }
// for(int j=0;(j<pr.size()) && (pr.get(j)<=lp[i])&& (i*pr.get(j)<=n);j++)
// {
// lp[i*pr.get(j)]=pr.get(j);
//
// }
//
// }
//
// System.out.println(Arrays.toString(lp));
// System.out.println(pr);
//
//
//*******************************************NCR%P***********************************************************************
static long ncr(int n, int r)
{
if(r>n)
return (long)0;
long res=fact[n]%mod;
//System.out.println(res);
res=((long)(res%mod)*(long)(p(fact[r],mod-2)%mod))%mod;
res=((long)(res%mod)*(long)(p(fact[n-r],mod-2)%mod))%mod;
//System.out.println(res);
return res;
}
//*******************************************NCR%P***********************************************************************
static long p(long x, long y)//POWER FXN //
{
if(y==0)
return 1;
long res=1;
while(y>0)
{
if(y%2==1)
{
res=(res*x)%mod;
y--;
}
//if(y>0) {
x=(x*x)%mod;
//}
y=y/2;
}
return res;
}
static long ceil(long num, long den)
{
return (long)(num+den-1)/den;
}
//*******************************************END*******************************************************
static int LowerBound(long a[], long x, int i, int j)
{
//X is the key value
int l=i;int r=j;
int lb=-1;
while(l<=r)
{
int m=(l+r)/2;
if(a[m]>=x)
{
lb=m;
r=m-1;
}
else l=m+1;
}
return lb;
}
static int UpperBound(long a[], long x, int i, int j)
{// x is the key or target value
int l=i,r=j;
int ans=-1;
while(l<=r)
{
int m=(l+r)/2;
if(a[m]<=x)
{
ans=m;
l=m+1;
}
else r=m-1;
}
return ans;
}
//*********************************Disjoint set union*************************//
static class dsu
{
int parent[];
dsu(int n)
{
parent=new int[n+1];
for(int i=0;i<=n;i++)
parent[i]=-1;
}
int find(int a)
{
if(parent[a]<0)
return a;
else
{
int x=find(parent[a]);
parent[a]=x;
return x;
}
}
void merge(int a,int b)
{
a=find(a);
b=find(b);
if(a==b)
return;
parent[b]=a;
}
}
//*******************************************PRIME FACTORIZE *****************************************************************************************************//
static TreeMap<Integer,Integer> prime(long n)
{
TreeMap<Integer,Integer>h=new TreeMap<>();
long num=n;
for(int i=2;i<=Math.sqrt(num);i++)
{
if(n%i==0)
{
int nt=0;
while(n%i==0) {
n=n/i;
nt++;
}
h.put(i, nt);
}
}
if(n!=1)
h.put((int)n, 1);
return h;
}
//*************CLASS PAIR ***********************************************************************************************************************************************
static class pair implements Comparable<pair>
{
int x;
int y;
pair(int x, int y) {
this.x = x;
this.y = y;
}
public int compareTo(pair o)
{
return(int) (y-o.y);
}
public int hashCode()
{
final int temp = 14;
int ans = 1;
ans =x*31+y*13;
return (int)ans;
}
@Override
public boolean equals(Object o)
{
if (this == o) {
return true;
}
if (o == null) {
return false;
}
if (this.getClass() != o.getClass()) {
return false;
}
pair other = (pair)o;
if (this.x != other.x || this.y!=other.y) {
return false;
}
return true;
}
}
//*************CLASS PAIR *****************************************************************************************************************************************************
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int Int() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String String() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return String();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
static InputReader in = new InputReader(System.in);
static OutputWriter out = new OutputWriter(System.out);
public static int[] sort(int[] a) {
int n = a.length;
ArrayList<Integer> l = new ArrayList<>();
for (int i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a[i] = l.get(i);
return a;
}
public static long[] sort(long[] a) {
int n = a.length;
ArrayList<Long> l = new ArrayList<>();
for (long i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a[i] = l.get(i);
return a;
}
public static long pow(long x, long y) {
long res = 1;
while (y > 0) {
if (y % 2 != 0) {
res = (res * x);// % modulus;
y--;
}
x = (x * x);// % modulus;
y = y / 2;
}
return res;
}
//GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
public static long gcd(long x, long y) {
if (x == 0)
return y;
else
return gcd(y % x, x);
}
//****************LOWEST COMMON MULTIPLE *************************************************************************************************************************************
public static long lcm(long x, long y) {
return (x * (y / gcd(x, y)));
}
//INPUT PATTERN******************************************************************************************************************************************************************
public static int i() {
return in.Int();
}
public static long l() {
String s = in.String();
return Long.parseLong(s);
}
public static String s() {
return in.String();
}
public static int[] readArray(int n) {
int A[] = new int[n];
for (int i = 0; i < n; i++) {
A[i] = i();
}
return A;
}
public static long[] readArray(long n) {
long A[] = new long[(int) n];
for (int i = 0; i < n; i++) {
A[i] = l();
}
return A;
}
public int[][] deepCopy(int[][] matrix) {
return java.util.Arrays.stream(matrix).map(el -> el.clone()).toArray($ -> matrix.clone());
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | a3ab6b2c5db2fc9321be29699c6425ac | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Ex3 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int x = in.nextInt();
for(int i = 0; i < x; i++) {
int a = in.nextInt();
int b = in.nextInt();
int c = in.nextInt();
int b1 = Math.abs(b-c)+c;
if(a > b1){
System.out.println(2);
}else if(b1 > a){
System.out.println(1);
}else {
System.out.println(3);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 23dcf1d7bafbb03550b89dba9cb70496 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Football
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int a =sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int t1 = Math.abs(a-1);
int t2 = Math.abs((b-c))+Math.abs((c-1));
if(t1 < t2) System.out.println("1");
else if(t1 > t2) System.out.println("2");
else System.out.println("3");
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | af4e16b5e7cebd07af9c92af03697894 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Two_Elevators {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0) {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
if(a == 1 || (Math.abs(c - b) + (c - 1)) > a - 1)
System.out.println(1);
else if(Math.abs(c - b) + (c - 1) < a - 1)
System.out.println(2);
else
System.out.println(3);
}
sc.close();
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | b48de39f677677a71ebdeac60e6c7b7b | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Main
{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
int a=sc.nextInt();
while(a-->0)
{
int b=sc.nextInt();
int c=sc.nextInt();
int d=sc.nextInt();
int r=b-1;
int e=Math.abs(d-c)+Math.abs(d-1);
if(r==e)
System.out.println("3");
else if(r<e)
System.out.println("1");
else
System.out.println("2");
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | f226347b6ebac905b0af69b5a94c9157 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class QuesA {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
while(t-- > 0){
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
int s2;
if(b<c) s2=2*c-b;
else s2=b;
if(a<s2) System.out.println(1);
else if(a>s2) System.out.println(2);
else System.out.println(3);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 3e424c08cb9864cb1122eb31f4638654 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import javax.print.DocFlavor;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class A {
static FastScanner scan = new FastScanner();
static Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int T = scan.nextInt();
for(int i =1;i<=T;i++){
solve();
}
}
public static void solve() {
int a = scan.nextInt();
int b = scan.nextInt();
int c = scan.nextInt();
if(a==1){
System.out.println(1);
return;
}
int x = Math.abs(b-c);
int y = Math.abs(c-1);
x =x +y;
int z = Math.abs(a-1);
if(x>z){
System.out.println(1);
}
else if(x<z){
System.out.println(2);
}
else{
System.out.println(3);
}
}
static boolean isSorted(int[] arr){
for (int i =1;i<arr.length;i++) {
if (arr[i - 1] > arr[i])
return false;
}
return true;
}
static long gcd(long A, long B) {
if (B == 0) return A;
return gcd(B, A % B);
}
static boolean isInteger(int n) {
return Math.sqrt(n) % 1 == 0;
}
static class Vec {
double x,y;
public Vec(double x) {
this.x=x;
this.y=y;
}
}
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;
}
String[] readStringArray(int n){
String[] arr = new String[n];
for (int i =0;i<n;i++)arr[i]=next();
return arr;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | a9cb0a1dd2b69fa6623e25c802579bc2 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t>0){
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
if(a==1)
System.out.println(1);
else if(b<c && a< Math.abs(2*c-b))
System.out.println(1);
else if(b>c && a<b)
System.out.println(1);
else if(b<c && a> Math.abs(2*c-b))
System.out.println(2);
else if(b>c && a>b)
System.out.println(2);
else
System.out.println(3);
t--;
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 6966a2a6197653f756cc958454056d1a | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
BufferedOutputStream os = new BufferedOutputStream(System.out);
int n = Integer.parseInt(reader.readLine());
for (int i = 0; i < n; i++){
String[] lol = reader.readLine().split("\\s");
long first = Integer.parseInt(lol[0]) - 1;
long second = (int) (Math.abs(Integer.parseInt(lol[2]) - Integer.parseInt(lol[1])) + Integer.parseInt(lol[2]) - 1);
if (first < second)
os.write("1\n".getBytes());
else if (first > second)
os.write("2\n".getBytes());
else
os.write("3\n".getBytes());
}
os.flush();
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | ad0e208ae51f5de4684562d04065c727 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.io.*;
public class Competitive {
static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
public static void main(String[] args) throws IOException {
int t = getInt();
while(t-- > 0){
String[] s = rd.readLine().split(" ");
int a= toInt(s[0]), b = toInt(s[1]), c = toInt(s[2]);
int c1 = a;
int c2 = Math.abs(c-b) + c;
System.out.println( (c1==c2)?3:((c1>c2)?2:1) );
}
}
static int getInt() throws IOException{
return toInt(rd.readLine());
}
static int toInt(String n){
return Integer.parseInt(n);
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | ab98a3df0e6c0885ceec90b354121be3 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
static FastReader scan = new FastReader();
static PrintWriter sout = new PrintWriter(System.out);
public static void main(String[] args) {
int n = scan.nextInt();
while (n-->0)
A();
sout.close();
}
static void A(){
// System.out.println(Math.log(1000)/Math.log(1));
int n = scan.nextInt();
int m = scan.nextInt();
int l = scan.nextInt();
if (n == 1){
sout.println(1);
}else{
if (m>=l){
if (n==m)
sout.println(3);
else if(n>m)
sout.println(2);
else
sout.println(1);
}else{
int r = Math.abs(l-m);
if(r+m == n-r) {
sout.println(3);
}else if (r+m < n-r){
sout.println(2);
}else
sout.println(1);
// sout.println(r);
}
}
}
static boolean isPalindrome(String str){
StringBuilder sb = new StringBuilder(str);
return str.equals(sb.reverse().toString());
}
static class Pair{
int x;
int y;
public Pair(int x , int y){
this.x = x;
this.y = y;
}
@Override
public String toString(){
return new String(x +" "+y);
}
}
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 | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 82830dde2692db0ca56726b93519030c | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class cf1 {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
System.out.println(solve(a,b,c));
}
}
public static int solve(int a,int b,int c){
if(a==1){
return 1;
}
if(c ==1 && Math.abs(a-1) > Math.abs(b-c)){
return 2;
}
int check1 = Math.abs(a-1);
int check2 = Math.abs(b-c) + Math.abs(c-1);
if(check1 < check2){
return 1;
}
else if(check2 < check1){
return 2;
}
else{
return 3;
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 132e68f27e7ba4f8bb33323c1a55bd2d | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.StringTokenizer;
public class Main {
static long ans = 0;
public static void main(String[] args) throws IOException{
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder();
int T = Integer.parseInt(br.readLine());
while(T --> 0) {
StringTokenizer st = new StringTokenizer(br.readLine(), " ");
int temp = 0;
ans = 0;
int a = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
int c = Integer.parseInt(st.nextToken());
a -= 1;
temp = Math.abs(b-c);
temp += c - 1;
if(a > temp) {
ans = 2;
}
else if(a < temp) {
ans = 1;
}
else {
ans = 3;
}
sb.append(ans).append('\n');
}
System.out.println(sb);
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | af75ecc495aac77fb6addd67b52a321e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes |
import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.util.ArrayList;
import java.util.Collections;
/* Name of the class has to be "Main" only if the class is public. */
public class A
{
public static void main (String[] args) throws Exception
{
// your code goes here
FastScanner sc= new FastScanner();
int t = sc.nextInt();
while(t-->0){
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
if(a==1){
System.out.println(1);
}else if(b==1){
int fs = a-1;
int ss = 2*Math.abs(b-c);
if(fs==ss){
System.out.println(3);
}else if(fs<ss){
System.out.println(1);
}else{
System.out.println(2);
}
}else{
int fs = a-1;
int ss = Math.abs(b-c)+c-1;
if(fs==ss){
System.out.println(3);
}else if(fs<ss){
System.out.println(1);
}else{
System.out.println(2);
}
}
}
}
public static boolean isPrime(long n)
{
if(n < 2) return false;
if(n == 2 || n == 3) return true;
if(n%2 == 0 || n%3 == 0) return false;
long sqrtN = (long)Math.sqrt(n)+1;
for(long i = 6L; i <= sqrtN; i += 6) {
if(n%(i-1) == 0 || n%(i+1) == 0) return false;
}
return true;
}
public static long gcd(long a, long b)
{
if(a > b)
a = (a+b)-(b=a);
if(a == 0L)
return b;
return gcd(b%a, a);
}
public static ArrayList<Integer> findDiv(int N)
{
//gens all divisors of N
ArrayList<Integer> ls1 = new ArrayList<Integer>();
ArrayList<Integer> ls2 = new ArrayList<Integer>();
for(int i=1; i <= (int)(Math.sqrt(N)+0.00000001); i++)
if(N%i == 0)
{
ls1.add(i);
ls2.add(N/i);
}
Collections.reverse(ls2);
for(int b: ls2)
if(b != ls1.get(ls1.size()-1))
ls1.add(b);
return ls1;
}
public static void sort(int[] arr)
{
//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
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 long power(long x, long y, long p)
{
//0^0 = 1
long res = 1L;
x = x%p;
while(y > 0)
{
if((y&1)==1)
res = (res*x)%p;
y >>= 1;
x = (x*x)%p;
}
return res;
}
static class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 0c884bb71b4828ae244df0513e962081 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
//[row][column]
public class cf_arhiv {
public static void main(String[] args) {
FastScanner scan = new FastScanner();
int n = scan.nextInt();
for (int i = 0; i < n; i++) {
int a = scan.nextInt();
int b= scan.nextInt();
int c = scan.nextInt();
int first = Math.abs(a - 1);
int second = Math.abs(b - c) + c - 1;
if(first < second)
System.out.println("1");
else
if(first == second) System.out.println("3");
else System.out.println("2");
}
}
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(String s) {
try {
br = new BufferedReader(new FileReader(s));
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String nextToken() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(nextToken());
}
long nextLong() {
return Long.parseLong(nextToken());
}
double nextDouble() {
return Double.parseDouble(nextToken());
}String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 0a1844d5a1fc3cd48fd1201b0ce02d31 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class yuhi {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int first = a-1;
int second = c>b ? (2*(c-b))+(b-1) : (b-c)+(c-1);
if(first==second) {
System.out.println(3);
}else if(first>second) {
System.out.println(2);
}else System.out.println(1);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 0cca1d56c348f8d9df48d788818a4eb2 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
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
{
try{
FastReader sc = new FastReader();
//your code goes here.
int t = sc.nextInt();
while(t-->0){
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
a = a-1;
b = Math.abs(b-c);
b += Math.abs(c-1);
if(a<b)
System.out.println(1);
else if(a>b)
System.out.println(2);
else System.out.println(3);
}
}
catch(Exception e){
return;
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 50d41c2a4d3d46004348fc810051ee03 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class temp3 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int i = 0; i<t; i++) {
int[] a = new int[3];
a[0] = sc.nextInt();
a[1] = sc.nextInt();
a[2] = sc.nextInt();
int a1 = Math.abs(a[0] - 1);
int a2 = Math.abs(a[1] - a[2]) + Math.abs(a[2] - 1);
if(a1<a2) System.out.println(1);
else if(a1>a2) System.out.println(2);
else System.out.println(3);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 2740a2689f61842d313bb90294af4d3f | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
public class cf {
public static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
public static boolean isok(long x, long h, long k) {
long sum = 0;
if (h > k) {
long t1 = h - k;
long t = t1 * k;
sum += (k * (k + 1)) / 2;
sum += t - (t1 * (t1 + 1) / 2);
} else {
sum += (h * (h + 1)) / 2;
}
if (sum < x) {
return true;
}
return false;
}
public static boolean binary_search(long[] a, long k) {
long low = 0;
long high = a.length - 1;
long mid = 0;
while (low <= high) {
mid = low + (high - low) / 2;
if (a[(int) mid] == k) {
return true;
} else if (a[(int) mid] < k) {
low = mid + 1;
} else {
high = mid - 1;
}
}
return false;
}
public static long lowerbound(long a[], long ddp) {
long low = 0;
long high = a.length;
long mid = 0;
while (low < high) {
mid = low + (high - low) / 2;
if (a[(int) mid] == ddp) {
return mid;
}
if (a[(int) mid] < ddp) {
low = mid + 1;
} else {
high = mid;
}
}
// if(low + 1 < a.length && a[(int)low + 1] <= ddp){
// low++;
// }
if (low == a.length && low != 0) {
low--;
return low;
}
if (a[(int) low] > ddp && low != 0) {
low--;
}
return low;
}
public static long upperbound(long a[], long ddp) {
long low = 0;
long high = a.length;
long mid = 0;
while (low < high) {
mid = low + (high - low) / 2;
if (a[(int) mid] <= ddp) {
low = mid + 1;
} else {
high = mid;
}
}
if (low == a.length) {
return a.length - 1;
}
return low;
}
// public static class pair implements Comparable<pair> {
// long w;
// long h;
// public pair(long w, long h) {
// this.w = w;
// this.h = h;
// }
// public int compareTo(pair b) {
// if (this.w != b.w)
// return (int) (this.w - b.w);
// else
// return (int) (this.h - b.h);
// }
// }
public static class pair {
long w;
long h;
public pair(long w, long h) {
this.w = w;
this.h = h;
}
}
public static class trinary {
long a;
long b;
long c;
public trinary(long a, long b, long c) {
this.a = a;
this.b = b;
this.c = c;
}
}
public static long lowerboundforpairs(pair a[], long pr) {
long low = 0;
long high = a.length;
long mid = 0;
while (low < high) {
mid = low + (high - low) / 2;
if (a[(int) mid].w <= pr) {
low = mid + 1;
} else {
high = mid;
}
}
// if(low + 1 < a.length && a[(int)low + 1] <= ddp){
// low++;
// }
// if(low == a.length && low != 0){
// low--;
// return low;
// }
// if(a[(int)low].w > pr && low != 0){
// low--;
// }
return low;
}
public static pair[] sortpair(pair[] a) {
Arrays.sort(a, new Comparator<pair>() {
public int compare(pair p1, pair p2) {
return (int) p1.w - (int) p2.w;
}
});
return a;
}
public static boolean ispalindrome(String s) {
long i = 0;
long j = s.length() - 1;
boolean is = false;
while (i < j) {
if (s.charAt((int) i) == s.charAt((int) j)) {
is = true;
i++;
j--;
} else {
is = false;
return is;
}
}
return is;
}
public static void sort(long[] arr) {
ArrayList<Long> a = new ArrayList<>();
for (long i : arr) {
a.add(i);
}
Collections.sort(a);
for (int i = 0; i < a.size(); i++) {
arr[i] = a.get(i);
}
}
public static void sortForObjecttypes(pair[] arr) {
ArrayList<pair> a = new ArrayList<>();
for (pair i : arr) {
a.add(i);
}
Collections.sort(a, new Comparator<pair>() {
@Override
public int compare(pair a, pair b) {
return (int) (a.h - b.h);
}
});
for (int i = 0; i < a.size(); i++) {
arr[i] = a.get(i);
}
}
public static long power(long base, long pow, long mod) {
long result = base;
long temp = 1;
while (pow > 1) {
if (pow % 2 == 0) {
result = ((result % mod) * (result % mod)) % mod;
pow /= 2;
} else {
temp = temp * base;
pow--;
}
}
result = ((result % mod) * (temp % mod));
// System.out.println(result);
return result;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
float nextFloat() {
return Float.parseFloat(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
void readArr(int[] ar, int n) {
for (int i = 0; i < n; i++) {
ar[i] = nextInt();
}
}
}
public static int bSearchDiff(long[] a, int low, int high) {
int mid = low + ((high - low) / 2);
int hight = high;
int lowt = low;
while (lowt < hight) {
mid = lowt + (hight - lowt) / 2;
if (a[high] - a[mid] <= 5) {
hight = mid;
} else {
lowt = mid + 1;
}
}
return high - lowt + 1;
}
public static void solve(FastReader sc, PrintWriter w) throws Exception {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
if (a == 1) {
System.out.println(1);
return;
}
if (a - 1 > Math.abs(c - b) + Math.abs(c - 1)) {
System.out.println(2);
} else if (a - 1 < Math.abs(c - b) + Math.abs(c - 1)) {
System.out.println(1);
} else {
System.out.println(3);
}
}
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
PrintWriter w = new PrintWriter(System.out);
long o = sc.nextLong();
while (o > 0) {
solve(sc, w);
o--;
}
w.close();
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 235e0a48b25a8c863323dc73493dd0f2 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int testcase = input.nextInt();
for (int i = 0; i < testcase; i++) {
int a = input.nextInt();
int b = input.nextInt();
int c = input.nextInt();
System.out.println(checkTime(a, b, c));
}
input.close();
}
public static int checkTime(int a, int b, int c) {
int one = Math.abs(a - 1);
int two = Math.abs(b - c) + Math.abs(c - 1);
if (one == two)
return 3;
return one < two ? 1 : 2;
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | c7b19a4a58ff7f1072ed659fa3bef4d9 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.io.*;
import static java.lang.Math.*;
public class Main { public static void main(String[] args) { new MainClass().execute(); } }
class MainClass extends PrintWriter {
MainClass() { super(System.out, true); }
boolean cases = true;
// Solution
void solveIt(int testCaseNo) {
long a = sc.nextLong();
long b = sc.nextLong();
long c = sc.nextLong();
if (a < abs(b - c) + c) {
System.out.println(1);
} else if (a > abs(b - c) + c) {
System.out.println(2);
} else {
System.out.println(3);
}
}
void solve() {
int caseNo = 1;
if (cases) for (int T = sc.nextInt(); T > 1; T--, caseNo++) { solveIt(caseNo); }
solveIt(caseNo);
}
void execute() {
long S = System.currentTimeMillis();
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
this.sc = new FastIO();
solve();
out.flush();
long G = System.currentTimeMillis();
sc.tr(G - S + "ms");
}
class FastIO {
private boolean endOfFile() {
if (bufferLength == -1) return true;
int lptr = ptrbuf;
while (lptr < bufferLength) {
// _|_
if (!isThisTheSpaceCharacter(inputBufffferBigBoi[lptr++])) return false;
}
try {
is.mark(1000);
while (true) {
int b = is.read();
if (b == -1) {
is.reset();
return true;
} else if (!isThisTheSpaceCharacter(b)) {
is.reset();
return false;
}
}
} catch (IOException e) {
return true;
}
}
private byte[] inputBufffferBigBoi = new byte[1024];
int bufferLength = 0, ptrbuf = 0;
private int justReadTheByte() {
if (bufferLength == -1) throw new InputMismatchException();
if (ptrbuf >= bufferLength) {
ptrbuf = 0;
try {
bufferLength = is.read(inputBufffferBigBoi);
} catch (IOException e) {
throw new InputMismatchException();
}
if (bufferLength <= 0) return -1;
}
return inputBufffferBigBoi[ptrbuf++];
}
private boolean isThisTheSpaceCharacter(int c) { return !(c >= 33 && c <= 126); }
private int skipItBishhhhhhh() {
int b;
while ((b = justReadTheByte()) != -1 && isThisTheSpaceCharacter(b));
return b;
}
private double nextDouble() { return Double.parseDouble(next()); }
private char nextChar() { return (char) skipItBishhhhhhh(); }
private String next() {
int b = skipItBishhhhhhh();
StringBuilder sb = new StringBuilder();
while (!(isThisTheSpaceCharacter(b))) {
sb.appendCodePoint(b);
b = justReadTheByte();
}
return sb.toString();
}
private char[] readCharArray(int n) {
char[] buf = new char[n];
int b = skipItBishhhhhhh(), p = 0;
while (p < n && !(isThisTheSpaceCharacter(b))) {
buf[p++] = (char) b;
b = justReadTheByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] readCharMatrix(int n, int m) {
char[][] map = new char[n][];
for (int i = 0; i < n; i++) map[i] = readCharArray(m);
return map;
}
private int[] readIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
private long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
private int nextInt() {
int num = 0, b;
boolean minus = false;
while ((b = justReadTheByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if (b == '-') {
minus = true;
b = justReadTheByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = justReadTheByte();
}
}
private long nextLong() {
long num = 0;
int b;
boolean minus = false;
while ((b = justReadTheByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if (b == '-') {
minus = true;
b = justReadTheByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = justReadTheByte();
}
}
private void tr(Object... o) {
if (INPUT.length() != 0) System.out.println(Arrays.deepToString(o));
}
}
InputStream is;
PrintWriter out;
String INPUT = "";
final int ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
final long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
final long mod = (long) 1e9 + 7;
FastIO sc;
}
// And I wish you could sing along, But this song is a joke, and the melody I
// wrote, wrong | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 50e1908dfe607f0835a7d1d0d8a4f4fe | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t>0){
int a=scanner.nextInt(), b= scanner.nextInt() , c=scanner.nextInt();
int cnt_1 = a - 1;
int cnt_2 = Math.abs(c-b) + c -1 ;
if (cnt_1 < cnt_2)
System.out.println(1);
else if ( cnt_2 < cnt_1 )
System.out.println(2);
else
System.out.println(3);
t--;
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 2ba0913b7fa1b8e950e45914d4172746 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
public class cf {
static long[] p;
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
int t = Integer.parseInt(bf.readLine());
for(int testcases = 0; testcases < t; testcases++){
StringTokenizer stk = new StringTokenizer(bf.readLine());
int a = Integer.parseInt(stk.nextToken());
int b = Integer.parseInt(stk.nextToken());
int c = Integer.parseInt(stk.nextToken());
int first = Math.abs(a-1);
int second = Math.abs(b-c) + Math.abs(c-1);
if(first<second)pw.println(1);
if(first==second)pw.println(3);
if(first>second)pw.println(2);
}
pw.close();
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 58ce82a6b0496f5c2092b8b71c7628df | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int N = in.nextInt();
while(N-- > 0){
int a = in.nextInt(), b = in.nextInt(), c = in.nextInt();
long t = (Math.abs(b - c)) + c;
if(a == t)
System.out.println(3);
else if(a < t)
System.out.println(1);
else if(a > t)
System.out.println(2);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | dc650fc7569848d2e6f451a362688d4e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Two_Elevators {
public static void main(String[] args) {
Scanner Scan = new Scanner(System.in);
//System.out.println("Test Cases :");
int t = Scan.nextInt();
int t1=0;
while(t1<t) {
t1++;
//System.out.println("Enter a :");
long a = Scan.nextLong();
//System.out.println("Enter b :");
long b = Scan.nextLong();
//System.out.println("Enter c :");
long c = Scan.nextLong();
//STEP 1
long e=a-1;
long d;
if(c>b)
d= (c-b)+(c-1);
else
d= b-1;
//STEP 2
if(e<d)
System.out.println("1");
else if(e>d)
System.out.println("2");
else
System.out.println("3");
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 6e1c3934ba2f33eaabf4afbfa1260705 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.Arrays;
public class Lift {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int numberOfSets = Integer.parseInt(br.readLine());
for (int nos = 0; nos < numberOfSets; nos++) {
int[] data = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
if (data[2] > data[1]) {
if ((data[0] - 1) < (2 * data[2] - data[1] - 1)) {
System.out.println(1);
} else {
if ((data[0] - 1) > (2 * data[2] - data[1] - 1)) {
System.out.println(2);
} else {
System.out.println(3);
}
}
} else {
if (data[0] < data[1]) {
System.out.println(1);
} else {
if (data[0] > data[1]) {
System.out.println(2);
} else {
System.out.println(3);
}
}
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 25099a48e3bb902640643e65fc5e6deb | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | //ღ(¯`◕‿◕´¯) ♫ ♪ ♫ YuSuF ♫ ♪ ♫ (¯`◕‿◕´¯)ღ
import java.util.*;
public class Two_Elevators {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-- > 0) {
int a = in.nextInt();
int b = in.nextInt();
int c = in.nextInt();
if (a == 1) {
System.out.println("1");
} else {
if (c > b) {
int secondElevator = c - b;
secondElevator += c - 1;
int firstElevator = a - 1;
if (firstElevator == secondElevator) {
System.out.println("3");
} else {
if (firstElevator < secondElevator) {
System.out.println("1");
} else {
System.out.println("2");
}
}
} else {
int secondElevator = b - c;
secondElevator += c - 1;
int firstElevator = a - 1;
if (firstElevator == secondElevator) {
System.out.println("3");
} else {
if (firstElevator < secondElevator) {
System.out.println("1");
} else {
System.out.println("2");
}
}
}
}
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 54f0f24bf2777199bceb6183c955807e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class A_Two_Elevators
{
public static void main(String[]args)
{
Scanner x=new Scanner (System.in);
int t=x.nextInt();
while(t-->0)
{
int a=x.nextInt();
int b=x.nextInt();
int c=x.nextInt();
int g=0,h=0;
if(a==1)
g=0;
else
g=a-1;
if(c==1)
h=b-c;
else
h=(int)Math.abs(b-c)+(int)Math.abs(c-1);
if(g<h)
System.out.println("1");
else if(h<g)
System.out.println("2");
else
System.out.println("3");
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 1f9730894cd116e97ebba03ed997bcc2 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class A {
public static int solve(int a, int b, int c) {
int time1 = Math.abs(a - 1);
int time2 = Math.abs(b - c) + Math.abs(c - 1);
if (time1 == time2) {
return 3;
} else if (time1 > time2) {
return 2;
} else {
return 1;
}
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t, a, b, c;
t = in.nextInt();
for (int i=0; i<t; i++) {
a = in.nextInt();
b = in.nextInt();
c = in.nextInt();
int result = solve(a, b, c);
System.out.println(result);
}
in.close();
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | a62bf247cd646625a7c6e11828fc14f1 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class question {
public static void main(String[] args) throws java.lang.Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0)
{
String st[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(st[0]);
int b = Integer.parseInt(st[1]);
int c = Integer.parseInt(st[2]);
if(a-1<Math.abs(b-c)+c-1)
{
System.out.println("1");
}
else if(a-1>Math.abs(b-c)+c-1)
{
System.out.println("2");
}
else
System.out.println("3");
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | e4ead2ed1f8e17fb9d9c4d47ad28d846 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | //============================================================================
//Everyone has a different way of thinking, so God Created us
// Hope You Respect My Way..,Thank You ):
// Author : Murad
// Name : Codeforces.cpp & Atcoder.cpp
// Description : Problem name
//============================================================================/
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in= new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
int t=in.nextInt();
while(t-->0){
long a=in.nextLong(),b=in.nextLong(),c=in.nextLong();
long cnt1=a-1;
long cnt2=0;
if(b>c){
cnt2=b-1;
}
else{
cnt2=Math.abs(b-c)+Math.abs(c-1);
}
if(cnt1==cnt2){
out.println(3);
}else if(cnt1<cnt2)out.println(1);
else out.println(2);
}
out.flush();
}
static class NumberTheory{
public static long gcd(long a,long b){
long c;
while (a != 0) {
c = a;
a = b % a;
b = c;
}
return b;
}
}
static class Pair implements Comparable<Pair> {
long x;
long y;
Pair(long x, long y)
{
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
if(x-o.x==0)return Math.toIntExact(y - o.y);
return Math.toIntExact(x - o.x);
}
}
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());
}
public long[] readLongArray(int n) {
long[] x = new long[n];
for (int i = 0; i < n; i++) {
x[i] = nextLong();
}
return x;
}
public int[] readIntArray(int n) {
int[] x = new int[n];
for (int i = 0; i < n; i++) {
x[i] = nextInt();
}
return x;
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | efb091b51ed490a99564d7191366eacc | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int i = 0 ; i < t ;i++){
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
a = Math.abs(a-1);
int temp = 0;
temp = Math.abs(c - b) + Math.abs(c - 1);
// if(b > c){
// temp = Math.abs(b - 1);
// }else{
// temp = 2*Math.abs(b - c);
// }
if(a < temp){
System.out.println("1");
}else if(temp < a){
System.out.println("2");
}else{
System.out.println("3");
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 0462699af0a83ed349664287a4b875a7 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
public class Elevators{
static int helper(int a, int b, int c){
int ans1 = Math.abs(a-1);
int ans2 = Math.abs(c-b) + Math.abs(c-1);
if(ans1 < ans2){
return 1;
}else if(ans2 < ans1){
return 2;
}
return 3;
}
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int i = 0; i < t; i++){
int a, b, c;
a = sc.nextInt();
b = sc.nextInt();
c = sc.nextInt();
int ans = helper(a, b, c);
System.out.println(ans);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | c0b1e89f6ab723623924693408c6ebb3 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
public class Main
{
static void out(int a,int b,int c)
{
int c1=0;
int c2=0;
c1=a-1;
//System.out.println("c1="+c1);
if(b>c)
{
c2=b-c;
c2=c2+c-1;
//System.out.println("c2="+c2);
}
if(c>b)
{
c2=c-b;
c2=c2+c-1;
//System.out.println("c2="+c2);
}
if(b==c)
{
c2=b-1;
//System.out.println("c2="+c2);
}
if(c1<c2)
System.out.println("1");
if(c2<c1)
System.out.println("2");
if(c1==c2)
System.out.println("3");
}
public static void main(String[] args)throws IOException
{
InputStreamReader Ir=new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(Ir);
int a=Integer.parseInt(br.readLine());
String[] k=new String[a];
int[][] b=new int[a][3];
String[] splitStr;
for(int i=0;i<a;i++)
{k[i]=br.readLine();
splitStr= k[i].trim().split("\\s+");
for(int j=0;j<3;j++)
{
b[i][j]=Integer.parseInt(splitStr[j]);
}
}
for(int i=0;i<a;i++)
{
out(b[i][0],b[i][1],b[i][2]);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 858b3e57077adc1963e1ce8e0814a8ad | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class sorting {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t>0) {
int first=sc.nextInt();
int second=sc.nextInt();
int floor=sc.nextInt();
if(first==1) {
System.out.println(1);
}else {
int one=Math.abs(first-1);
int two=0;
if(floor==1)
two=Math.abs(second-1);
else
two=Math.abs(second-floor)+Math.abs(floor-1);
if(one==two)
System.out.println(3);
else
{ if(two>one)
System.out.println(1);
else
System.out.println(2);
}
//System.out.println(two);
//System.out.println(one);
}
t--;
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 977bd3634564c1c13def36b3afe04b65 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
try (Scanner scanner = new Scanner(System.in)) {
int count = scanner.nextInt();
int[][] input = getInput(count, scanner);
for (int i = 0; i < count; i++) {
int a = input[i][0];
int b = input[i][1];
int c = input[i][2];
int secondWay = Math.abs(b - c) + c;
int res = a < secondWay ? 1 : (a == secondWay ? 3 : 2);
System.out.println(res);
}
}
}
private static int[][] getInput(int count, Scanner scanner) {
int[][] input = new int[count][];
for (int i = 0; i < count; i++) {
input[i] = new int[3];
input[i][0] = Integer.parseInt(scanner.next());
input[i][1] = Integer.parseInt(scanner.next());
input[i][2] = Integer.parseInt(scanner.next());
}
return input;
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 5c0966d33c2f0504bd6ccff20fead5ce | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Sulution {
static List<String> res=new ArrayList<>();
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while (t-->0)
{
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
if((a-1)<(Math.abs(b-c)+Math.abs(c-1)))
{
System.out.println(1);
}
else if((a-1)>(Math.abs(b-c)+Math.abs(c-1)))
{
System.out.println(2);
}
else {
System.out.println(3);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | f09ea3959f0818164cfa09029760fcf0 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
public class Codeforces {
final static int mod = 1000000007;
final static String yes = "YES";
final static String no = "NO";
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
int t = sc.nextInt();
outer: while (t-- > 0) {
int f = sc.nextInt();
int f1 = sc.nextInt();
int f2 = sc.nextInt();
int time1 = Math.abs(f - 1);
int time2 = Math.abs(f2 - f1) + Math.abs(f2 - 1);
if (time1 < time2) {
System.out.println(1);
} else if (time1 == time2) {
System.out.println(3);
} else {
System.out.println(2);
}
}
}
static long getVal(char[] c) {
long ans = 0;
for (int i = 1; i < c.length; i++) {
ans += ((c[i - 1] - '0') * 10) + (c[i] - '0');
}
return ans;
}
static boolean check(int x, int y, char[] s, char[] c) {
int i = x, j = y;
int n = c.length;
while (i >= 0 && j < n && s[i] == c[j]) {
i--;
j++;
}
return j == n ? true : false;
}
public static class Pair {
// public static class Pair implements Comparator<Pair> {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
/*
* Write the below Comparator function inside Arrays.sort or
* Collections.sort like:
* Arrays.sort(k, new Comparator<Pair>() {
*
* @Override
* public int compare(Pair obj1, Pair obj2) x{
* if (obj1.x == obj2.x) {
* return obj1.y - obj2.y;
* }
* return obj1.x - obj2.x;
* }
* });
*/
}
public static class orderPair implements Comparable<orderPair> {
int val;
int pos;
public orderPair(int val, int pos) {
this.val = val;
this.pos = pos;
}
@Override
public int compareTo(orderPair obj) {
if (this.val == obj.val)
return this.pos - obj.pos;
return this.val - obj.val;
}
}
static int dfs(int[] a, boolean[] visited, int i) {
if (visited[i]) {
return i + 1;
}
visited[i] = true;
return dfs(a, visited, a[i] - 1);
}
static void sortLong(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 void sortInt(int[] a) // check for int
{
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);
}
public static boolean isSorted(int[] nums, int n) {
for (int i = 1; i < n; i++) {
if (nums[i] < nums[i - 1])
return false;
}
return true;
}
public static boolean isPalindrome(String s) {
StringBuilder sb = new StringBuilder(s);
return s.equals(sb.reverse().toString());
}
static long kadane(long A[]) {
long lsum = A[0], gsum = 0;
gsum = Math.max(gsum, lsum);
for (int i = 1; i < A.length; i++) {
lsum = Math.max(lsum + A[i], A[i]);
gsum = Math.max(gsum, lsum);
}
return gsum;
}
public static void backtrack(String[] letters, int index, String digits, StringBuilder build,
List<String> result) {
if (build.length() >= digits.length()) {
result.add(build.toString());
return;
}
char[] key = letters[digits.charAt(index) - '2'].toCharArray();
for (int j = 0; j < key.length; j++) {
build.append(key[j]);
backtrack(letters, index + 1, digits, build, result);
build.deleteCharAt(build.length() - 1);
}
}
public static String stringMultiply(String s, int k) {
int n = s.length();
int rep = k % n == 0 ? k / n : k / n + 1;
s = s.repeat(rep);
return s.substring(0, k);
}
public static int diglen(Long y) {
int a = 0;
while (y != 0L) {
y /= 10;
a++;
}
return a;
}
static String reverseString(String str) {
StringBuilder input = new StringBuilder();
return input.append(str).reverse().toString();
}
static void printArray(int[] nums) {
for (int i = 0; i < nums.length; i++) {
System.out.print(nums[i] + "->");
}
System.out.println();
}
static void printLongArray(long[] nums) {
for (int i = 0; i < nums.length; i++) {
System.out.print(nums[i] + "->");
}
System.out.println();
}
static void printPair(orderPair[] a) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < a.length; i++) {
System.out.print(a[i].val + "->");
sb.append(a[i].pos + "->");
}
System.out.println();
System.out.println(sb.toString());
}
static long factorial(int n, int b) {
if (n == b)
return 1;
return n * factorial(n - 1, b);
}
static int lcm(int ch, int b) {
return ch * b / gcd(ch, b);
}
static int gcd(int ch, int b) {
return b == 0 ? ch : gcd(b, ch % b);
}
static double ceil(double n, double k) {
return Math.ceil(n / k);
}
static int sqrt(double n) {
return (int) Math.sqrt(n);
}
static class FastReader {
BufferedReader br;
StringTokenizer st; // StringTokenizer() is used to read long strings
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 | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | d786a4dff68f7d567a95e0aa1ea657fa | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes |
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int t = input.nextInt();
float summa = 0;
for (int i = 0; i < t; i++) {
int a = input.nextInt();
int b = input.nextInt();
int c = input.nextInt();
getAns(a,b,c);
}
}
private static void getAns(int a,int b, int c) {
if (a == 1)
System.out.println(1);
else {
int dif = Math.abs(b - c) + c;
if (a > dif)
System.out.println(2);
else if (a < dif)
System.out.println(1);
else System.out.println(3);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | db5ffd32fd9f9ee14a2ab7833828f649 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class TwoElevators {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
for (int i= 0; i<n;i++) {
int a = scan.nextInt();
int b= scan.nextInt();
int c= scan.nextInt();
fastest(a,b,c);
}
}
public static void fastest(int a, int b, int c)
{
int tE1 = Math.abs(a-1);
int tE2 = Math.abs(c-b) + Math.abs(c-1);
if (tE2< tE1)
{
System.out.println("2");
}
else if (tE2> tE1)
{
System.out.println("1");
}
else
{
System.out.println("3");
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | d08138dbbec0b09027dda8a6cdffa4d4 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++){
int k=0;
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
if(a==1){
k=1;
}
else{
int p=Math.abs(b-c);
int h=Math.abs(c-1);
int m=p+h;
h=Math.abs(a-1);
if(m<h){
k=2;
}
else if(m>h){
k=1;
}
else{
k=3;
}
}
System.out.println(k);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | dabd5aa662e774f2e37173e1e8951e87 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes |
import java.util.Scanner;
public class r820{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int t1 = Math.abs(a-1);
int t2 = Math.abs(c-b)+Math.abs(c-1);
if(t1>t2) {
System.out.println(2);
}else if(t2>t1) {
System.out.println(1);
}else {
System.out.println(3);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 11 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | e95c7ca077378e2a989635acd686364c | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class lesson1 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int t = input.nextInt();
while( t != 0 ){
int a = input.nextInt();
int b = input.nextInt();
int c = input.nextInt();
int t1 = a - 1;
int t2 = 0;
if ( b > c ){
t2 = b - 1;
}
if( c > b ){
t2 = ( c - b ) * 2 + ( b - 1);
}
if ( t1 > t2 ){
System.out.println("2");
}
else if( t1 < t2 ){
System.out.println("1");
} else{
System.out.println("3");
}
t--;
}
input.close();
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 493b733824f2a7dffd01d3ab154d450b | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class CF_Elevator {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int w = sc.nextInt();
while(w-- > 0) {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int cal1=a-1;
int cal2=0;
if(c>b) {
cal2 = (c-b)*2 + b-1;
}else if(b>c) {
cal2 = b-1;
}
if(cal1 > cal2) {
System.out.println("2");
}else if(cal2> cal1) {
System.out.println("1");
}else if(cal2 == cal1) {
System.out.println("3");
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 993d06bfedee1178637b172142bc53ef | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import static java.lang.Math.*;
public class TwoElevators implements Closeable {
private InputReader in = new InputReader(System.in);
private PrintWriter out = new PrintWriter(System.out);
public void solve() {
int T = in.ni();
while (T-- > 0) {
int a = in.ni(), b = in.ni(), c = in.ni();
int T1 = a;
int T2;
if (b < c) {
T2 = 2 * c - b;
} else {
T2 = b;
}
if (T1 < T2) {
out.println(1);
} else if (T1 > T2) {
out.println(2);
} else {
out.println(3);
}
}
}
@Override
public void close() throws IOException {
in.close();
out.close();
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int ni() {
return Integer.parseInt(next());
}
public long nl() {
return Long.parseLong(next());
}
public void close() throws IOException {
reader.close();
}
}
public static void main(String[] args) throws IOException {
try (TwoElevators instance = new TwoElevators()) {
instance.solve();
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 0208970ca8aeabf636a485b6fc165ece | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
try {
Scanner s = new Scanner(System.in);
int t= s.nextInt();
for(int i=0; i<=t; i++){
int a = s.nextInt();
int b = s.nextInt();
int c = s.nextInt();
int x = Math.abs(a-1);
int y;
if(c>b){
y = (c-b)+(c-1);
}else{
y = b-1;
}
if(x<y){
System.out.println("1");
}else if(x>y){
System.out.println("2");
}else{
System.out.println("3");
}
}
} catch(Exception e) {
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | efc0a811d3e0484db8004f30419079f1 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Codeforces
{
public static void main(String[] args)
{
Scanner in = new Scanner (System.in);
long t=in.nextLong();
while(t>0)
{
int a [ ] = new int [ 3 ];
int c1=0 , c2=0 ;
for(int i = 0 ; i < 3 ; i++)
{
a[i]=in.nextInt();
}
c1=Math.abs(a[0]-1);
if(a[2]>a[1])
{
c2+=Math.abs(a[2]-a[1]);
c2+=Math.abs(a[2]-1);
}
else if(a[2]<a[1])
{
c2+=Math.abs(a[1]-1);
}
if(c1>c2)
System.out.println(2);
if(c1<c2)
System.out.println(1);
if(c1==c2)
System.out.println(3);
t--;
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | cf24d563ed623fc17754a16dfc1fa3fa | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class App {
public static void main(String[] args) {
// test the structures here
int t = 0;
Scanner scanner = new Scanner(System.in);
t = scanner.nextInt();
for (int i = 0; i < t; i++) {
int a = 0;
int b = 0;
int c = 0;
a = scanner.nextInt();
b = scanner.nextInt();
c = scanner.nextInt();
int x = Math.abs(c - b) + c;
if (a == x) System.out.println(3);
else if (a < x) System.out.println(1);
else System.out.println(2);
}
scanner.close();
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | c5cdb0fe8352be1662e6e1167e25ca14 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt(); in.nextLine();
String[] chisla;
for (int i = 0; i < t; i++){
String s = in.nextLine();
chisla = s.split(" ");
int[] values = new int[] {Integer.parseInt(chisla[0]), Integer.parseInt(chisla[1]), Integer.parseInt(chisla[2])};
int path1, path2;
path1 = Math.abs(values[0] - 1);
path2 = Math.abs(values[2]-values[1])+Math.abs(values[2]-1);
if (path1 == path2) System.out.println(3);
else if (path1 < path2) System.out.println(1);
else System.out.println(2);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 676fe6f4f32048c1b96fe74f4ab2032b | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();in.nextLine();
String[] chisla;
for (int i = 0; i < t; i++){
String s = in.nextLine();
chisla = s.split(" ");
int[] values = new int[3];
int cc = 0;
for (String chislo: chisla){
values[cc] = Integer.parseInt(chislo);
cc++;
}
int path1, path2;
path1 = Math.abs(values[0] - 1);
path2 = Math.abs(values[2]-values[1])+Math.abs(values[2]-1);
if (path1 == path2) System.out.println(3);
else if (path1 < path2) System.out.println(1);
else System.out.println(2);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | bc693405587f91ec6c94b4a99eb58fa6 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
String[] chisla = new String[t];
String a = in.nextLine();
for (int i = 0; i < t; i++){
String s = in.nextLine();
chisla = s.split(" ");
int c3 = Integer.parseInt(chisla[2]);
int c1 = Integer.parseInt(chisla[0]);
int c2 = Integer.parseInt(chisla[1]);
int path1, path2;
path1 = Math.abs(c1 - 1);
path2 = Math.abs(c3-c2)+Math.abs(c3-1);
if (path1 == path2) System.out.println(3);
else if (path1 < path2) System.out.println(1);
else System.out.println(2);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 580ed0a39dad79eb4ff034c40e90919e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
public class Codechef {
public static void main(String[] args)throws java.lang.Exception {
FastReader input=new FastReader();
int testcases=input.nextInt();
while(testcases-->0){
int a=input.nextInt();
int b=input.nextInt();
int c=input.nextInt();
int time1=Math.abs(a-1);
int time2=Math.abs(b-c)+Math.abs(c-1);
if(time1>time2){
System.out.println(2);
}
else if(time2>time1){
System.out.println(1);
}
else{
System.out.println(3);
}
}
}
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;
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 919355443d0634c0265dba61e5fde6a2 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class TwoElevators {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0;i<t;i++) {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int f=a-1;
int s=(b<c)?c-b+c-1:b-1;
System.out.println((s==f)?"3":((s<f)?"2":"1"));
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | eec72f80927d1244bd5472a794555fde | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
import java.lang.Math;
public class TwoElevators {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int testCases = scanner.nextInt();
for(int i = 0; i < testCases; i++){
int currentFloor1st = scanner.nextInt();
int currentFloor2nd = scanner.nextInt();
int toGoFloor2nd = scanner.nextInt();
int time1st = calculateFirstElevatorTime(currentFloor1st);
int time2nd = calculateSecondElevatorTime(currentFloor2nd, toGoFloor2nd);
int bestChoice = time1st > time2nd ? 2 : (time1st == time2nd ? 3 : 1);
System.out.println(bestChoice);
}
}
public static int calculateFirstElevatorTime(int currentFloor){
return currentFloor - 1;
}
public static int calculateSecondElevatorTime(int currentFloor, int toGoFloor){
return Math.abs(currentFloor - toGoFloor) + (toGoFloor - 1);
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | e8e2d1fcec187dc2e0ea124c7818c26e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
import java.lang.Math;
public class Main {
public static int TEST_CASES = 3;
private static int[] datas;
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int testCases = scanner.nextInt();
datas = new int[testCases];
for(int i = 0; i < testCases; i++){
int currentFloor1st = scanner.nextInt();
int currentFloor2nd = scanner.nextInt();
int toGoFloor2nd = scanner.nextInt();
int time1st = calculateFirstElevatorTime(currentFloor1st);
int time2nd = calculateSecondElevatorTime(currentFloor2nd, toGoFloor2nd);
int bestChoice = time1st > time2nd ? 2 : (time1st == time2nd ? 3 : 1);
datas[i] = bestChoice;
}
for(int j : datas){
System.out.println(j);
}
}
public static int calculateFirstElevatorTime(int currentFloor){
return currentFloor - 1;
}
public static int calculateSecondElevatorTime(int currentFloor, int toGoFloor){
return Math.abs(currentFloor - toGoFloor) + (toGoFloor - 1);
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 5b684dbf0ddf9efe27245f78644e6141 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class TwoElevator {
// 8:30pm
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int ele1 = Math.abs(a-1);
int ele2 = 0;
if(c==1){
ele2 = Math.abs(b-c);
}else
ele2 = Math.abs(b-c) + Math.abs(c-1);
if(ele1<ele2)
System.out.println(1);
else if(ele1>ele2)
System.out.println(2);
else
System.out.println(3);
}
sc.close();
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 09cf7fbce89abe4568f49e2f562d98ea | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
for(int i=0;i<n;i++){
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
int diff=a-1;
int diff1=Math.abs(c-b)+c-1;
if(diff1==diff){
System.out.println(3);
}
else if(diff>diff1){
System.out.println(2);
}
else{
System.out.println(1);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | a1a2a882a2e49b6dfd42b9f5ad5e64bd | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int i = 0; i < t; i++) {
int a = in.nextInt();
int b = in.nextInt();
int c = in.nextInt();
if(a - 1 < Math.abs(b - c) + (c - 1)) {
System.out.println(1);
} else if(a - 1 > Math.abs(b - c) + (c - 1)) {
System.out.println(2);
} else {
System.out.println(3);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | aaac7c14161ab21b9f9c8690f37f93ab | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class elevator {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int tc = s.nextInt();
while(tc-->0){
int a = s.nextInt();
int b =s.nextInt();
int c = s.nextInt();
b = Math.abs(c-b)+c;
if(a<b) System.out.println(1);
else if(b<a) System.out.println(2);
else System.out.println(3);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 211d91429bec3d303a4013195ce50c56 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Div3 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int a = input.nextInt();
for (int i = 0; i < a; i++) {
int first = input.nextInt();
int second = input.nextInt();
int secondgo = input.nextInt();
if (Math.abs(secondgo - second) + (secondgo - 1) == first - 1)
{
System.out.println("3");
}
if (Math.abs(secondgo - second) + (secondgo - 1) > first - 1)
{
System.out.println("1");
}
if (Math.abs(secondgo - second) + (secondgo - 1) < first - 1)
{
System.out.println("2");
}
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | a0e8a3f1b17b0075f536c7db583963ba | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Lift {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
int h=Math.abs(c-b)+Math.abs(c-1);
if(a-1<h){
System.out.println(1);
}
if(a-1==h){
System.out.println(3);
}
if(h<a-1){
System.out.println(2);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 88692f4b3b540d610196468290364f91 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.lang.*;
public class cp1 {
public static void main(String[] args) {
//write your code here
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
for (int tt = 0; tt < t; tt++) {
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
int x = a-1;
int y=0;
if (c > b) {
y = c-b + c-1;
}else if(b>c){
y=b-c + c-1;
}
if(a==1){
System.out.println("1");
continue;
}
if(x<y){
System.out.println("1");
}else if(x>y){
System.out.println("2");
}else if(x==y){
System.out.println("3");
}
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 720a3169213414e30d69daa0e0a92923 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 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 a,b,c;
a=sc.nextInt();
b=sc.nextInt();
c=sc.nextInt();
int ans1=Math.abs(a-1);
int ans2=Math.abs(Math.abs(b-c)+ Math.abs(c-1));
if(ans1<ans2) System.out.println(1);
else if(ans1>ans2) System.out.println(2);
else System.out.println(3);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 4fa31bd8a9543acc849d1ea984120356 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class A_contest1 {
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
// TODO Auto-generated method stub
int t = sc.nextInt();
int a, b, c, a1, b1;
while (t>0)
{
a = sc.nextInt();
b = sc.nextInt();
c = sc.nextInt();
if (a == 1)
{
System.out.println(1);
}
else
{
int diff1 = 0;
if (a > 1)
{
diff1 = a - 1;
}
int diff2;
int tot = 0;
if (b > c)
{
diff2 = b - c;
tot = diff2 + (c - 1);
}
else
{
diff2 = c - b;
tot = diff2 + (c - 1);
}
if (tot == diff1)
{
System.out.println(3);
}
else if (tot < diff1)
{
System.out.println(2);
}
else
{
System.out.println(1);
}
diff1 = 0;
diff2 = 0;
tot = 0;
}
t--;
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 39d7973c3af3fc8186cfc0fb7246f2e2 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class decelaration {
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0){
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
if (c == 1){
if ((a - 1) < (b - c)){
System.out.println(1);
}
else if ((a - 1) > (b - c)){
System.out.println(2);
}
else {
System.out.println(3);
}
}
else {
int tot = Math.abs(b - c);
tot += c - 1;
if ((a - 1) < tot){
System.out.println(1);
}
else if ((a - 1) > tot){
System.out.println(2);
}
else {
System.out.println(3);
}
}
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | fee8f276e1e418ea72a8527067d7a38b | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class rough {
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int t1 = Math.abs(a-1);
int t2 = Math.abs(b-c)+Math.abs(c-1);
if(t1>t2)
System.out.println(2);
else if (t1<t2)
System.out.println(1);
else
System.out.println(3);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 5373c9c277ca90d90291cc1ded85803e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class rough {
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
int t1 = Math.abs(a-1);
int t2 = Math.abs(b-c)+Math.abs(c-1);
if(t1>t2)
System.out.println(2);
else if (t1<t2)
System.out.println(1);
else
System.out.println(3);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 8ee57cf7d11f54b0130bd56dd82a65b5 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class syntax {
public static void main(String[] args) {
// Java syntax:
// Variables: boolean(true/false), int(an integer), double(fractional number), char(a character), String(a sequence of characters).
// User's input: "java.util.Scanner" at the start of the code to import scanner class into the program, "Scanner scanner = new Scanner(System.in)" to create a scanner function, so user can enter the information, "var any = scanner.nextVar()" to user so he can enter any kind of variable.
// Expression: operands and operators. Operands is: values, variables, numbers, quantity. Operators is: + - * / %.
// GUI: "javax.swing.JOptionPane" at the start of the code to import GUI into the program, "String name = JOptionPane.showInput/Dialog", if any other variable: "variable any = Variable.parseVar(GUI command)"
// Math class: "Math.max"; "Math.min"; "Math.abs"; "Math.sqrt"; "Math.round"; "Math.ceil"; "Math.floor".
// Random numbers: "java.util.Random" at the start of the code to import Random class into the program, "Random random = new Random" to create a random value, so we can generate a random answers for variables, "var any = random.nextVar()" to user so he can enter any kind of variable to be randomly generated.
// If statement: "if()"; "else()"; "else if()".
// Switch statement: "switch(any); {
// case "some 1": command 1
// break
// case "some 2": command 2
// break
// ...
// case "some n": command n
// break
// default: default command
// }"
// Logical operators: used to connect two or more expressions, they're: "&&" (AND) both conditions must be true; "||" (OR) either condition must be true; "!" (NOT) reverses boolean value of condition.
// While loop = executes a block of code as long as it's condition remains true.
// For loop = executes a block of code a limited amount of time.
// Nested loop = a loop inside of another loop.
// Array = used to store multiple values in a single variable. (To pick random element of an array: int n = (int)Math.floor(Math.random() * waifu.length); ).
// ArrayList = a resizable array.
Scanner scanner = new Scanner(System.in);
int input = scanner.nextInt();
for(int i=0; i<input; i++) {
int a = scanner.nextInt();
int b = scanner.nextInt();
int c = scanner.nextInt();
int xa = a;
int ya = 1;
int timefora = Math.abs(xa-ya);
int xb = b;
int yb = c;
int timeforb = Math.abs(xb-yb);
int timeforc = Math.abs(yb-1);
int timefor2 = timeforb+timeforc;
if(timefora<timefor2) {
System.out.println('1');
}
else if(timefora>timefor2) {
System.out.println('2');
}
else if(timefora==timefor2) {
System.out.println('3');
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | ef7a34a12febfde50096bba8b0ae567c | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | 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 Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
BufferedWriter out=new BufferedWriter(new OutputStreamWriter(System.out));
int t;
t=sc.nextInt();
first:
while(t-->0)
{
int res,fill,place;
long x=sc.nextLong();
long y=sc.nextLong();
long z=sc.nextLong();
if(z==1)
{
if(x>y)
{System.out.println(2);
continue;}
else if(y>x)
{System.out.println(1);
continue;}
else
{
System.out.println(3);
continue;
}
}
if(x-1>Math.abs(y-z)+z-1)
{
System.out.println(2);
continue;
}
else if(x-1<Math.abs(y-z)+z-1)
{
System.out.println(1);
continue;
}
System.out.println(3);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | cc16fdd08dda3e177b2a60b3aafdce37 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.math.*;
public class Solution
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
int a1=Math.abs(b-c)+Math.abs(c-1);
int a2=Math.abs(a-1);
if(a1>a2)
System.out.println("1");
else if(a2>a1)
System.out.println("2");
else
System.out.println("3");
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 7644ba8161fffc9a3797942c87464886 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes |
import java.util.Scanner;
public class Main {
public static void main(String[]args)
{
Scanner sc=new Scanner(System.in);
long n1,n2;
int n=sc.nextInt();
for(int i=1;i<=n;i++)
{
long a=sc.nextLong();
long b=sc.nextLong();
long c=sc.nextLong();
n1=Math.abs(b-c)+Math.abs(c-1);
n2=Math.abs(a-1);
if(n1>n2)
{
System.out.println("1");
}
else if(n2>n1)
{
System.out.println("2");
}
else
{
System.out.println("3");
}
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 331afcb5872f3e17c37b969037f999c7 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class q1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
int t1=0;
int t2=0;
if(a==1)
t1=0;
else
t1=a-1;
if(b>c)
t2=b-1;
else
t2=(c-b)+(c-1);
if(t1>t2)
System.out.println(2);
else if(t1<t2)
System.out.println(1);
else
System.out.println(3);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 27f66844e1e87df7d34df208b9d8d411 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.lang.System.*;
public class TwoElevators implements Runnable {
private final void process() throws IOException {
// InputReader input = new InputReader(new FileReader(System.getenv("INPUT")));
// PrintWriter output = new PrintWriter(new BufferedOutputStream(new FileOutputStream(System.getenv("OUTPUT"))));
InputReader input = new InputReader(in);
PrintWriter output = new PrintWriter(new BufferedOutputStream(out));
short test = input.nextShort();
while (test-- > 0) {
int a = input.nextInt() - 1, b = input.nextInt(), c = input.nextInt();
int optB = b < c ? (c - b) + (c - 1) : b - 1;
if (a == optB) {
output.println(3);
} else if (a < optB) {
output.println(1);
} else {
output.println(2);
}
}
input.close();
output.close();
}
@Override
public void run() {
try {
process();
} catch (IOException ignored) {}
}
public static void main(String... args) throws IOException {
new Thread(null, new TwoElevators(), "Main", 1 << 26).start();
}
private final void printArr(PrintWriter output, short...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private final void printArr(PrintWriter output, int...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private final void printArr(PrintWriter output, double...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private final void printArr(PrintWriter output, long...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private final void printArr(PrintWriter output, char...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private final void printArr(PrintWriter output, String...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
}
class InputReader {
private StringTokenizer token;
private BufferedReader buffer;
public InputReader(InputStream stream) {
buffer = new BufferedReader(new InputStreamReader(stream));
}
public InputReader(FileReader reader) throws FileNotFoundException {
buffer = new BufferedReader(reader);
}
public final String next() throws IOException {
while (token == null || !token.hasMoreTokens())
token = new StringTokenizer(buffer.readLine());
return token.nextToken();
}
public final String nextLine() throws IOException {
return buffer.readLine();
}
public final byte nextByte() throws IOException {
return Byte.parseByte(next());
}
public final short nextShort() throws IOException {
return Short.parseShort(next());
}
public final int nextInt() throws IOException {
return Integer.parseInt(next());
}
public final long nextLong() throws IOException {
return Long.parseLong(next());
}
public final double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public final char nextChar() throws IOException {
return next().charAt(0);
}
public final boolean nextBoolean() throws IOException {
return Boolean.parseBoolean(next());
// return Boolean.getBoolean(next());
// return Boolean.valueOf(next());
}
public int[] readIntArray(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
public int[] readIntArray() throws IOException {
return java.util.Arrays.stream(nextLine().split("\\s+")).
mapToInt(Integer::parseInt).toArray();
}
public long[] readLongArray(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
public long[] readLongArray() throws IOException {
return java.util.Arrays.stream(nextLine().split("\\s+")).
mapToLong(Long::parseLong).toArray();
}
public double[] readDoubleArray(int n) throws IOException {
double[] arr = new double[n];
for (int i = 0; i < n; i++)
arr[i] = nextDouble();
return arr;
}
public double[] readDoubleArray() throws IOException {
return java.util.Arrays.stream(nextLine().split("\\s+")).
mapToDouble(Double::parseDouble).toArray();
}
public char[] readCharArray() throws IOException {
return nextLine().toCharArray();
}
public String[] readStringArray(int n) throws IOException {
String[] arr = new String[n];
for (int i = 0; i < n; i++)
arr[i] = next();
return arr;
}
public String[] readStringArray() throws IOException {
return nextLine().split("\\s+");
}
public boolean ready() throws IOException {
return buffer.ready();
}
public void close() throws IOException {
buffer.close();
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 80632fef1a1243b36d9ae6bcf96007b0 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class elevator
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
int a=sc.nextInt(),b=sc.nextInt(),c=sc.nextInt();
if(a==1)
{
System.out.println("1");
}
else
{
int f,s;
f=Math.abs(a-1);
if(b<c)
s=Math.abs(c-b)+Math.abs(c-1);
else
s=Math.abs(b-1);
if(f<s)
System.out.println("1");
else if(s<f)
System.out.println("2");
else
System.out.println("3");
}
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | a4008188475c7a9a2cacf00ac77864b0 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
public class main2 {
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 IOException
{
Reader s = new Reader();
int test =s.nextInt();
for(int i = 0;i<test;i++)
{
long a=s.nextInt();
long b=s.nextInt();
long c=s.nextInt();
long d=Math.abs(b-c);
long e=Math.abs(c-1);
if(Math.abs(a-1)==Math.abs(d+e))
System.out.println("3");
else if(Math.abs(d+e)>Math.abs(a-1))
System.out.println("1");
else
System.out.println("2");
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | bc615a80dfb37fae716b7603973db836 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | //package com.company;
import java.util.*;
public class Codeforces {
static int gcd(int a,int b){
if (a == 0){
return b;
}
return gcd(b % a,a);
}
static int lcm(int a,int b){
return ((a * b)/gcd(a,b));
}
public static void main(String[] args) {
Scanner lta = new Scanner(System.in);
int t = lta.nextInt();
while (t-- >0) {
int a = lta.nextInt();
int b = lta.nextInt();
int c = lta.nextInt();
if (t < 0) {
if (a == 1) {
System.out.println(1);
} else if (a == 3 && b == 1 && c == 2) {
System.out.println(3);
} else {
System.out.println(2);
}
}
else {
int i = Math.abs(b - c) + Math.abs(c - 1);
if (i < Math.abs(a - 1)){
System.out.println(2);
}
else if (i > Math.abs(a - 1)){
System.out.println(1);
}
else {
System.out.println(3);
}
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 4bc26629ba9650418de31ecd4b1d4da6 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public final class CF{
public static void main(String[] args)
{
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t != 0)
{
int a = s.nextInt();
int b = s.nextInt();
int c = s.nextInt();
int k = Math.abs(1 - c) + Math.abs(c - b);
int m = Math.abs(1 - a);
if(m > k)
System.out.println(2);
else if(k > m)
System.out.println(1);
else System.out.println(3);
t--;
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 11355cdf9818f0b702f8cb9aec17b201 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import static java.lang.System.*;
import static java.lang.System.out;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Arrays.sort;
import static java.util.Collections.shuffle;
import static java.util.Collections.sort;
public class Solution {
private final static FastScanner scanner = new FastScanner();
private final static int mod = (int) 1e9+7;
private final static int max_value = Integer.MAX_VALUE;
private final static int min_value = Integer.MIN_VALUE;
private final static String endl = "\n";
private static void solve() {
int a = ii(), b = ii(), c = ii();
int x = b<c?(c+c-b):b;
if (a<x) pl(1);
else if (a == x) pl(3);
else pl(2);
}
public static void main(String[] args) {
int t = ii();
while (t-->0) solve();
}
private static void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
private static void swap(int[] a, int[] b, int i) {
int temp = a[i];
a[i] = b[i];
b[i] = temp;
}
private static int f(int x) {
String a = x+"";
return a.length();
}
private static void iota(int[] a, int x) {
for (int i = 0; i<a.length; i++) a[i] = x++;
}
private static void reverse(int[] a) {
int[] b = new int[a.length];
int k = 0;
for (int i = a.length-1; i>=0; i--) {
b[k++] = a[i];
}
System.arraycopy(b, 0, a, 0, b.length);
}
private static int[] rai(int n) {
return scanner.readArray(n);
}
private static String[] ras(int n) {
return IntStream.range(0, n).mapToObj(i -> s()).toArray(String[]::new);
}
private static int lcm (int a, int b) {
return a / gcd (a, b) * b;
}
private static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
private static int ii() {
return scanner.nextInt();
}
private static long l() {
return scanner.nextLong();
}
private static double d() {
return scanner.nextDouble();
}
private static String s() {
return scanner.next();
}
private static int toInt(String s) {
return Integer.parseInt(s);
}
private static ArrayList<Integer> list() {
return new ArrayList<>();
}
private static HashSet<Integer> set() {
return new HashSet<>();
}
private static HashMap<Integer, Integer> map() {
return new HashMap<>();
}
private static int toInt(char c) {
return Integer.parseInt(c+"");
}
private static<K> void pl(K a) {
p(a+endl);
}
private static<K> void p(K a) {
out.print(a);
}
private static void yes() {
pl("YES");
}
private static void no() {
pl("NO");
}
private static int max_a(int[] a) {
int max = a[0];
for (int j : a) {
if (j > max) {
max = j;
}
}
return max;
}
private static int min_a(int[] a) {
int min = a[0];
for (int j : a) {
if (j < min) {
min = j;
}
}
return min;
}
private static long sum(int[] a) {
return stream(a).asLongStream().sum();
}
private static void pl() {
out.println();
}
private static void printArray(long[] a) {
StringBuilder builder = new StringBuilder();
for (long i : a)
builder.append(i).append(' ');
pl(builder);
}
private static void printArray(int[] a) {
StringBuilder builder = new StringBuilder();
for (int i : a)
builder.append(i).append(' ');
pl(builder);
}
private static<K> void printArray(K[] a) {
StringBuilder builder = new StringBuilder();
for (K i : a)
builder.append(i).append(' ');
pl(builder);
}
private static int reverseInteger(int k) {
String a = k+"", res = "";
for (int i = a.length()-1; i>=0; i--) {
res+=a.charAt(i);
}
return toInt(res);
}
private static int phi(int n) {
int result = n;
for (int i=2; i*i<=n; i++)
if (n % i == 0) {
while (n % i == 0)
n /= i;
result -= result / i;
}
if (n > 1)
result -= result / n;
return result;
}
private static int pow(int a, int n) {
if (n == 0)
return 1;
if (n % 2 == 1)
return pow(a, n-1) * a;
else {
int b = pow (a, n/2);
return b * b;
}
}
private static boolean isPrimeLong(long n) {
BigInteger a = BigInteger.valueOf(n);
return a.isProbablePrime(10);
}
private static boolean isPrime(long n) {
return IntStream.iterate(2, i -> i * i <= n, i -> i + 1).noneMatch(i -> n % i == 0);
}
private static List<Integer> primes(int N) {
int[] lp = new int[N+1];
List<Integer> pr = new ArrayList<>();
for (int i=2; i<=N; ++i) {
if (lp[i] == 0) {
lp[i] = i;
pr.add(i);
}
for (int j = 0; j<pr.size() && pr.get(j)<=lp[i] && i*pr.get(j)<=N; ++j)
lp[i * pr.get(j)] = pr.get(j);
}
return pr;
}
private static Set<Integer> toSet(int[] a) {
return stream(a).boxed().collect(Collectors.toSet());
}
private static int linearSearch(int[] a, int key) {
return IntStream.range(0, a.length).filter(i -> a[i] == key).findFirst().orElse(-1);
}
private static<K> int linearSearch(K[] a, K key) {
return IntStream.range(0, a.length).filter(i -> a[i].equals(key)).findFirst().orElse(-1);
}
static int upper_bound(int[] arr, int key) {
int index = binarySearch(arr, key);
int n = arr.length;
if (index < 0) {
int upperBound = abs(index) - 1;
if (upperBound < n)
return upperBound;
else return -1;
}
else {
while (index < n) {
if (arr[index] == key)
index++;
else {
return index;
}
}
return -1;
}
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(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());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
private static void stableSort(int[] a) {
List<Integer> list = stream(a).boxed().sorted().toList();
setAll(a, list::get);
}
private static void psort(int[] arr, int n) {
int min = min_a(arr);
int max = max_a(arr);
int range = max-min+1, i, j, index = 0;
int[] count = new int[range];
for(i = 0; i<n; i++)
count[arr[i] - min]++;
for(j = 0; j<range; j++)
while(count[j]-->0)
arr[index++]=j+min;
}
private static void csort(int[] a, int n) {
int max = max_a(a);
int min = min_a(a);
int range = max - min + 1;
int[] count = new int[range];
int[] output = new int[n];
for (int i = 0; i < n; i++) {
count[a[i] - min]++;
}
for (int i = 1; i < range; i++) {
count[i] += count[i - 1];
}
for (int i = n - 1; i >= 0; i--) {
output[count[a[i] - min] - 1] = a[i];
count[a[i] - min]--;
}
arraycopy(output, 0, a, 0, n);
}
private static void csort(char[] arr) {
int n = arr.length;
char[] output = new char[n];
int[] count = new int[256];
for (int i = 0; i < 256; ++i)
count[i] = 0;
for (char c : arr) ++count[c];
for (int i = 1; i <= 255; ++i)
count[i] += count[i - 1];
for (int i = n - 1; i >= 0; i--) {
output[count[arr[i]] - 1] = arr[i];
--count[arr[i]];
}
arraycopy(output, 0, arr, 0, n);
}
// Java Program to show segment tree operations like construction,
// query and update
static class SegmentTree
{
int[] st; // The array that stores segment tree nodes
/* Constructor to construct segment tree from given array. This
constructor allocates memory for segment tree and calls
constructSTUtil() to fill the allocated memory */
SegmentTree(int[] arr, int n)
{
// Allocate memory for segment tree
//Height of segment tree
int x = (int) (Math.ceil(Math.log(n) / Math.log(2)));
//Maximum size of segment tree
int max_size = 2 * (int) Math.pow(2, x) - 1;
st = new int[max_size]; // Memory allocation
constructSTUtil(arr, 0, n - 1, 0);
}
// A utility function to get the middle index from corner indexes.
int getMid(int s, int e) {
return s + (e - s) / 2;
}
/* A recursive function to get the sum of values in given range
of the array. The following are parameters for this function.
st --> Pointer to segment tree
si --> Index of current node in the segment tree. Initially
0 is passed as root is always at index 0
ss & se --> Starting and ending indexes of the segment represented
by current node, i.e., st[si]
qs & qe --> Starting and ending indexes of query range */
int getSumUtil(int ss, int se, int qs, int qe, int si)
{
// If segment of this node is a part of given range, then return
// the sum of the segment
if (qs <= ss && qe >= se)
return st[si];
// If segment of this node is outside the given range
if (se < qs || ss > qe)
return 0;
// If a part of this segment overlaps with the given range
int mid = getMid(ss, se);
return getSumUtil(ss, mid, qs, qe, 2 * si + 1) +
getSumUtil(mid + 1, se, qs, qe, 2 * si + 2);
}
/* A recursive function to update the nodes which have the given
index in their range. The following are parameters
st, si, ss and se are same as getSumUtil()
i --> index of the element to be updated. This index is in
input array.
diff --> Value to be added to all nodes which have i in range */
void updateValueUtil(int ss, int se, int i, int diff, int si)
{
// Base Case: If the input index lies outside the range of
// this segment
if (i < ss || i > se)
return;
// If the input index is in range of this node, then update the
// value of the node and its children
st[si] = st[si] + diff;
if (se != ss) {
int mid = getMid(ss, se);
updateValueUtil(ss, mid, i, diff, 2 * si + 1);
updateValueUtil(mid + 1, se, i, diff, 2 * si + 2);
}
}
// The function to update a value in input array and segment tree.
// It uses updateValueUtil() to update the value in segment tree
void updateValue(int arr[], int n, int i, int new_val)
{
// Check for erroneous input index
if (i < 0 || i > n - 1) {
System.out.println("Invalid Input");
return;
}
// Get the difference between new value and old value
int diff = new_val - arr[i];
// Update the value in array
arr[i] = new_val;
// Update the values of nodes in segment tree
updateValueUtil(0, n - 1, i, diff, 0);
}
// Return sum of elements in range from index qs (query start) to
// qe (query end). It mainly uses getSumUtil()
int getSum(int n, int qs, int qe)
{
// Check for erroneous input values
if (qs < 0 || qe > n - 1 || qs > qe) {
System.out.println("Invalid Input");
return -1;
}
return getSumUtil(0, n - 1, qs, qe, 0);
}
// A recursive function that constructs Segment Tree for array[ss..se].
// si is index of current node in segment tree st
int constructSTUtil(int arr[], int ss, int se, int si)
{
// If there is one element in array, store it in current node of
// segment tree and return
if (ss == se) {
st[si] = arr[ss];
return arr[ss];
}
// If there are more than one elements, then recur for left and
// right subtrees and store the sum of values in this node
int mid = getMid(ss, se);
st[si] = constructSTUtil(arr, ss, mid, si * 2 + 1) +
constructSTUtil(arr, mid + 1, se, si * 2 + 2);
return st[si];
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 716468ba8a5cbd0d8ee00c78a7d9346f | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int i = 1; i <= t; i++) {
int a=in.nextInt(),b=in.nextInt(),c=in.nextInt();
if(a<Math.abs(c-b)+c) System.out.println(1);
else{
if(a==Math.abs(c-b)+c) System.out.println("3");
else System.out.println("2");
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | feca03a547400ad5e5e4fd1fe7fc9d26 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.io.*;
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.hasMoreTokens()){
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
public static void main(String[] args) {
try {
FastReader in=new FastReader();
FastWriter out = new FastWriter();
int testCases=in.nextInt();
while(testCases-- > 0){
int a = in.nextInt();
int b = in.nextInt();
int c = in.nextInt();
int time = Math.abs(b-c);
int secondElevator = time + c;
int min = Math.min(a,secondElevator);
if(a < secondElevator){
System.out.println("1");
}
else if(secondElevator < a){
System.out.println("2");
}
else{
System.out.println("3");
}
}
out.close();
} catch (Exception e) {
return;
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 09e7b3ae1d830b89d7d16f14d2529f05 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | // ronin
import java.io.*;
import java.util.*;
public class A {
public void solve(int a, int b, int c, PrintWriter out) {
int res = a - (Math.abs(c - b) + c);
int ans = 3;
if (res > 0) {
ans = 2;
}
if (res < 0) {
ans = 1;
}
out.println(ans);
}
public void read(FastScanner fastScanner, PrintWriter out) {
int a, b, c, d, e, n;
a = fastScanner.nextInt();
b = fastScanner.nextInt();
c = fastScanner.nextInt();
solve(a, b, c, out);
}
// int n = fastScanner.nextInt();
// while (n--> 0) {
// }
///////////////////////////////////////////////
///////////////////////////////////////////////
///////////////////////////////////////////////
///////////////////////////////////////////////
///////////////////////////////////////////////
public static boolean isLocal = false;
public static void main(String[] args) {
FastScanner fastScanner = new FastScanner();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int N = fastScanner.nextInt();
while (N-- > 0) {
new A().read(fastScanner, out);
}
out.println();
out.close();
if (isLocal) {
try {
new A().buildFile();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
}
static class FastScanner {
BufferedReader br;
StringTokenizer st = new StringTokenizer("");
public FastScanner() {
if (isLocal) {
try {
File f = new File("src/B/input.txt");
this.br = new BufferedReader(new InputStreamReader(new FileInputStream(f)));
} catch (FileNotFoundException e) {
throw new RuntimeException(e);
}
} else {
this.br = new BufferedReader(new InputStreamReader(System.in));
}
}
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 void buildFile() throws IOException {
File thisFile = new File("src/B/A.java");
String s = "";
try (FileReader reader = new FileReader(thisFile)) {
int c;
StringBuilder stringBuilder = new StringBuilder();
while ((c = reader.read()) != -1) {
stringBuilder.append((char) c);
}
s = stringBuilder.toString();
}
/////////////////////////////////////////////////////////////////////////////
s = s.replace("// cout", "// cout");
s = s.replace("// System.out.print", "// System.out.print");
s = s.replace("// cout", "// cout");
s = s.replace("// ronin", "// ronin");
s = s.replace("public static boolean isLocal = false;",
"public static boolean isLocal = false;");
//////////////////////////////////////////////////////////////////////////////////////////////////////////
File file = new File("src/B/LastBuild.txt");
if (!file.exists()) {
file.createNewFile();
}
try (FileWriter writer = new FileWriter(file, false)) {
writer.write(s);
writer.flush();
}
}
///////
public <T> void sout(List<T> s) {
for (T o : s) {
// System.out.print(o + " ");
}
}
public <T, V> void sout(Map<T, V> s) {
for (T t : s.keySet()) {
// System.out.println(t + "=" + s.get(s));
}
}
/////////
/////////
/////////
/////////
class Pair<V, T> {
private V a;
private T b;
public Pair(V a, T b) {
this.a = a;
this.b = b;
}
public V getA() {
return a;
}
public T getB() {
return b;
}
public Pair<V, T> setA(V a) {
this.a = a;
return this;
}
public Pair<V, T> setB(T b) {
this.b = b;
return this;
}
@Override
public String toString() {
return new StringJoiner(", ", Pair.class.getSimpleName() + "[", "]")
.add("a=" + a)
.add("b=" + b)
.toString();
}
}
//algs
private boolean isEven(int a) {
return (a & 1) == 0;
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | bdeffd93413993076d15e254e1326e8e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
public static int sol(int a, int b, int c) {
int res = a - (Math.abs(c - b) + c);
if (res>0) {
return 2;
}
if (res<0) {
return 1;
}
return 3;
}
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 main(String[] args) {
FastScanner fastScanner = new FastScanner();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int n = fastScanner.nextInt();
for (int i = 0; i < n; i++) {
int a, b, c;
a = fastScanner.nextInt();
b = fastScanner.nextInt();
c = fastScanner.nextInt();
out.println(sol(a, b, c));
}
out.close();
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | cabc3816bc1c95a98ba2cc5cea996a5e | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.lang.*;
import java.util.*;
public class Elevators {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
int a,b,c;
while(t-->0) {
a = in.nextInt();
b = in.nextInt();
c = in.nextInt();
if(Math.abs(a-1)>Math.abs(c-b)+Math.abs(1-c))
System.out.println(2);
else if(Math.abs(a-1)<Math.abs(c-b)+Math.abs(1-c))
System.out.println(1);
else
System.out.println(3);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | a3dd8f0760eb372f112e7f2297a7ca69 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int num = scanner.nextInt();
List<List<Integer>> list = new ArrayList<>();
for (int i = 0; i < num; i++) {
List <Integer> tempList = new ArrayList<>();
for (int j = 0; j < 3; j++) {
tempList.add(scanner.nextInt());
}
list.add(tempList);
}
for (List<Integer> integers : list) {
int n1 = integers.get(0) - 1;
int n2 = Math.abs(integers.get(2) - integers.get(1)) + integers.get(2) - 1;
if (n1 < n2) {
System.out.println(1);
} else if (n2 < n1) {
System.out.println(2);
} else {
System.out.println(3);
}
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 6a10bba3bc561234fba15484a936c7e6 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class Te2
{
void elevator(int[]a1)
{
int[]ar=a1.clone();
int x=1;
int a= ar[0];
int b= ar[1];
int c= ar[2];
if(b>c)
{
if((a-x)<((b-c)+(c-x)))
{
System.out.println("1");
}
else if((a-x)>((b-c)+(c-x)))
{
System.out.println("2");
}
else
{
System.out.println("3");
}
}
if(b<c)
{
if((a-x)<((c-b)+(c-x)))
{
System.out.println("1");
}
else if((a-x)>((c-b)+(c-x)))
{
System.out.println("2");
}
else
{
System.out.println("3");
}
}
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
//System.out.println("Enter the number of test cases:");
int t=sc.nextInt();
for(int i=0;i<t;i++)
{
int[] arr = new int[3];
for(int j=0;j<3;j++)
{
//System.out.println("Enter the floors:");
arr[j]=sc.nextInt();
}
Te2 t1=new Te2();
t1.elevator(arr);
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 7b9766ec477676f8cf30728569795e16 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Two_Elevator {
static Scanner sc = new Scanner(System.in);
public static int solve(int[] arr){
int a = arr[0];
int b = arr[1];
int c = arr[2];
if(a == 1) return a;
int count1 = a - 1;
int count2 = 0;
if(b >= c){
count2 = b - 1;
}
else{
count2 += c + (c - b) - 1;
}
if(count1 < count2) return 1;
else if(count1 > count2) return 2;
else return 3;
}
public static void main(String[] args)
{
int t = Integer.parseInt(sc.nextLine());
int[] arr = new int[3];
for(int i = 0; i < t; i++)
{
arr[0] = sc.nextInt();
arr[1] = sc.nextInt();
arr[2] = sc.nextInt();
System.out.println(solve(arr));
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | e30010b2aa6f3f3320e72dea361af32d | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
public class practice
{
public static void main(String[] args)
{
Scanner get = new Scanner(System.in);
int a = get.nextInt();
for(int b=1;b<=a;b++)
{
long x = get.nextLong();
long y = get.nextLong();
long z = get.nextLong();
long t1 = x-1;
long m=(x-1);
long d=Math.abs(y-z)+Math.abs(z-1);
if(m>d)
System.out.println(2);
else if(m<d)
System.out.println(1);
else
System.out.println(3);
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 018820ed90f7ebf771085b0d82135109 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | // A. Two Elevators
import java.util.Scanner;
import java.util.ArrayList;
import java.util.List;
public class Main {
public static void main(String args[]){
Scanner scanner = new Scanner(System.in);
int numTimes = scanner.nextInt();
// Will contain Lists of abc
List<List<Integer>> array = new ArrayList<>();
// Input
scanner.nextLine();
for (int i = 0; i < numTimes; i++) {
String line = scanner.nextLine();
List<Integer> abc = getABC(line);
array.add(abc);
}
// Output
for (int i = 0; i < numTimes; i++) {
System.out.println(getFastestElevator(array.get(i)));
}
}
static List<Integer> getABC(String line){
List<Integer> list = new ArrayList<>();
String tempStr = "";
for (int i = 0; i < line.length(); i++) {
Character ch = line.charAt(i);
if (ch.compareTo(' ') == 0) {
list.add(Integer.parseInt(tempStr));
tempStr = "";
} else if (i == line.length()-1) {
list.add(Integer.parseInt(tempStr+line.charAt(i)));
} else {
tempStr += line.charAt(i);
}
}
return list;
}
static int getFastestElevator(List<Integer> abc){
int a = abc.get(0);
int b = abc.get(1);
int c = abc.get(2);
int firstElevatorDist = a - 1;
int secondElevatorDist = Math.abs(b - c) + c - 1;
if (firstElevatorDist < secondElevatorDist) {
return 1;
} else if (firstElevatorDist > secondElevatorDist) {
return 2;
} else {
return 3;
}
}
} | Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | 938a283188bf527e1c23647de85efdc3 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.Scanner;
public class Ab {
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
int t = sc.nextInt();
while (t-- > 0) {
solve();
}
}
private static void solve() {
int a = sc.nextInt();
int b = sc.nextInt();
int c = sc.nextInt();
a--;
b = Math.abs(b - c) + Math.abs(c - 1);
if (a < b) {
System.out.println("1");
} else if (b < a) {
System.out.println("2");
} else {
System.out.println("3");
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output | |
PASSED | eeb651c9dbb04b2807745c76caaa9191 | train_110.jsonl | 1662993300 | Vlad went into his appartment house entrance, now he is on the $$$1$$$-th floor. He was going to call the elevator to go up to his apartment.There are only two elevators in his house. Vlad knows for sure that: the first elevator is currently on the floor $$$a$$$ (it is currently motionless), the second elevator is located on floor $$$b$$$ and goes to floor $$$c$$$ ($$$b \ne c$$$). Please note, if $$$b=1$$$, then the elevator is already leaving the floor $$$1$$$ and Vlad does not have time to enter it. If you call the first elevator, it will immediately start to go to the floor $$$1$$$. If you call the second one, then first it will reach the floor $$$c$$$ and only then it will go to the floor $$$1$$$. It takes $$$|x - y|$$$ seconds for each elevator to move from floor $$$x$$$ to floor $$$y$$$.Vlad wants to call an elevator that will come to him faster. Help him choose such an elevator. | 256 megabytes | import java.util.*;
import java.io.*;
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.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
// to read an array from input
int[] nextIntArray() {
return Arrays.stream(nextLine().split("\\s")).mapToInt(i -> Integer.parseInt(i)).toArray();
}
long[] nextLongArray() {
return Arrays.stream(nextLine().split("\\s")).mapToLong(i -> Long.parseLong(i)).toArray();
}
double[] nextDoubleArray() {
return Arrays.stream(nextLine().split("\\s")).mapToDouble(i -> Double.parseDouble(i)).toArray();
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
public static void main(String[] args) {
try {
FastReader in = new FastReader();
FastWriter out = new FastWriter();
int testCases = in.nextInt();
while (testCases-- > 0) {
int a = in.nextInt();
int b = in.nextInt();
int c = in.nextInt();
int timeByOne = 0;
int timeByTwo = Math.abs(c-b);
timeByOne += a-1;
timeByTwo += c-1;
if (timeByOne < timeByTwo)
out.println(1);
else if (timeByOne > timeByTwo)
out.println(2);
else
out.println(3);
}
out.close();
} catch (Exception e) {
return;
}
}
}
| Java | ["3\n\n1 2 3\n\n3 1 2\n\n3 2 1"] | 1 second | ["1\n3\n2"] | NoteIn the first test case of the example, the first elevator is already on the floor of $$$1$$$.In the second test case of the example, when called, the elevators would move as follows: At the time of the call, the first elevator is on the floor of $$$3$$$, and the second one is on the floor of $$$1$$$, but is already going to another floor; in $$$1$$$ second after the call, the first elevator would be on the floor $$$2$$$, the second one would also reach the floor $$$2$$$ and now can go to the floor $$$1$$$; in $$$2$$$ seconds, any elevator would reach the floor $$$1$$$. In the third test case of the example, the first elevator will arrive in $$$2$$$ seconds, and the second in $$$1$$$. | Java 17 | standard input | [
"math"
] | aca2346553f9e7b6e944ca2c74bb0b3d | The first line of the input contains the only $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. This is followed by $$$t$$$ lines, three integers each $$$a$$$, $$$b$$$ and $$$c$$$ ($$$1 \le a, b, c \le 10^8$$$, $$$b \ne c$$$) — floor numbers described in the statement. | 800 | Output $$$t$$$ numbers, each of which is the answer to the corresponding test case. As an answer, output: $$$1$$$, if it is better to call the first elevator; $$$2$$$, if it is better to call the second one; $$$3$$$, if it doesn't matter which elevator to call (both elevators will arrive in the same time). | standard output |
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