exec_outcome stringclasses 1 value | code_uid stringlengths 32 32 | file_name stringclasses 111 values | prob_desc_created_at stringlengths 10 10 | prob_desc_description stringlengths 63 3.8k | prob_desc_memory_limit stringclasses 18 values | source_code stringlengths 117 65.5k | lang_cluster stringclasses 1 value | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_time_limit stringclasses 27 values | prob_desc_sample_outputs stringlengths 2 796 | prob_desc_notes stringlengths 4 3k ⌀ | lang stringclasses 5 values | prob_desc_input_from stringclasses 3 values | tags listlengths 0 11 | src_uid stringlengths 32 32 | prob_desc_input_spec stringlengths 28 2.37k ⌀ | difficulty int64 -1 3.5k ⌀ | prob_desc_output_spec stringlengths 17 1.47k ⌀ | prob_desc_output_to stringclasses 3 values | hidden_unit_tests stringclasses 1 value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
PASSED | 4bd498219066a221a33b93b04d7e58cb | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.stream.IntStream;
/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) throws IOException
{
// your code goes here
FastReader sc = new FastReader();
int t=sc.nextInt();
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
while(t--!=0){
int n=sc.nextInt();
int arr[]=new int[n];
boolean ans=true;
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
for(int i=0;i<n;i++){
if(arr[i]%arr[0]!=0){ ans=false; break;}
}
if(ans){
System.out.println("yes");
}else{
System.out.println("no");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 1abbcf0a037ade1cc0225745bcb26a4f | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static long mod = (int)1e9+7;
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc = new FastReader();
int t = sc.nextInt();
// int t = 1;
O: while(t-->0)
{
int n=sc.nextInt();
int a[]=sc.readArray(n);
for(int i=1;i<n;i++)
{
if(a[i]%a[0] !=0)
{
System.out.println("NO");
continue O;
}
}
System.out.println("YES");
}
}
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;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
static void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
static boolean isPalindrome(String str)
{
// Pointers pointing to the beginning
// and the end of the string
int i = 0, j = str.length() - 1;
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str.charAt(i) != str.charAt(j))
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;
}
static boolean palindrome_array(char arr[], int n)
{
// Initialise flag to zero.
int flag = 0;
// Loop till array size n/2.
for (int i = 0; i <= n / 2 && n != 0; i++) {
// Check if first and last element are different
// Then set flag to 1.
if (arr[i] != arr[n - i - 1]) {
flag = 1;
break;
}
}
// If flag is set then print Not Palindrome
// else print Palindrome.
if (flag == 1)
return false;
else
return true;
}
static boolean allElementsEqual(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
static boolean allElementsDistinct(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]!=arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
public static void reverse_Long(long[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
long temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
static boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
static boolean isReverseSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] < a[i + 1]) {
return false;
}
}
return true;
}
static int[] rearrangeEvenAndOdd(int arr[], int n)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
int[] array = list.stream().mapToInt(i->i).toArray();
return array;
}
static long[] rearrangeEvenAndOddLong(long arr[], int n)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
long[] array = list.stream().mapToLong(i->i).toArray();
return array;
}
static boolean isPrime(long 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 (long i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static int getSum(int n)
{
int sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n/10;
}
return sum;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcdLong(long a, long b)
{
if (b == 0)
return a;
return gcdLong(b, a % b);
}
static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static int countDigit(long n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | ca0a498e410970c8eea9f82513817acf | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner sc= new Scanner(System.in);
int t=sc.nextInt();
while(t--!=0)
{
int n= sc.nextInt();
int arr[]= new int[n];
for(int i=0;i<n;i++)
{
arr[i]= sc.nextInt();
}
boolean flag=false;
for(int i=n-1;i>0;i--)
{
if(arr[i]%arr[0]!=0)
{
System.out.println("NO");
flag=true;
break;
}
}
if(!flag)
System.out.println("YES");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | bfd24d0af416ed667755b45ef5a81a1a | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class Main{
static Scanner sc = new Scanner(new BufferedInputStream(System.in));
static int sc(){return sc.nextInt();}
static void print(Object o){out.print(o);}
static void println(Object o){ out.println(o); }
static void println(){out.println(); }
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
static final int N = 500,P = 131,MOD = (int)1e9+7;
static int n,t;
static int INF= 0x3f3f3f3f;
public static void main(String[] args) throws Exception{
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
t = Integer.parseInt(bufferedReader.readLine()) ;
while(t-->0) {
n = Integer.parseInt(bufferedReader.readLine());
String[] strings = bufferedReader.readLine().split(" ");
int [] arr =new int[n+1];
boolean bo = true;
arr[0] = Integer.parseInt(strings[0]);
int sum = arr[0];
for(int i = 1;i<n;i++) {
arr[i] = Integer.parseInt(strings[i]);
sum+=arr[i];
if(arr[i]%arr[0]!=0) {
bo = false;
}
}
if(bo) {println("YES");}
else {println("NO");}
}
bufferedReader.close();
out.close();
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 0b1a3e7fa443b4f1b56227564c3f34a2 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Solution
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int k=0;k<t;k++){
int n=sc.nextInt();
int a[]=new int[n];
for(int j=0;j<n;j++){
a[j]=sc.nextInt();
}
int c=0;
if(a[0]==1) System.out.println("YES");
else{
for(int i=1;i<n;i++){
if(a[i]%a[0]!=0){
c=1;
break;
}
}
if(c==1) System.out.println("NO");
else System.out.println("YES");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | df2f8b6e88ff89cd218bac7362b22654 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class a
{
public static void main (String[] args) throws IOException
{
Scanner sc=new Scanner (System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
long []a=new long[n];boolean f=true;
a[0]=sc.nextLong();
for(int i=1;i<n;i++){
a[i]=sc.nextLong();
if(a[i]%a[0]!=0)f=false;
}
if(f)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 80c533838341a93215133233f8b72e43 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class B {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
mainLoop:
for(int i = 0; i < t; i++) {
int n = Integer.parseInt(br.readLine());
StringTokenizer st = new StringTokenizer(br.readLine());
ArrayList<Integer> list = new ArrayList<>();
for(int j = 0; j < n; j++) {
list.add(Integer.parseInt(st.nextToken()));
}
int h = list.get(0);
for(int j = 1; j < n; j++) {
if(Math.floor((double)list.get(j)/h) != Math.ceil((double)list.get(j)/h)) {
pw.println("NO");
continue mainLoop;
}
}
pw.println("YES");
}
pw.close();
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f1cc3ac0d7d5ca2a35a617d3c6ffeadb | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static FastReader in = new FastReader();
public static void main(String args[]) throws IOException {
int t = in.nextInt();
boolean flag = true;
Main ob = new Main();
int cse = 1;
loop: while (t-- > 0) {
flag = true;
int a[] = in.readintarray(in.nextInt());
for (int i = 1; i < a.length; i++) {
if (a[i] % a[0] != 0) {
flag = false;
break;
}
}
if (flag) {
System.out.println("yes");
} else {
System.out.println("no");
}
}
}
static boolean negCheck(int a[]) {
for (int i : a)
if (i < 0)
return false;
return true;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readintarray(int n) {
int res[] = new int[n];
for (int i = 0; i < n; i++)
res[i] = nextInt();
return res;
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 11 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 80077ad21aa745ed9892011c262d1a39 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.util.Scanner;
public class Problem1 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
boolean f = true;
for (int j = 0; j < t; j++) {
int n = scan.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = scan.nextInt();
}
for (int k = 1; k < n; k++) {
if(arr[k]%arr[0]==0){
f=true;
}else{
f=false;
break;
}
}
if(f){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 46582d8af3317909c79b8c93ccf76e2f | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | // package com.algorithms;
import java.util.*;
public class Pset6 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int length = sc.nextInt();
int[] arr = new int[length];
int fixed = 0;
for (int i = 0; i < length; i++) {
if (i == 0) {
fixed = sc.nextInt();
} else {
arr[i] = sc.nextInt();
}
}
boolean flag = true;
for (int i = 1; i < length; i++) {
if (arr[i] % fixed != 0) {
flag = false;
break;
}
}
if (flag) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
sc.close();
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 98c4da912973518bcdee5bcab66528a8 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class MyClass {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int tc = sc.nextInt();
while(tc-- > 0){
int n = sc.nextInt();
int[] arr = new int[n];
boolean ans = true;
for(int i = 0; i < n; i++){
arr[i] = sc.nextInt();
if(i != 0){
if(arr[i]%arr[0] != 0){
ans = false;
}
}
}
if(ans) System.out.println("YES");
else System.out.println("NO");
}
sc.close();
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 6a7f7e03679e0a0e456d9a4bea1b7e8b | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class test {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int j = 0; j < t; j++) {
int n = sc.nextInt();
int A[] = new int[n];
for (int i = 0; i < n; i++) {
A[i] = sc.nextInt();
}
// Arrays.sort(A);
int min = A[0];
boolean bool=true;
for (int i = 1; i < n; i++) {
if (A[i] % min != 0) {
// System.out.println("NO");
bool=false;
}
}
if(bool) System.out.println("YES");
else System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 23ea72249ffac81d8cf85d037e34c670 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Main{
public static void main(String[] args){
int i,j,k,n,m,t;
Scanner sc = new Scanner(System.in);
t = sc.nextInt();
while(t>0)
{
t-=1;
n = sc.nextInt();
int[] arr = new int[n];
for(i=0;i<n;i++)
arr[i] = sc.nextInt();
boolean ans = true;
for(i=1;i<n && ans;i++)
{
if(arr[i]%arr[0]!=0)
ans = false;
}
if(ans)
System.out.println("YES");
else
System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 94519a81aebb887392a638e4b1a46c17 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Random;
import java.util.Set;
import java.util.StringTokenizer;
public class Main {
public static final boolean MULTIPLE_TESTCASES = true;
static class Solution {
public void solve() {
var n = in.nextInt();
var first = in.nextInt();
var increasing = true;
var isOne = first == 1;
var isMiltiple = true;
for (int i = 1; i < n; ++i) {
var cur = in.nextInt();
if (cur < first) {
increasing = false;
first = cur;
continue;
}
if (cur == 1) {
isOne = true;
first = cur;
continue;
}
if (cur % first != 0) {
isMiltiple = false;
first = cur;
continue;
}
}
if (!increasing) {
out.println("NO");
return;
}
if (isOne) {
out.println("YES");
return;
}
if (isMiltiple) {
out.println("YES");
} else {
out.println("NO");
}
}
private void swap(char[] arr, int i, int j) {
var tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
private void shuffle(int[] a) {
Random r = new Random();
for (var i = 0; i < a.length; i++) {
var oi = r.nextInt(a.length);
var temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
}
// Initialise Solution class
private final FastIOAdapter in;
private final PrintWriter out;
public Solution(FastIOAdapter in, PrintWriter out) {
this.in = in;
this.out = out;
}
}
public static void main(String[] args) throws Exception {
try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
int count = 1;
if (MULTIPLE_TESTCASES) {
count = ioAdapter.nextInt();
}
Solution solution = new Solution(ioAdapter, ioAdapter.out);
while (count-- > 0) {
solution.solve();
}
}
}
static class FastIOAdapter implements AutoCloseable {
private final BufferedReader br;
private final PrintWriter out;
private StringTokenizer st = new StringTokenizer("");
public FastIOAdapter() {
this.br = new BufferedReader(new InputStreamReader(System.in));
this.out =
new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
}
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
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[] readArrayLong(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());
}
@Override
public void close() throws Exception {
out.flush();
out.close();
br.close();
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | e181f8f4ced3b2ec3043446a146e587e | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class CF1708A {
// For fast input output
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
try {
br = new BufferedReader(
new FileReader("input.txt"));
PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
System.setOut(out);
} catch (Exception e) {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
// end of fast i/o code
public static boolean isPalindrome(String str) {
int i = 0;
int j = str.length() - 1;
int flag = 0;
while (i <= j) {
if (str.charAt(i) != str.charAt(j)) {
flag = 1;
break;
}
i++;
j--;
}
return flag == 1 ? false : true;
}
public static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
class Pair {
int x1;
int x2;
public Pair(int x1, int x2) {
this.x1 = x1;
this.x2 = x2;
}
}
static boolean isAssending(int nums[]){
for(int i =1;i<nums.length;i++){
if(nums[i-1]>nums[i])
return false;
}
return true;
}
public static void main(String[] args) {
FastReader fs = new FastReader();
int t =fs.nextInt();
while(t-->0){
int n =fs.nextInt();
int nums[]=fs.readArray(n);
boolean flag=true;
for(int i =1;i<n;i++){
if(nums[i]%nums[0]!=0){
flag=false;
break;
}
}
if (flag)
System.out.println("YES");
else
System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f46022d6abf8b2ecdb2610b8746946d8 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner inp = new Scanner(System.in);
int t = inp.nextInt();
while (t-- > 0) {
int n = inp.nextInt();
int temp = inp.nextInt();
boolean r = true;
for (int i = 1; i < n; i++) {
if (inp.nextInt() % temp != 0)
r = false;
}
System.out.println(r ? "YES" : "NO");
} // end while
} // end main
} // end class | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | ba548e04e8deba215a1795552dccefb5 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.util.*;
public class DifferenceOperations {
public static void solve(Scanner sc) {
int n = sc.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
int ans = 0;
for(int i = 1; i < n; i++) {
if(arr[i] % arr[0] != 0) {
ans = 1;
}
}
if(ans == 0) {
System.out.println("YES");
}
else
System.out.println("NO");
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = Integer.parseInt(sc.nextLine());
while(t-->0) {
solve(sc);
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f7571b536cb048613eaed821a145b3f5 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class ReduceToZero {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for( int i=0; i < t; i++){
int n = sc.nextInt();
sc.nextLine();
String[] val = sc.nextLine().split(" ");
int[] nums = new int[n];
for( int j=0; j < n; j++){
nums[j] = Integer.parseInt( val[j]);
}
process( nums);
}
sc.close();
}
public static void process( int[] nums){
int n= nums.length;
boolean flag1 = true;
for( int i=1; i < n; i++ ){
boolean flag2 = false;
for( int j=0; j < i; j++){
if( nums[j] <= nums[i] && nums[i] % nums[j] == 0 )
flag2 = true;
}
if( flag2 == false)
flag1 = false;
}
if( flag1 == true)
System.out.println("YES");
else
System.out.println("NO");
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f6b825025f27c50826eb761032af5aef | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
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 n = sc.nextInt();
sc.nextLine();
String[] values = sc.nextLine().split(" ");
int[] arr = new int[n];
int j = 0;
for (String val : values) {
arr[j++] = Integer.parseInt(val);
}
System.out.println(solve(arr));
}
sc.close();
}
public static String solve(int[] numbers) {
for (int i = 1; i < numbers.length; i++) {
if (numbers[i] % numbers[0] != 0)
return "NO";
}
return "YES";
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | fc6a6f256e73f2528c52f21002550aa6 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
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 n = sc.nextInt();
int[] arr = new int[n];
for (int j = 0; j < n; j++) {
arr[j] = sc.nextInt();
}
System.out.println(solve(arr));
}
sc.close();
}
static String solve(int[] numbers) {
for (int i = 1; i < numbers.length; i++) {
if (numbers[i] % numbers[0] != 0)
return "NO";
}
return "YES";
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 5bcc523e2c4db3cdeef7f8bea7881c51 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.util.Scanner;
public class DifferenceOperations {
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while (t != 0) {
int length = scan.nextInt();
int[] array = new int[length];
for (int i = 0; i < length; i++) {
array[i] = scan.nextInt();
}
System.out.println(function(array));
t--;
}
}
public static String function(int[] array){
for(int i = 0; i < array.length; i++){
if(array[i] % array[0] != 0){
return "NO";
}
}
return "YES";
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 0946280cc5f73dfaaebc526adc16f294 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.lang.*;
public class Try {
public static void main(String[] args){
Scanner input=new Scanner(System.in);
int testcases=input.nextInt();
while(testcases-->0){
int size=input.nextInt();
int[]arr=new int[size];
for(int a=0;a<size;a++){
arr[a]=input.nextInt();
}
Boolean ok=true;
for(int j=1;j<arr.length;j++){
if(arr[j]%arr[0]!=0){
ok=false;
break;
}
}
if(ok){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | eda77e0566e5a3a8bf4d942ff6ca8e11 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class a {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t --> 0) {
int n = sc.nextInt();
int [] ara = new int[n];
ara[0] = sc.nextInt();
int count = 0;
for (int i = 1; i < n; i++) {
ara[i] = sc.nextInt();
if (ara[i] % ara[0] == 0) count++;
}
if (count == n - 1) System.out.println("YES");
else System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 7b78347935a97ca97c1f8e3183c94922 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.*;
public class R808D2A {
static class InputReader {
public final BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader() {
reader = new BufferedReader(new InputStreamReader(System.in), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public int[] readIntArray(int n) {
int[] x = new int[n];
for (int i = 0; i < n; i++) {
x[i] = nextInt();
}
return x;
}
public int[] readBitString() {
char[] chars = next().toCharArray();
int[] x = new int[chars.length];
for (int i = 0; i < x.length; i++) {
x[i] = chars[i] - '0';
}
return x;
}
public Integer[] readIntegerArray(int n) {
Integer[] x = new Integer[n];
for (int i = 0; i < n; i++) {
x[i] = nextInt();
}
return x;
}
public long[] readLongArray(int n) {
long[] x = new long[n];
for (int i = 0; i < n; i++) {
x[i] = nextLong();
}
return x;
}
public Long[] readLongObjectArray(int n) {
Long[] x = new Long[n];
for (int i = 0; i < n; i++) {
x[i] = nextLong();
}
return x;
}
public double[] readDoubleArray(int n) {
double[] x = new double[n];
for (int i = 0; i < n; i++) {
x[i] = nextDouble();
}
return x;
}
}
static class FastWriter {
BufferedWriter output;
public FastWriter() {
output = new BufferedWriter(
new OutputStreamWriter(System.out));
}
void write(String s) {
try {
output.write(s);
} catch (IOException e) { e.printStackTrace(); }
}
void endLine() {
try {
output.write("\n");
output.flush();
} catch (IOException e) { e.printStackTrace(); }
}
void writeInt(int x) {
write(Integer.toString(x));
endLine();
}
void writeLong(long x) {
write(Long.toString(x));
endLine();
}
void writeLine(String line) {
write(line);
endLine();
}
void writeIntArray(int[] a) {
write(Integer.toString(a[0]));
for (int i = 1; i < a.length; i++) {
write(" "+Integer.toString(a[i]));
}
endLine();
}
void writeLongArray(long[] a) {
write(Long.toString(a[0]));
for (int i = 1; i < a.length; i++) {
write(" "+Long.toString(a[i]));
}
endLine();
}
}
public static class Utility {
public static void safeSort(int[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void safeSort(long[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void shuffle(int[] a) {
Random r = new Random();
for(int i = 0; i < a.length-2; i ++) {
int j = i + r.nextInt(a.length - i);
int t = a[i];
a[i] = a[j];
a[j] = t;
}
}
public static void shuffle(long[] a) {
Random r = new Random();
for(int i = 0; i < a.length-2; i ++) {
int j = i + r.nextInt(a.length - i);
long t = a[i];
a[i] = a[j];
a[j] = t;
}
}
public static HashMap<Integer,Integer> primeFactorize(int x) {
HashMap<Integer,Integer> factors = new HashMap();
int f = 2;
int max = (int)Math.sqrt(x);
while(x > 1 && f <= max) {
int c = 0;
while(x%f == 0) {
x /= f;
c += f%2 + 1;
}
if(c > 0)
factors.put(f,c);
f ++;
}
if(x > 1)
factors.put(x,1);
return factors;
}
}
public static void main(String[] args)
{
InputReader r = new InputReader();
FastWriter w = new FastWriter();
int cases = r.nextInt();
for(int test = 0; test < cases; test ++) {
//INPUT
int n = r.nextInt();
//int k = r.nextInt();
int[] a = r.readIntArray(n);
//CODE
String output = "YES";
for(int i = 1; i < n; i ++) {
if(a[i]%a[0] != 0)
output = "NO";
}
//END CODE
//OUTPUT
w.writeLine(output);
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | db646fbf22bc3a19d4a096df6ac8769d | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
//import org.apache.commons.lang3.ArrayUtils;
public class DifferenceOperations
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t>0)
{
int n = sc.nextInt();
//int x = sc.nextInt();
int ar[] = new int[n];
boolean flag = true;
for(int i = 0; i<n; ++i)
{
ar[i]= sc.nextInt();
}
//if(ArrayUtils.isSorted(array1) == false)
// System.out.println("NO");
//else
for(int i = 1; i<n; ++i)
{
if(ar[i]%ar[0] != 0)
{
flag = false;
break;
}
}
if(flag)
System.out.println("Yes");
else
System.out.println("No");
--t;
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 0381b49ce459560fd2754b6f8054de42 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
import java.lang.Math;
import java.math.BigInteger;
public class A {
static Scanner keyboard = new Scanner(System.in);
public static void main(String[] args) {
int t = keyboard.nextInt();
for (int i = 0; i < t; i++) {
solve();
}
}
public static void solve() {
int n = keyboard.nextInt();
int first = 0;
String ret = "";
for (int i = 0; i < n; i++) {
int next = keyboard.nextInt();
if (first == 0) {
first = next;
}
else {
if (next % first != 0){
ret = "NO";
}
}
}
if (ret.equals("")) {
ret = "YES";
}
System.out.println(ret);
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | b3bbcfb8dc86da2206ae906edba1d8be | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
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 sc = new Reader();
long t=sc.nextLong();
while(t-->0){
long n=sc.nextLong();
boolean flag=false;
long []arr= new long[(int)n];
for(int i=0; i<n; i++){
arr[i]=sc.nextLong();
}
for (int i = 1; i < arr.length; i++) {
if(arr[i]%arr[0]==0) flag=true;
else{
flag=false;
break;
}
}
if(flag) System.out.println("YES");
else System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 7a03285efdd636aa80326bad334df985 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class problemA{
public static void main(String[] args) {
Scanner sh = new Scanner(System.in);
int t = sh.nextInt();
while(t>0){
int n = sh.nextInt();
int a[] = new int[n];
Set<Integer> s = new HashSet<>();
for(int i = 0; i<n; i++){
a[i] = sh.nextInt();
}
boolean check = true;
int min = a[0];
for(int i = 1; i<n; i++){
if(a[i]%min!=0){
check = false;
}
}
if(check){
System.out.println("YES");
}else{
System.out.println("NO");
}
t--;}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | bf3e7bab2c96be35399be882e458d76e | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Arrays;
public class First {
public static void solve(Scanner sc) {
int n = sc.nextInt();
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
list.add(sc.nextInt());
}
boolean flag = true;
for (int i = 1; i < n; i++) {
if (list.get(i) % list.get(0) != 0) {
flag = false;
break;
}
}
if (flag == true) {
System.out.println("YES");
}else {
System.out.println("NO");
}
}
public static void main( String[] args ) {
Scanner sc = new Scanner(System.in);
int tt = sc.nextInt();
while (tt-- > 0) {
solve(sc);
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | d2d78efd1d52b3fb2d26b70a17beb854 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{
static final PrintWriter out =new PrintWriter(System.out);
static final FastReader sc = new FastReader();
//I invented a new word!Plagiarism!
//Did you hear about the mathematician who’s afraid of negative numbers?He’ll stop at nothing to avoid them
//What do Alexander the Great and Winnie the Pooh have in common? Same middle name.
//I finally decided to sell my vacuum cleaner. All it was doing was gathering dust!
//ArrayList<Integer> a=new ArrayList <Integer>();
//char[] a = s.toCharArray();
public static void main (String[] args) throws java.lang.Exception
{ int tes=sc.nextInt();
while(tes-->0)
{
String ans[]={"NO","YES"};
int n=sc.nextInt();
int a[]=new int[n];
int i,f=1;;
for(i=0;i<n;i++)
a[i]=sc.nextInt();
for(i=1;i<n;i++)
{
if(a[i]%a[0]!=0)
{
f=0;
break;
}
}
System.out.println(ans[f]);
}
}
public static int lis(int[] arr) {
int n = arr.length;
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(arr[0]);
for(int i = 1 ; i<n;i++) {
int x = al.get(al.size()-1);
if(arr[i]>=x) {
al.add(arr[i]);
}else {
int v = upper_bound(al, 0, al.size(), arr[i]);
al.set(v, arr[i]);
}
}
return al.size();
}
public static int lower_bound(ArrayList<Integer> ar,int lo , int hi , int k)
{
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (ar.get(mid) <k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==ar.size())
{
return -1;
}
return s;
}
public static int upper_bound(ArrayList<Integer> ar,int lo , int hi, int k)
{
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (ar.get(mid) <=k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==ar.size())
{
return -1;
}
return s;
}
static boolean isPrime(long N)
{
if (N<=1) return false;
if (N<=3) return true;
if (N%2 == 0 || N%3 == 0) return false;
for (int i=5; i*i<=N; i=i+6)
if (N%i == 0 || N%(i+2) == 0)
return false;
return true;
}
static int countBits(long a)
{
return (int)(Math.log(a)/Math.log(2)+1);
}
static long fact(long N)
{
long mod=1000000007;
long n=2;
if(N<=1)return 1;
else
{
for(int i=3; i<=N; i++)n=(n*i)%mod;
}
return n;
}
private static boolean isInteger(String s) {
try {
Integer.parseInt(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
private static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static boolean isPalindrome(String str) {
int i = 0, j = str.length() - 1;
while (i < j)
if (str.charAt(i++) != str.charAt(j--))
return false;
return true;
}
private static String reverseString(String str) {
StringBuilder sb = new StringBuilder(str);
return sb.reverse().toString();
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 4e0f73324250d9d937d160ee48f6631b | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;
public class A1708 {
public static PrintWriter out;
public static MyScanner scanner;
private static void code() {
int testcase = scanner.nextInt();
while (testcase-- > 0) {
int n=scanner.nextInt();
ArrayList<Long> arrayList=new ArrayList<>(n+10);
for (int i = 0; i <n; i++) {
arrayList.add(scanner.nextLong());
}
int val=1;
for (int i = 1; i < n; i++) {
if(arrayList.get(i)%arrayList.get(0)!=0)
{
val=-1;
break;
}
}
out.println(val==-1?"NO":"YES");
}
}
public static void main(String[] args) throws FileNotFoundException {
// scanner = new MyScanner("./input.txt");
scanner = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out), true);
// out=new java.io.PrintWriter(new java.io.BufferedOutputStream(new java.io.FileOutputStream("./output.text")),true);
code();
out.close();
}
static class MyScanner {
private final BufferedReader bufferedReader;
private StringTokenizer stringTokenizer;
public MyScanner(String path) throws FileNotFoundException {
bufferedReader = new BufferedReader(new InputStreamReader(new FileInputStream(path)));
}
public MyScanner() {
bufferedReader = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
try {
stringTokenizer = new StringTokenizer(bufferedReader.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return stringTokenizer.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 = bufferedReader.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 602deab0a95199a8176f8d8f72cc5f8a | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Main{
public static void main(String args[]){
Scanner input=new Scanner(System.in);
int testcase=input.nextInt();
input.nextLine();
for(int k=1;k<=testcase;k++){
int n=input.nextInt();
input.nextLine();
int[] x= new int[n];
for(int i=0;i<n;i++){
int a =input.nextInt();
x[i]=a;
}
boolean y=true;
for(int j=1;j<x.length;j++){
if(x[j]%x[0]!=0)y=false;
}
if(y)System.out.println("yes");
else System.out.println("no");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 33f56b5f047494de1d99395dd2fc1384 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.BufferedReader;
import java.io.FileOutputStream;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.math.BigInteger;
import java.util.StringTokenizer;
import java.io.PrintWriter;
import java.util.*;
public class Main {
//+++++++++++++++++): Solving Problem :(+++++++++++++++++++++//
// greedy, dp, pre-computation, map, list, pair, tree map //
//-----------------------------------------------------------//
private static void solve() {
int n = in.nextInt();
int[] arr = new int[n];
for(int i=0; i<n; i++) arr[i] = in.nextInt();
for(int i=1; i<n; i++){
if(arr[i]%arr[0]!=0){
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
public static void main(String[] args) {
int t = 1;
t = in.nextInt();
while(t-- > 0){
solve();
}
}
//###################################################################
private static PrintWriter pw = new PrintWriter(System.out);
private static FastReader in = new FastReader();
static class FastReader{
private BufferedReader br;
private StringTokenizer st;
public FastReader(){
try{
br = new BufferedReader(new FileReader("input.txt"));
PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
System.setOut(out);
}
catch(Exception exception){
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());
}
BigInteger nextBigInt(){
return new BigInteger(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
private static class Pair<U extends Comparable<U>, V extends Comparable<V> > implements Comparable<Pair<U, V>> {
public U first;
public V second;
public Pair(U first, V second){
this.first = first;
this.second = second;
}
public Pair(){
//default constructor
}
public Pair<U, V> makePair(U first, V second){
return new Pair<>(first, second);
}
@Override
public int compareTo(Pair<U, V> p){
int t = first.compareTo(p.first);
if(t == 0) return second.compareTo(p.second);
return t;
}
}
private static long gcd(long a, long b){
if(b==0) return a;
return gcd(b, a%b);
}
private static long lcm(long a, long b){
return (a/gcd(a, b))*b;
}
private static boolean isEven(long n){
return ((n&1)==0)?true:false;
}
private static boolean isOdd(long n){
return ((n&1)==1)?true:false;
}
private static boolean isPowerOfTwo(long n){
return (n&(n-1))==0?true:false;
}
private static long[] swap(long a, long b){
a^=b;
b^=a;
a^=b;
return new long[]{a, b};
}
private static int totalDigitsInNumber(long n){
return (int) Math.floor(Math.log10(n)) + 1;
}
private static long log2(long n){
assert n >= 1;
int log = 0;
if (n > 0xffff) {
n >>>= 16;
log = 16;
}
if (n > 0xff) {
n >>>= 8;
log |= 8;
}
if (n > 0xf) {
n >>>= 4;
log |= 4;
}
if (n > 0b11) {
n >>>= 2;
log |= 2;
}
return log + (n >>> 1);
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | ee2e85326410551c45823048e4d16474 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Difference_Operations {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int t=in.nextInt();
for(int i=1;i<=t;i++) {
int n=in.nextInt();
int a[]=new int[n];
boolean check=true;
for(int j=0;j<n;j++) a[j]=in.nextInt();
for(int k=1;k<n;k++) {
if(a[k]%a[0]!=0) check=false;
}
if(check) System.out.println("YES");
else System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f84611ef28a8257eb8c1ffd7ec8db929 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
public class DifferentOperations {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
int n = sc.nextInt();
int mark = 0;
int[] a = new int[n];
for (int j = 0; j < n; j++) {
a[j] = sc.nextInt();
if (a[j] % a[0] != 0) {
mark = 1;
}
}
if (mark == 0) {
System.out.println("YES");
} else System.out.println("NO");
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 4c935f3d86decd826c285e2fcc42ccff | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | /******************************************************************************
Online Java Compiler.
Code, Compile, Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.
*******************************************************************************/
import java.util.*;
public class Main
{
public static boolean issorted(int []arr){
for(int i = 1;i<arr.length;i++){
if(arr[i]<arr[i-1]){
return false;
}
}
return true;
}
public static long sum(int []arr){
long sum = 0;
for(int i = 0;i<arr.length;i++){
sum+=arr[i];
}
return sum;
}
public static class pair implements Comparable<pair>{
int x;
int y;
pair(int x,int y){
this.x = x;
this.y = y;
}
public int compareTo(pair o){
return this.x - o.x; // sort increasingly on the basis of x
// return o.x - this.x // sort decreasingly on the basis of x
}
}
public static void swap(int []arr,int i,int j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static int gcd(int a, int b){
if (b == 0)
return a;
return gcd(b, a % b);
}
static int []parent = new int[1000];
static int []size = new int[1000];
public static void make(int v){
parent[v] = v;
size[v] = 1;
}
public static int find(int v){
if(parent[v]==v){
return v;
}
else{
return parent[v] = find(parent[v]);
}
}
public static void union(int a,int b){
a = find(a);
b = find(b);
if(a!=b){
if(size[a]>size[b]){
parent[b] = parent[a];
size[b]+=size[a];
}
else{
parent[a] = parent[b];
size[a]+=size[b];
}
}
}
static boolean []visited = new boolean[1000];
public static void dfs(int vertex,ArrayList<ArrayList<Integer>>graph){
if(visited[vertex] == true){
return;
}
System.out.println(vertex);
for(int child : graph.get(vertex)){
// work to be done before entering the child
dfs(child,graph);
// work to be done after exitting the child
}
}
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-->0){
int n = scn.nextInt();
int []arr = new int[n];
for(int i = 0;i<n;i++){
arr[i] = scn.nextInt();
}
int val =arr[0];
boolean f = true;
for(int i = 1;i<n;i++){
if(arr[i]%val!=0){
f = false;
}
}
if(f){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 9f1728f7b6f36e2980dadf31f0c898e1 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
public class A_Difference_Operations {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int i = 0; i < t; i++){
boolean flag = true;
int n = s.nextInt();
long[] arr = new long[n];
for(int j = 0; j < n; j++){
arr[j] = s.nextLong();
}
for(int j = 1; j < n; j++){
if((arr[j])%(arr[0]) != 0){
flag = false;
}
}
if(flag){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
s.close();
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | fc7be07f067628501a8f618068817b02 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class HelloWorld {
public static void main(String[] args) {
//System.out.println("Hello, World!");
Scanner sc=new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int size = sc.nextInt();
int[] arr = new int[size];
for(int i =0;i<size;i++){
arr[i]=sc.nextInt();
}
int var = arr[0];
boolean div = true;
for(int i =1;i<size;i++){
if(arr[i]%var != 0){
div=false;
break;
}
}
if(div==true){
System.out.println("YES");
}
else System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | df12abad630c9f5be67e8853135471aa | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = Integer.parseInt(sc.nextLine());
while (t>0){
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
boolean ans = true;
for (int i = n-1 ; i > 0 ; i--) {
a[i] = a[i] % a[i-1];
}
for (int i = n - 1; i > 0; i--) {
if (a[i]>0){
int j = i-1;
while (j >= 0 ){
if (a[j] != 0 && a[i] % a[j] == 0) {
a[i] = 0;
break;
}
j--;
}
if (a[i] != 0) {
ans = false;
break;
}
}
}
if (ans) System.out.println("YES");
else System.out.println("NO");
t--;
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f9bf7c36897bb972087bcbe4354f4644 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
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.sort;
public class Solution {
private final static FastScanner scanner = new FastScanner();
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 n = ii();
var a = readArrayInt(n);
for (int i = 1; i<n; i++) {
if (a[i]%a[0]!=0) {
no();
return;
}
}
yes();
}
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[] 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];
}
return b;
}
private static int[] readArrayInt(int n) {
return scanner.readArray(n);
}
private static String[] readArrayString(int n) {
return IntStream.range(0, n).mapToObj(i -> s()).toArray(String[]::new);
}
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 println(K a) {
print(a+endl);
}
private static<K> void print(K a) {
out.print(a);
}
private static void yes() {
println("YES");
}
private static void no() {
println("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 println() {
out.println();
}
private static void printArray(int[] a) {
stream(a).mapToObj(i -> i + " ").forEachOrdered(Solution::print);
println();
}
private static<K> void printArray(K[] a) {
stream(a).map(k -> k + " ").forEachOrdered(Solution::print);
println();
}
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 long phi(long n) {
long result = n;
for (long 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 isPrime(int 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(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
public 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]--;
}
System.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]];
}
System.arraycopy(output, 0, arr, 0, n);
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | df41c92a7ff37b977b7240f691105c01 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | /*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
//package Codeforces;
import java.util.Scanner;
/**
*
* @author Vinay Jain
*/
public class cf_808_a {
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0){
int n = sc.nextInt();
int ar[] = new int[n];
int ele = sc.nextInt();
boolean flag = false;
for(int i = 1 ; i < n ; i++){
ar[i] = sc.nextInt();
if(!flag){
if(ar[i] % ele != 0)
flag = true;
}
}
if(flag == true)
System.out.println("NO");
else
System.out.println("YES");
/*if(n == 2){
if(ar[1] - ar[0] == ar[0])
System.out.println("YES");
else
System.out.println("NO");
continue;
}
boolean flag = true;
int last = ar[1] - ar[0];
for(int i = 2 ; i < n-1 ; i++){
if(ar[i] - ar[i-1] != last){
flag = false;
break;
}
}
if(flag)
System.out.println("YES");
else
System.out.println("NO");
*/
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 1ec81f983ed9491fd54d2fb91833ed30 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
public class codeforces_2022_07_16_A {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
for (int numCases = scanner.nextInt(); numCases > 0; --numCases) {
int n = scanner.nextInt();
int[] arr = new int[n];
boolean doable = true;
for (int i = 0; i < n; i++) {
arr[i] = scanner.nextInt();
if (arr[i] % arr[0] != 0) {
doable = false;
}
}
System.out.println(doable ? "YES": "NO");
}
scanner.close();
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 531fe5fec560652324fa870a7a4cf8e4 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | //package com.company;
import java.util.*;
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-- > 0){
int n = in.nextInt();
int[] arr = new int[n];
int flag = 0;
for (int i = 0; i < n; i++) {
arr[i] = in.nextInt();
int temp = arr[0];
if(arr[i]%temp != 0){
flag = 1;
}
}
if(flag == 1){
System.out.println("NO");
}else{
System.out.println("YES");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | a57333caff92402b3fa91b6a3843494c | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
public class Difference_Operations {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t >= 1 && t<=100) {
int n = sc.nextInt();
int[] a = new int[n];
boolean tf = false;
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
if (a[i] % a[0] != 0) tf = true;
}
if (!tf) System.out.println("YES");
else System.out.println("NO");
t--;
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 017a78aae5845d86f6477cd7939292d3 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.util.*;
public class main {
public static void main(String[] args) throws Exception {
Scanner scan = new Scanner(System.in);
StringBuilder str = new StringBuilder();
int tc = scan.nextInt();
while (tc-->0) {
int n = scan.nextInt();
int first = scan.nextInt();
boolean isFound = false;
for (int i = 0; i < n-1; i++) {
if (scan.nextInt() % first != 0 && !isFound) {
str.append("NO\n");
isFound = true;
}
}
if (!isFound) str.append("YES\n");
}
System.out.println(str.toString());
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | ae5c5de64faa45c6883d25cafec7fc2c | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Objects;
/**
* You are given an array a consisting of n positive integers.
*
* You are allowed to perform this operation any number of times (possibly, zero):
*
* choose an index i (2≤i≤n), and change [ai] to [ai]−[ai−1].
* Is it possible to make [ai]=0 for all 2≤i≤n?
*/
public class DifferenceOperations {
private static final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
public static void main(String[] args) throws IOException {
int testCases = Integer.parseInt(reader.readLine());
for (int i = 0; i < testCases; i++) {
check();
}
}
private static int arrayLength;
private static void check() throws IOException {
arrayLength = Integer.parseInt(reader.readLine());
BigInteger[] array = Arrays.stream(reader.readLine().split(" "))
.map(s -> BigInteger.valueOf(Long.parseLong(s))).toArray(BigInteger[]::new);
BigInteger commonGcd = gcdOfArray(array);
if (Objects.equals(array[0], commonGcd)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
private static BigInteger gcdOfArray(BigInteger[] array) {
BigInteger gcd = array[0];
for (int i = 1; i < arrayLength; i++) {
gcd = gcd.gcd(array[i]);
}
return gcd;
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 0c80ca47dec0c0c806eaf4fc50c54708 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Difference_Operations {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int t=in.nextInt();
for(int i=1;i<=t;i++) {
int n=in.nextInt();
int a[]=new int[n];
boolean check=true;
for(int j=0;j<n;j++) a[j]=in.nextInt();
for(int k=1;k<n;k++) {
if(a[k]%a[0]!=0) check=false;
}
if(check) System.out.println("YES");
else System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 4fef00cbb88b0d0df6c465bd44ba08e0 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Codechef
{
public static void main(String[]args){
Scanner sc=new Scanner(System.in);
int test=sc.nextInt();
while(test-->0)
{
System.out.println(solve(sc)); //solve(sc);
}
sc.close();
}
public static String solve(Scanner sc){
int n=sc.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++)
arr[i]=sc.nextInt();
boolean f=false;
int temp=arr[0];
for(int i=1;i<n;i++)
{
if(arr[i]%temp!=0)
{
f=true;
break;
}
}
if(f==true)
return "NO";
return "YES";
}
}
class define{
public static final long power( long a,long b)
{
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = (res * a);
a = a * a;
b >>= 1;
}
return res;
}
public static final int power( int a,int b)
{
int res = 1;
while (b > 0) {
if ((b & 1)==1)
res = (res * a);
a = a * a;
b >>= 1;
}
return res;
}
public static final void printarray( long a[])
{
for(long i:a)
System.out.print(i+" ");
System.out.println();
}
public static final void printarray( int a[])
{
for(int i:a)
System.out.print(i+" ");
System.out.println();
}
public static final void swap( long a,long b)
{
a=a^b;
b=a^b;
a=a^b;
System.out.println(a+" "+b);
}
public static final void swap( int a,int b)
{
a=a^b;
b=a^b;
a=a^b;
System.out.println(a+" "+b);
}
public static final int maxxorbetweentwonumber( int a,int b)
{
int store=a^b;
int count=0;
while(store>0)
{
count++;
store=store>>1;
}
int res=1;
int ans=0;
while(count-->0)
{
ans+=res;
res=res<<1;
}
return ans;
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 2102b0040ada0f4216fd0cb577beb011 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int[] arr = new int[n];
for(int i = 0 ; i < n; ++i)
arr[i] = sc.nextInt();
boolean t_f = true;
int mod = arr[0];
for(int i = 1; i < n; ++i){
if(arr[i] % mod != 0){
t_f = false;
break;
}
}
if(t_f)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 66a39e19f7de92f9c852faff91394160 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.*;
public class operations {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
for (int i = 0; i < t; i++) {
int n = Integer.parseInt(br.readLine());
StringTokenizer st = new StringTokenizer(br.readLine());
int initial = Integer.parseInt(st.nextToken());
boolean valid = true;
for (int j = 0; j < n - 1; j++) {
if (Integer.parseInt(st.nextToken()) % initial != 0) {
valid = false;
}
}
if (valid) {
pw.println("YES");
}
else {
pw.println("NO");
}
}
pw.close();
br.close();
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | e6122d3322f989f45dc45fc45ab36168 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
import static java.lang.System.out;
public class Problem12 {
static Scanner scanner = new Scanner(System.in);
static void solve() {
int n = scanner.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = scanner.nextInt();
if (isPossible(arr, n)) out.println("YES");
else out.println("NO");
}
static boolean isPossible(int[] arr, int n) {
for (int i = 1; i < n; i++) if (arr[i] % arr[0] != 0) return false;
return true;
}
public static void main(String[] args) {
int T = scanner.nextInt();
while (T-- > 0) solve();
scanner.close();
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 9bee7da8d0c22d6ffc32f25e124418cd | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;
public class pp {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for(int i=0;i<n;i++)
arr[i]=sc.nextInt();
boolean a = false;
for(int i=1; i<n;i++){
a=false;
for(int j=0;j<i;j++){
if(arr[i]%arr[j]==0){
a=true;
break;
}
}
if(!a)
break;
}
out.println(a?"YES":"NO");
}
out.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public boolean hasNext() {
return st.hasMoreTokens();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public boolean ready() throws IOException {
return br.ready();
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 6800cbbd7b81bf6c9230b8506c4e0740 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.text.*;
import java.util.*;
/**
* Provide prove of correctness before implementation. Implementation can cost a lot of time.
* Anti test that prove that it's wrong.
* <p>
* Do not confuse i j k g indexes, upTo and length. Do extra methods!!! Write more informative names to simulation
* <p>
* Will program ever exceed limit?
* Try all approaches with prove of correctness if task is not obvious.
* If you are given formula/rule: Try to play with it.
* Analytic solution/Hardcoded solution/Constructive/Greedy/DP/Math/Brute force/Symmetric data
* Number theory
* Game theory (optimal play) that consider local and global strategy.
* Start writing the hardest code first
*/
public class A {
final boolean ONLINE_JUDGE = java.lang.System.getProperty("ONLINE_JUDGE") != null;
final boolean ANTI_TEST_FINDER_MODE = false;
final Random random = new Random(42);
private int solveOne(int testCase) {
int n = nextInt();
int[] a = nextIntArr(n);
boolean ok = true;
for(int i = 1; i < n; i++) {
ok &= a[i] % a[0] == 0;
}
System.out.println(ok ? "YES" : "NO");
return 0;
}
private int solveOneNaive(int testCase) {
return 0;
}
private void solve() {
if (ANTI_TEST_FINDER_MODE) {
int t = 100_000;
for (int testCase = 0; testCase < t; testCase++) {
int expected = solveOneNaive(testCase);
int actual = solveOne(testCase);
if (expected != actual) {
throw new AssertionRuntimeException(
this.getClass().getSimpleName(),
testCase,
expected,
actual);
}
}
} else {
int t = nextInt();
for (int testCase = 0; testCase < t; testCase++) {
solveOne(testCase);
}
}
}
class AssertionRuntimeException extends RuntimeException {
AssertionRuntimeException(String testName,
int testCase,
Object expected,
Object actual, Object... input) {
super("Testcase: " + testCase + "\n expected = " + expected + ",\n actual = " + actual + ",\n " + Arrays.deepToString(input));
}
}
private void assertThat(boolean b) {
if (!b) throw new RuntimeException();
}
private void assertThat(boolean b, String s) {
if (!b) throw new RuntimeException(s);
}
private int assertThatInt(long a) {
assertThat(Integer.MIN_VALUE <= a && a <= Integer.MAX_VALUE,
"Integer overflow long = [" + a + "]" + " int = [" + (int) a + "]");
return (int) a;
}
void _______debug(String str, Object... os) {
if (!ONLINE_JUDGE) {
System.out.println(MessageFormat.format(str, os));
System.out.flush();
}
}
void _______debug(Object o) {
if (!ONLINE_JUDGE) {
_______debug("{0}", String.valueOf(o));
}
}
private int nextInt() {
return System.in.readInt();
}
private long nextLong() {
return System.in.readLong();
}
private String nextString() {
return System.in.readString();
}
private int[] nextIntArr(int n) {
return System.in.readIntArray(n);
}
private long[] nextLongArr(int n) {
return System.in.readLongArray(n);
}
public static void main(String[] args) {
new A().run();
}
static class System {
private static FastInputStream in;
private static FastPrintStream out;
}
private void run() {
// final long startTime = java.lang.System.currentTimeMillis();
final boolean USE_IO = ONLINE_JUDGE;
if (USE_IO) {
System.in = new FastInputStream(java.lang.System.in);
System.out = new FastPrintStream(java.lang.System.out);
solve();
System.out.flush();
} else {
final String nameIn = "input.txt";
final String nameOut = "output.txt";
try {
System.in = new FastInputStream(new FileInputStream(nameIn));
System.out = new FastPrintStream(new PrintStream(nameOut));
solve();
System.out.flush();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
// final long endTime = java.lang.System.currentTimeMillis();
// _______debug("Execution time: {0}", endTime - startTime);
}
private static class FastPrintStream {
private static final int BUF_SIZE = 8192;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastPrintStream() {
this(java.lang.System.out);
}
public FastPrintStream(OutputStream os) {
this.out = os;
}
public FastPrintStream(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastPrintStream print(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE) innerFlush();
return this;
}
public FastPrintStream print(char c) {
return print((byte) c);
}
public FastPrintStream print(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerFlush();
}
return this;
}
public FastPrintStream print(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerFlush();
});
return this;
}
//can be optimized
public FastPrintStream print0(char[] s) {
if (ptr + s.length < BUF_SIZE) {
for (char c : s) {
buf[ptr++] = (byte) c;
}
} else {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerFlush();
}
}
return this;
}
//can be optimized
public FastPrintStream print0(String s) {
if (ptr + s.length() < BUF_SIZE) {
for (int i = 0; i < s.length(); i++) {
buf[ptr++] = (byte) s.charAt(i);
}
} else {
for (int i = 0; i < s.length(); i++) {
buf[ptr++] = (byte) s.charAt(i);
if (ptr == BUF_SIZE) innerFlush();
}
}
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000) return 10;
if (l >= 100000000) return 9;
if (l >= 10000000) return 8;
if (l >= 1000000) return 7;
if (l >= 100000) return 6;
if (l >= 10000) return 5;
if (l >= 1000) return 4;
if (l >= 100) return 3;
if (l >= 10) return 2;
return 1;
}
public FastPrintStream print(int x) {
if (x == Integer.MIN_VALUE) {
return print((long) x);
}
if (ptr + 12 >= BUF_SIZE) innerFlush();
if (x < 0) {
print((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L) return 19;
if (l >= 100000000000000000L) return 18;
if (l >= 10000000000000000L) return 17;
if (l >= 1000000000000000L) return 16;
if (l >= 100000000000000L) return 15;
if (l >= 10000000000000L) return 14;
if (l >= 1000000000000L) return 13;
if (l >= 100000000000L) return 12;
if (l >= 10000000000L) return 11;
if (l >= 1000000000L) return 10;
if (l >= 100000000L) return 9;
if (l >= 10000000L) return 8;
if (l >= 1000000L) return 7;
if (l >= 100000L) return 6;
if (l >= 10000L) return 5;
if (l >= 1000L) return 4;
if (l >= 100L) return 3;
if (l >= 10L) return 2;
return 1;
}
public FastPrintStream print(long x) {
if (x == Long.MIN_VALUE) {
return print("" + x);
}
if (ptr + 21 >= BUF_SIZE) innerFlush();
if (x < 0) {
print((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastPrintStream print(double x, int precision) {
if (x < 0) {
print('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
print((long) x).print(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
print((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastPrintStream println(char c) {
return print(c).println();
}
public FastPrintStream println(int x) {
return print(x).println();
}
public FastPrintStream println(long x) {
return print(x).println();
}
public FastPrintStream println(String x) {
return print(x).println();
}
public FastPrintStream println(Object x) {
return print(x.toString()).println();
}
public FastPrintStream println(double x, int precision) {
return print(x, precision).println();
}
public FastPrintStream println() {
return print((byte) '\n');
}
public FastPrintStream printf(String format, Object... args) {
return print(String.format(format, args));
}
private void innerFlush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerFlush");
}
}
public void flush() {
innerFlush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
}
private static class FastInputStream {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public FastInputStream(InputStream stream) {
this.stream = stream;
}
public double[] readDoubleArray(int size) {
double[] array = new double[size];
for (int i = 0; i < size; i++) {
array[i] = readDouble();
}
return array;
}
public String[] readStringArray(int size) {
String[] array = new String[size];
for (int i = 0; i < size; i++) {
array[i] = readString();
}
return array;
}
public char[] readCharArray(int size) {
char[] array = new char[size];
for (int i = 0; i < size; i++) {
array[i] = readCharacter();
}
return array;
}
public String readText() {
StringBuilder result = new StringBuilder();
while (true) {
int character = read();
if (character == '\r') {
continue;
}
if (character == -1) {
break;
}
result.append((char) character);
}
return result.toString();
}
public long[] readLongArray(int size) {
long[] array = new long[size];
for (int i = 0; i < size; i++) {
array[i] = readLong();
}
return array;
}
public int[] readIntArray(int size) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = readInt();
}
return array;
}
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 peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int peekNonWhitespace() {
while (isWhitespace(peek())) {
read();
}
return peek();
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public char[] readStringAsCharArray() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
char[] resArr = new char[res.length()];
res.getChars(0, res.length(), resArr, 0);
return resArr;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public char readCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return readString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 66150710d7010e97dd76641e7fe3c7dd | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
public class Mavenproject3 {
public static void main(String[] args) {
Scanner in =new Scanner(System.in);
int x=in.nextInt();
for(int i=1;i<=x;i++){
int y=in.nextInt();
int [] z=new int[y];
boolean fact=true;
for(int g=0;g<y;g++){
z[g]=in.nextInt();
}
for(int j=0;j<y;j++){
if(z[j]%z[0]!=0)
fact=false;
}
if(fact){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 1b91992c8c81e8b65078fd5874024d35 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
import java.text.*;
public class Main{
public static void main(String args[]) throws IOException{
Read sc=new Read();
int n=sc.nextInt();
for(int i=0;i<n;i++){
int m=sc.nextInt();
int a=sc.nextInt();
String ok="YES";
for(int j=0;j<m-1;j++){
int b=sc.nextInt();
if(b%a>0){
ok="NO";
}
}
sc.println(ok);
}
//sc.print(0);
sc.bw.flush();
sc.bw.close();
}
}
class Read{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public Read(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public byte nextByte() throws IOException{
return Byte.parseByte(next());
}
public short nextShort() throws IOException{
return Short.parseShort(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public void println(int a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
return;
}
public void print(int a) throws IOException{
bw.write(String.valueOf(a));
return;
}
public void println(String a) throws IOException{
bw.write(a);
bw.newLine();
return;
}
public void print(String a) throws IOException{
bw.write(a);
return;
}
public void println(long a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
return;
}
public void print(long a) throws IOException{
bw.write(String.valueOf(a));
return;
}
public void println(double a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
return;
}
public void print(double a) throws IOException{
bw.write(String.valueOf(a));
return;
}
public void print(BigInteger a) throws IOException{
bw.write(a.toString());
return;
}
public void print(char a) throws IOException{
bw.write(String.valueOf(a));
return;
}
public void println(char a) throws IOException{
bw.write(String.valueOf(a));
bw.newLine();
return;
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f9c3e437a00fcd5bd2aa4970944a7762 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.abs;
import java.util.*;
import java.math.*;
public class A_Difference_Operations {
public static void solve(Scanner in){
int n = in.nextInt();
int arr[] = new int[n];
for(int i = 0; i<n ; i++){
arr[i] = in.nextInt();
}
int mn = arr[0];
boolean flag = false;
for(int i = 1; i<n; i++){
if(arr[i] % mn != 0){
flag = true;
}
}
if(flag == false)
System.out.println("YES");
else
System.out.println("NO");
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int i = 0; i<t; i++){
solve(in);
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 574407b6aaafcb5cb7a125eaf4c6d8a5 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class try_sol {
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int test = sc.nextInt();
for(int t=0; t<test ;t++)
{
int n = sc.nextInt();
int[] arr = new int[n];
for(int i=0; i<n ;i++)
arr[i] = sc.nextInt();
boolean flag = false;
int count=0;
for(int i=1; i<n ;i++)
{
if(arr[i]%arr[0] == 0)
{
count++;
}
}
if(count == n-1)
System.out.println("YES");
else
System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | a19b03c57ab34bc0139a1295059c7c29 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int tc = 0; tc < t; ++tc) {
int n = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < a.length; ++i) {
a[i] = sc.nextInt();
}
System.out.println(solve(a) ? "YES" : "NO");
}
sc.close();
}
static boolean solve(int[] a) {
return IntStream.range(1, a.length).allMatch(i -> a[i] % a[0] == 0);
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 30d54e1287023c9ab1cbd63d3da29548 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Test {
public static void main(String[] args) throws IOException {
InputReader a = new InputReader(System.in);
int testCases = a.nextInt();
int arraySize;
int[] array;
while (testCases > 0) {
arraySize = a.nextInt();
array = a.readIntArray(arraySize);
while (arraySize > 1) {
if ( array[arraySize-1] == 0 || array[arraySize-1] % array[0] != 0) {
System.out.println("NO");
break;
}
arraySize--;
}
if (arraySize == 1) {
System.out.println("YES");
}
testCases--;
}
}
}
class InputReader {
public final 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 double nextDouble() {
return Double.parseDouble(next());
}
public int[] readIntArray(int n) {
int[] x = new int[n];
for (int i = 0; i < n; i++) {
x[i] = nextInt();
}
return x;
}
public int[] readBitString() {
char[] chars = next().toCharArray();
int[] x = new int[chars.length];
for (int i = 0; i < x.length; i++) {
x[i] = chars[i] - '0';
}
return x;
}
public Integer[] readIntegerArray(int n) {
Integer[] x = new Integer[n];
for (int i = 0; i < n; i++) {
x[i] = nextInt();
}
return x;
}
public long[] readLongArray(int n) {
long[] x = new long[n];
for (int i = 0; i < n; i++) {
x[i] = nextLong();
}
return x;
}
public Long[] readLongObjectArray(int n) {
Long[] x = new Long[n];
for (int i = 0; i < n; i++) {
x[i] = nextLong();
}
return x;
}
public double[] readDoubleArray(int n) {
double[] x = new double[n];
for (int i = 0; i < n; i++) {
x[i] = nextDouble();
}
return x;
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 2acad1d08d946b156b7131b4391d621d | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.stream.IntStream;
public class TaskA {
public static void main(String[] args) {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
TaskA s = new TaskA();
s.solve(in, out);
out.flush();
}
void solveOne(FastScanner in, PrintWriter out) {
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
boolean can = IntStream.range(1, n)
.allMatch(i -> a[i] % a[0] == 0);
out.println(can ? "YES" : "NO");
}
void solve(FastScanner in, PrintWriter out) {
int t = 1;
t = in.nextInt();
for (int tc = 1; tc <= t; tc++) {
// out.printf("Case #%d: ", tc);
solveOne(in, out);
}
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
float nextFloat() { return Float.parseFloat(next()); }
double nextDouble() { return Double.parseDouble(next()); }
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | f2d72959539a69e62e58fc6f02216267 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.sql.SQLOutput;
import java.util.*;
import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.abs;
import static java.lang.Math.sqrt;
public class Main
{
private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public static void main(String[] args) throws IOException {
int t = getSingle();
while(t --> 0) {
int n = getSingle();
if(solve(n)) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}
public static boolean solve(int n) throws IOException {
StringTokenizer st = new StringTokenizer(br.readLine());
int first = getNextInt(st);
for(int i = 1; i < n; ++i) {
if(getNextInt(st) % first != 0) {
return false;
}
}
return true;
}
public static int[] getInts(int n, BufferedReader br, StringTokenizer st) {
int[] arr = new int[n];
for(int i=0; i < n; ++i) {
arr[i] = Integer.parseInt(st.nextToken());
}
return arr;
}
public static void printArr(int[] a) {
for(int i : a) {
System.out.print(i + " ");
}
System.out.println();
}
public static void sort(int[] a) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : a) {
ls.add(x);
}
Collections.sort(ls);
for (int i = 0; i < a.length; i++) {
a[i] = ls.get(i);
}
}
public static long pow(long x, long y, long p) {
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;
}
public static long getSingleLong() throws IOException {
return Long.parseLong(br.readLine());
}
public static int getSingle() throws IOException {
return Integer.parseInt(br.readLine());
}
public static int getNextInt(StringTokenizer st) {
return Integer.parseInt(st.nextToken());
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 33ec643f6bd0e7ac79e614481b591ece | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.*;
public class A {
Input in;
PrintWriter out;
public A() {
in = new Input(System.in);
out = new PrintWriter(System.out);
}
public static void main(String[] args) {
A solution = new A();
for (int t = solution.in.nextInt(); t > 0; t--) {
solution.solve();
}
solution.out.close();
}
void solve() {
int n = in.nextInt();
int[] a = in.nextIntArray(n);
for (int i=1; i<n; i++) {
if (a[i] % a[i-1] > 0) {
out.println("NO");
return;
} else {
a[i] = Math.min(a[i], a[i-1]);
}
}
out.println("YES");
}
static class Input {
BufferedReader br;
StringTokenizer st;
public Input(InputStream in) {
br = new BufferedReader(new InputStreamReader(in));
st = new StringTokenizer("");
}
String nextString() {
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(nextString());
}
long nextLong() {
return Long.parseLong(nextString());
}
int[] nextIntArray(int size) {
int[] ans = new int[size];
for (int i = 0; i < size; i++) {
ans[i] = nextInt();
}
return ans;
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | c36c892bf3517f03b54e01a1f41c39c3 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.math.*;
import java.util.*;
import java.io.*;
public class A
{
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
if(st.hasMoreTokens())
{
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static int gcd(int a, int b)
{
BigInteger b1 = BigInteger.valueOf(a);
BigInteger b2 = BigInteger.valueOf(b);
BigInteger gcd = b1.gcd(b2);
return gcd.intValue();
}
public static long gcd(long a, long b)
{
BigInteger b1 = BigInteger.valueOf(a);
BigInteger b2 = BigInteger.valueOf(b);
BigInteger gcd = b1.gcd(b2);
return gcd.longValue();
}
static FastReader Sc = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args)
{
int t = Sc.nextInt();
while(t --> 0)
{
int n = Sc.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++)
{
arr[i] = Sc.nextInt();
}
boolean check = true;
for(int i = 1; i < arr.length; i++)
{
if(arr[i] % arr[0] != 0)
{
check = false;
break;
}
}
if(check)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | e6163cc9d1488cdddc3abc44dda6dd59 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | /******************************************************************************
Online Java Compiler.
Code, Compile, Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.
*******************************************************************************/
import java.util.*;
public class Main
{
public static boolean issorted(int []arr){
for(int i = 1;i<arr.length;i++){
if(arr[i]<arr[i-1]){
return false;
}
}
return true;
}
public static long sum(int []arr){
long sum = 0;
for(int i = 0;i<arr.length;i++){
sum+=arr[i];
}
return sum;
}
public static class pair implements Comparable<pair>{
int x;
int y;
pair(int x,int y){
this.x = x;
this.y = y;
}
public int compareTo(pair o){
return this.x - o.x; // sort increasingly on the basis of x
// return o.x - this.x // sort decreasingly on the basis of x
}
}
public static void swap(int []arr,int i,int j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static int gcd(int a, int b){
if (b == 0)
return a;
return gcd(b, a % b);
}
static int []parent = new int[1000];
static int []size = new int[1000];
public static void make(int v){
parent[v] = v;
size[v] = 1;
}
public static int find(int v){
if(parent[v]==v){
return v;
}
else{
return parent[v] = find(parent[v]);
}
}
public static void union(int a,int b){
a = find(a);
b = find(b);
if(a!=b){
if(size[a]>size[b]){
parent[b] = parent[a];
size[b]+=size[a];
}
else{
parent[a] = parent[b];
size[a]+=size[b];
}
}
}
static boolean []visited = new boolean[1000];
public static void dfs(int vertex,ArrayList<ArrayList<Integer>>graph){
if(visited[vertex] == true){
return;
}
System.out.println(vertex);
for(int child : graph.get(vertex)){
// work to be done before entering the child
dfs(child,graph);
// work to be done after exitting the child
}
}
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-->0){
int n = scn.nextInt();
int []arr = new int[n];
for(int i = 0;i<n;i++){
arr[i] = scn.nextInt();
}
int val =arr[0];
boolean f = true;
for(int i = 1;i<n;i++){
if(arr[i]%val!=0){
f = false;
}
}
if(f){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | c678daeb4cfb2e60d42fd676730132c0 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 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 java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int[] arr = new int[n];
for(int i=0; i<n; i++){
arr[i] = sc.nextInt();
}
boolean f = false;
for(int i=0; i<n; i++){
if((arr[i] % arr[0])!=0)
f = true;
}
if(f==false)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 4cb4ec88ed6e720c9862e6944567e71a | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.lang.System.*;
public class DifferenceOperations implements Runnable {
final String[] possible = {"YES", "NO"};
private final void solution() 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));
byte test = input.nextByte();
while (test-- > 0) {
int n = input.nextInt();
long[] arr = input.readLongArray(n);
boolean poss = true;
for (int i = n - 1; i >= 1; i--) {
if (arr[i] % arr[0] != 0) {
poss = false;
break;
}
; }
int idx = poss ? 0 : 1;
output.println(possible[idx]);
}
output.close();
input.close();
}
@Override
public void run() {
try {
solution();
} catch (IOException ignore) {}
}
public static void main(String... args) throws IOException {
new Thread(null, new DifferenceOperations(), "Main", 1 << 26).start();
}
private static final void printArr(PrintWriter output, int...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private static final void printArr(PrintWriter output, double...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private static final void printArr(PrintWriter output, long...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private static final void printArr(PrintWriter output, String...arr) {
output.println(Arrays.toString(arr).replaceAll("\\[|]|, ", " ").trim());
}
private static final boolean isPowerofTwo(long n){
if (n <= 0) return false;
int bit = (int) (Math.log(n) / Math.log(2));
long expec = 1 << bit;
return n % expec == 0;
}
}
class InputReader {
private StringTokenizer st;
private BufferedReader br;
public InputReader(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public InputReader(FileReader s) throws FileNotFoundException {
br = new BufferedReader(s);
}
public final String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public final String nextLine() throws IOException {
return br.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 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 br.ready();
}
public void close() throws IOException {
br.close();
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | a4ce0dd43a0d11ca45a8451d9a60f02b | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int size = in.nextInt();
for (int i = 0; i < size; i++) {
int size_arr = in.nextInt();
int[] arr = new int[size_arr];
for (int j = 0; j < arr.length; j++) {
arr[j] = in.nextInt();
}
boolean flag = false;
for (int j = 1; j < arr.length; j++) {
if(arr[j] % arr[0] != 0) {
flag = true;
break;
}
}
if(!flag) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | e64fc11d364455d3377125175688122f | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.Scanner;
public class solution33 {
public static void main(String[] arg) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < arr.length; i++) {
arr[i] = sc.nextInt();
}
int c = 0;
for (int j : arr) {
if (j % arr[0] == 0)
c ++;
}
if(c==n)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 88f860e83d043d1870ab03bc96c0ba05 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class A_Difference_Operations{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- != 0){
int n = sc.nextInt();
int arr[] = new int[n];
for(int i = 0; i < n; i++){
arr[i] = sc.nextInt();
}
int mod = arr[0];
boolean flag = true;
for(int i = 0; i < n; i++){
if(arr[i] % mod != 0){
System.out.println("NO");
flag = false;
break;
}
}
if(flag)
System.out.println("YES");
}
sc.close();
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 93459a0f4fa04f1f7e70f2164e44c211 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int[] arr = new int[n];
for(int i = 0 ; i < n; ++i)
arr[i] = sc.nextInt();
boolean flag = true;
int mod = arr[0];
for(int i = 1; i < n; ++i){
if(arr[i] % mod != 0){
flag = false;
break;
}
}
if(flag)
System.out.println("YES");
else
System.out.println("NO");
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 66b9e71aafbb9e4d7843162a2a669f8a | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.StringTokenizer;
public class Ant {
public static void main(String [] args)
{
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int Test = fs.nextInt();
// int Test = 1;
for(int test =0;test<Test;test++)
{
/*#### START WRITING CODE ####*/
int n = fs.nextInt();
int [] a = fs.readA(n);
boolean check = true;
int p = a[0];
for(int i=0;i<n;i++)
{
if(a[i]%p!=0)
{
out.println("NO");
check = false;
break;
}else {
continue;
}
}
if(check == true)
{
out.println("YES");
}
/*#### STOP WRITING CODE ####*/
}
out.close();
}
/*###################### FAST SCANNER JAVA CP ###############################*/
static class FastScanner{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while(!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
}catch(IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
int [] readA(int n)
{
int [] nums = new int[n];
for(int i=0;i<n;i++)
{
nums[i]=nextInt();
}
return nums;
}
}
static void debugA(int [] arr)
{
for(int i=0;i<arr.length;i++)
{
System.out.print(arr[i]+" ");
}
}
static void sortA(int [] arr)
{
List<Integer> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list);
for(int i=0;i<list.size();i++)
{
arr[i]=list.get(i);
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | ae580868fdc055043d720ef5273405d7 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes |
import java.util.*;
import java.io.*;
public class A {
public static void main(String[] args) throws IOException {
Soumit sc = new Soumit();
int tc = sc.nextInt();
StringBuilder sb = new StringBuilder();
while (tc-->0){
int n = sc.nextInt();
int[] arr = sc.nextIntArray(n);
boolean flag = true;
for(int i=1;i<n;i++){
if(arr[i]%arr[0] != 0){
flag = false;
break;
}
}
if(flag){
sb.append("YES\n");
}
else{
sb.append("NO\n");
}
}
System.out.println(sb);
sc.close();
}
static class Soumit {
final private int BUFFER_SIZE = 1 << 18;
final private DataInputStream din;
final private byte[] buffer;
private PrintWriter pw;
private int bufferPointer, bytesRead;
StringTokenizer st;
public Soumit() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Soumit(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public void streamOutput(String file) throws IOException {
FileWriter fw = new FileWriter(file);
BufferedWriter bw = new BufferedWriter(fw);
pw = new PrintWriter(bw);
}
public void println(String a) {
pw.println(a);
}
public void print(String a) {
pw.print(a);
}
public String readLine() throws IOException {
byte[] buf = new byte[3000064]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public void sort(int[] arr) {
ArrayList<Integer> arlist = new ArrayList<>();
for (int i : arr)
arlist.add(i);
Collections.sort(arlist);
for (int i = 0; i < arr.length; i++)
arr[i] = arlist.get(i);
}
public void sort(long[] arr) {
ArrayList<Long> arlist = new ArrayList<>();
for (long i : arr)
arlist.add(i);
Collections.sort(arlist);
for (int i = 0; i < arr.length; i++)
arr[i] = arlist.get(i);
}
public int[] nextIntArray(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextLong();
}
return arr;
}
public double[] nextDoubleArray(int n) throws IOException {
double[] arr = new double[n];
for (int i = 0; i < n; i++) {
arr[i] = nextDouble();
}
return arr;
}
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;*/
if (din != null) din.close();
if (pw != null) pw.close();
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 335b8c0bd5a99775141d10829c134294 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
/**
* @Author Awiiiiii~
* @Data 2022/7/16 0:56
* @Version 1.0
*/
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 len = sc.nextInt();
boolean f = true;
int first = sc.nextInt();
for (int j = 0; j < len - 1; j++) {
f &= (sc.nextInt() % first == 0);
}
System.out.println(f ? "YES" : "NO");
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 2fdd92f64a4f137f6329e3097fc5e819 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
ADifferenceOperations solver = new ADifferenceOperations();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++) {
solver.solve(i, in, out);
}
out.close();
}
static class ADifferenceOperations {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
for (int i = 1; i < n; i++) {
if (a[i] % a[0] != 0) {
out.println("NO");
return;
}
}
out.println("YES");
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new UnknownError();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new UnknownError();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new UnknownError();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return nextString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 414284290a6e99a6e39bdfeccda1972f | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class A {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}
catch(IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() throws IOException{
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;
}
int[] nextArray(int n){
int[] a = new int[n];
for(int i = 0; i < n; i++){
a[i] = Integer.parseInt(next());
}
return a;
}
}
public static void main(String[] args) throws IOException
{
// System.setIn(new FileInputStream(new File("/Users/pluo/Desktop/CF/input.txt")));
// System.setOut(new PrintStream(new File("/Users/pluo/Desktop/CF/output.txt")));
FastReader sc = new FastReader();
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int tt = sc.nextInt();
for(int a0 = 0; a0 < tt; a0++){
int n = sc.nextInt();
int[] a = sc.nextArray(n);
boolean ok = true;
for(int i = 1; i < n; i++){
if(a[i] < a[0] || a[i] % a[0] != 0){
ok = false;
break;
}
}
out.write((ok) ? "YES\n" : "NO\n");
}
out.flush();
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 679fe16015d23eb61588d37ebf1db4ea | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.*;
import java.util.*;
public class main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++) arr[i] = sc.nextInt();
boolean temp = true;
for(int i = 1; i < n; i++) {
if(arr[i] % arr[0] != 0) {
temp = false;
break;
}
}
System.out.println(temp ? "Yes" : "No");
}
sc.close();
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 00e8efb4d5eaaed051ba04a9a636e5b7 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | /*----------- ---------------*
Author : Ryan Ranaut
__Hope is a big word, never lose it__
------------- --------------*/
import java.io.*;
import java.util.*;
public class Codeforces1 {
static PrintWriter out = new PrintWriter(System.out);
static final int mod = 1_000_000_007;
static long min = (long) (-1e16);
static final int max = (int)1e6;
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());
}
int[] readIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
/*--------------------------------------------------------------------------*/
//Try seeing general case
//Minimization Maximization - BS..... Connections - Graphs.....
//Greedy not worthy - Try DP
//Think edge cases
public static void main(String[] args)
{
FastReader s = new FastReader();
int t = s.nextInt();
while(t-->0)
{
int n = s.nextInt();
long[] a = s.readLongArray(n);
long gcd = a[0];
for(int i=1;i<n;i++)
gcd = gcd(a[i], gcd);
if(gcd == a[0])
out.println("YES");
else
out.println("NO");
}
out.close();
}
public static void find()
{
}
/*----------------------------------End of the road--------------------------------------*/
static class DSU {
int[] parent;
int[] ranks;
int[] groupSize;
int size;
public DSU(int n) {
size = n;
parent = new int[n];//0 based
ranks = new int[n];
groupSize = new int[n];//Size of each component
for (int i = 0; i < n; i++) {
parent[i] = i;
ranks[i] = 1;
groupSize[i] = 1;
}
}
public int find(int x)//Path Compression
{
if (parent[x] == x)
return x;
else
return parent[x] = find(parent[x]);
}
public void union(int x, int y)//Union by rank
{
int x_rep = find(x);
int y_rep = find(y);
if (x_rep == y_rep)
return;
if (ranks[x_rep] < ranks[y_rep]) {
parent[x_rep] = y_rep;
groupSize[y_rep] += groupSize[x_rep];
} else if (ranks[x_rep] > ranks[y_rep]) {
parent[y_rep] = x_rep;
groupSize[x_rep] += groupSize[y_rep];
} else {
parent[y_rep] = x_rep;
ranks[x_rep]++;
groupSize[x_rep] += groupSize[y_rep];
}
size--;//Total connected components
}
}
public static int gcd(int x, int y) {
return y == 0 ? x : gcd(y, x % y);
}
public static long gcd(long x, long y) {
return y == 0L ? x : gcd(y, x % y);
}
public static int lcm(int a, int b) {
return (a * b) / gcd(a, b);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static long pow(long a, long b) {
if (b == 0L)
return 1L;
long tmp = 1;
while (b > 1L) {
if ((b & 1L) == 1)
tmp *= a;
a *= a;
b >>= 1;
}
return (tmp * a);
}
public static long modPow(long a, long b, long mod) {
if (b == 0L)
return 1L;
long tmp = 1;
while (b > 1L) {
if ((b & 1L) == 1L)
tmp *= a;
a *= a;
a %= mod;
tmp %= mod;
b >>= 1;
}
return (tmp * a) % mod;
}
static long mul(long a, long b) {
return a * b;
}
static long fact(int n) {
long ans = 1;
for (int i = 2; i <= n; i++) ans = mul(ans, i);
return ans;
}
static long fastPow(long base, long exp) {
if (exp == 0) return 1;
long half = fastPow(base, exp / 2);
if (exp % 2 == 0) return mul(half, half);
return mul(half, mul(half, base));
}
static void debug(int... a) {
for (int x : a)
out.print(x + " ");
out.println();
}
static void debug(long... a) {
for (long x : a)
out.print(x + " ");
out.println();
}
static void debugMatrix(int[][] a) {
for (int[] x : a)
out.println(Arrays.toString(x));
}
static void debugMatrix(long[][] a) {
for (long[] x : a)
out.println(Arrays.toString(x));
}
static void reverseArray(int[] a) {
int n = a.length;
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void sort(int[] a) {
ArrayList<Integer> ls = new ArrayList<>();
for (int x : a) ls.add(x);
Collections.sort(ls);
for (int i = 0; i < a.length; i++)
a[i] = ls.get(i);
}
static void sort(long[] a) {
ArrayList<Long> ls = new ArrayList<>();
for (long x : a) ls.add(x);
Collections.sort(ls);
for (int i = 0; i < a.length; i++)
a[i] = ls.get(i);
}
static class Pair {
long x, y;
Pair(long x, long y) {
this.x = x;
this.y = y;
}
}
} | Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 56da19b94bec020c9aad32af4315b888 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.util.*;
import java.util.function.*;
import java.io.*;
// you can compare with output.txt and expected out
public class Round808A {
MyPrintWriter out;
MyScanner in;
// final static long FIXED_RANDOM;
// static {
// FIXED_RANDOM = System.currentTimeMillis();
// }
final static String IMPOSSIBLE = "IMPOSSIBLE";
final static String POSSIBLE = "POSSIBLE";
final static String YES = "YES";
final static String NO = "NO";
private void initIO(boolean isFileIO) {
if (System.getProperty("ONLINE_JUDGE") == null && isFileIO) {
try{
in = new MyScanner(new FileInputStream("input.txt"));
out = new MyPrintWriter(new FileOutputStream("output.txt"));
}
catch(FileNotFoundException e){
e.printStackTrace();
}
}
else{
in = new MyScanner(System.in);
out = new MyPrintWriter(new BufferedOutputStream(System.out));
}
}
public static void main(String[] args){
// Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
Round808A sol = new Round808A();
sol.run();
}
private void run() {
boolean isDebug = false;
boolean isFileIO = true;
boolean hasMultipleTests = true;
initIO(isFileIO);
int t = hasMultipleTests? in.nextInt() : 1;
for (int i = 1; i <= t; ++i) {
int n = in.nextInt();
int[] a = in.nextIntArray(n);
if(isDebug){
out.printf("Test %d\n", i);
}
boolean ans = solve(a);
out.printlnAns(ans);
if(isDebug)
out.flush();
}
in.close();
out.close();
}
private boolean solve(int[] a) {
int n = a.length;
a[1] %= a[0];
if(a[1] == 0)
a[1] = a[0];
else
return false;
for(int i=2; i<n; i++)
if(a[i] % a[1] != 0)
return false;
return true;
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
// 32768?
public MyScanner(InputStream is, int bufferSize) {
br = new BufferedReader(new InputStreamReader(is), bufferSize);
}
public MyScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
// br = new BufferedReader(new InputStreamReader(System.in));
// br = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
}
public void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[][] nextTreeEdges(int n, int offset){
int[][] e = new int[n-1][2];
for(int i=0; i<n-1; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[][] nextMatrix(int n, int m) {
return nextMatrix(n, m, 0);
}
int[][] nextMatrix(int n, int m, int offset) {
int[][] mat = new int[n][m];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[i][j] = nextInt()+offset;
}
}
return mat;
}
int[][] nextPairs(int n){
return nextPairs(n, 0);
}
int[][] nextPairs(int n, int offset) {
int[][] xy = new int[2][n];
for(int i=0; i<n; i++) {
xy[0][i] = nextInt() + offset;
xy[1][i] = nextInt() + offset;
}
return xy;
}
int[][] nextGraphEdges(){
return nextGraphEdges(0);
}
int[][] nextGraphEdges(int offset) {
int m = nextInt();
int[][] e = new int[m][2];
for(int i=0; i<m; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[] nextIntArray(int len) {
return nextIntArray(len, 0);
}
int[] nextIntArray(int len, int offset){
int[] a = new int[len];
for(int j=0; j<len; j++)
a[j] = nextInt()+offset;
return a;
}
long[] nextLongArray(int len) {
return nextLongArray(len, 0);
}
long[] nextLongArray(int len, int offset){
long[] a = new long[len];
for(int j=0; j<len; j++)
a[j] = nextLong()+offset;
return a;
}
}
public static class MyPrintWriter extends PrintWriter{
public MyPrintWriter(OutputStream os) {
super(os);
}
public void printlnAns(long ans) {
println(ans);
}
public void printlnAns(int ans) {
println(ans);
}
public void printlnAns(boolean ans) {
if(ans)
println(YES);
else
println(NO);
}
public void printAns(long[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(long[] arr){
printAns(arr);
println();
}
public void printAns(int[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(int[] arr){
printAns(arr);
println();
}
public <T> void printAns(ArrayList<T> arr){
if(arr != null && arr.size() > 0){
print(arr.get(0));
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i));
}
}
}
public <T> void printlnAns(ArrayList<T> arr){
printAns(arr);
println();
}
public void printAns(int[] arr, int add){
if(arr != null && arr.length > 0){
print(arr[0]+add);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]+add);
}
}
}
public void printlnAns(int[] arr, int add){
printAns(arr, add);
println();
}
public void printAns(ArrayList<Integer> arr, int add) {
if(arr != null && arr.size() > 0){
print(arr.get(0)+add);
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i)+add);
}
}
}
public void printlnAns(ArrayList<Integer> arr, int add){
printAns(arr, add);
println();
}
public void printlnAnsSplit(long[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public void printlnAnsSplit(int[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public <T> void printlnAnsSplit(ArrayList<T> arr, int split){
if(arr != null && !arr.isEmpty()){
for(int i=0; i<arr.size(); i+=split){
print(arr.get(i));
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr.get(j));
}
println();
}
}
}
}
static private void permutateAndSort(long[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
long temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private void permutateAndSort(int[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
int temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private int[][] constructChildren(int n, int[] parent, int parentRoot){
int[][] childrens = new int[n][];
int[] numChildren = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
numChildren[parent[i]]++;
}
for(int i=0; i<n; i++) {
childrens[i] = new int[numChildren[i]];
}
int[] idx = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
childrens[parent[i]][idx[parent[i]]++] = i;
}
return childrens;
}
static private int[][][] constructDirectedNeighborhood(int n, int[][] e){
int[] inDegree = new int[n];
int[] outDegree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outDegree[u]++;
inDegree[v]++;
}
int[][] inNeighbors = new int[n][];
int[][] outNeighbors = new int[n][];
for(int i=0; i<n; i++) {
inNeighbors[i] = new int[inDegree[i]];
outNeighbors[i] = new int[outDegree[i]];
}
int[] inIdx = new int[n];
int[] outIdx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outNeighbors[u][outIdx[u]++] = v;
inNeighbors[v][inIdx[v]++] = u;
}
return new int[][][] {inNeighbors, outNeighbors};
}
static private int[][] constructNeighborhood(int n, int[][] e) {
int[] degree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
degree[u]++;
degree[v]++;
}
int[][] neighbors = new int[n][];
for(int i=0; i<n; i++)
neighbors[i] = new int[degree[i]];
int[] idx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
neighbors[u][idx[u]++] = v;
neighbors[v][idx[v]++] = u;
}
return neighbors;
}
static private void drawGraph(int[][] e) {
makeDotUndirected(e);
try {
final Process process = new ProcessBuilder("dot", "-Tpng", "graph.dot")
.redirectOutput(new File("graph.png"))
.start();
} catch (IOException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
}
static private void makeDotUndirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict graph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "--" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
static private void makeDotDirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict digraph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "->" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 171de43479ffb3dffd10a6ed36375e53 | train_109.jsonl | 1657982100 | You are given an array $$$a$$$ consisting of $$$n$$$ positive integers.You are allowed to perform this operation any number of times (possibly, zero): choose an index $$$i$$$ ($$$2 \le i \le n$$$), and change $$$a_i$$$ to $$$a_i - a_{i-1}$$$. Is it possible to make $$$a_i=0$$$ for all $$$2\le i\le n$$$? | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
ADifferenceOperations solver = new ADifferenceOperations();
solver.solve(1, in, out);
out.close();
}
static class ADifferenceOperations {
public void solve(int testNumber, InputReader in, OutputWriter out) {
var tc = in.nextInt();
for (int i = 0; i < tc; i++) {
solution(i, in, out);
}
}
void solution(int testNumber, InputReader in, OutputWriter out) {
int n = in.nextInt();
int[] a = in.nextIntArray(n);
for (int i = 1; i < n; i++) {
if (a[i] % a[0] != 0) {
out.println("NO");
return;
}
}
out.println("YES");
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
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 println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
}
| Java | ["4\n\n2\n\n5 10\n\n3\n\n1 2 3\n\n4\n\n1 1 1 1\n\n9\n\n9 9 8 2 4 4 3 5 3"] | 1 second | ["YES\nYES\nYES\nNO"] | NoteIn the first test case, the initial array is $$$[5,10]$$$. You can perform $$$2$$$ operations to reach the goal: Choose $$$i=2$$$, and the array becomes $$$[5,5]$$$. Choose $$$i=2$$$, and the array becomes $$$[5,0]$$$. In the second test case, the initial array is $$$[1,2,3]$$$. You can perform $$$4$$$ operations to reach the goal: Choose $$$i=3$$$, and the array becomes $$$[1,2,1]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,1,1]$$$. Choose $$$i=3$$$, and the array becomes $$$[1,1,0]$$$. Choose $$$i=2$$$, and the array becomes $$$[1,0,0]$$$. In the third test case, you can choose indices in the order $$$4$$$, $$$3$$$, $$$2$$$. | Java 17 | standard input | [
"greedy",
"math"
] | 1c597da89880e87ffe791dd6b9fb2ac7 | The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of the test cases follows. The first line contains one integer $$$n$$$ ($$$2 \le n \le 100$$$) — the length of array $$$a$$$. The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le 10^9$$$). | 800 | For each test case, print "YES" (without quotes), if it is possible to change $$$a_i$$$ to $$$0$$$ for all $$$2 \le i \le n$$$, and "NO" (without quotes) otherwise. You can print letters in any case (upper or lower). | standard output | |
PASSED | 016b5c82e69138e8c23b8ed95ed1ed7e | train_109.jsonl | 1657982100 | Kawashiro Nitori is a girl who loves competitive programming. One day she found a rooted tree consisting of $$$n$$$ vertices. The root is vertex $$$1$$$. As an advanced problem setter, she quickly thought of a problem.Kawashiro Nitori has a vertex set $$$U=\{1,2,\ldots,n\}$$$. She's going to play a game with the tree and the set. In each operation, she will choose a vertex set $$$T$$$, where $$$T$$$ is a partial virtual tree of $$$U$$$, and change $$$U$$$ into $$$T$$$.A vertex set $$$S_1$$$ is a partial virtual tree of a vertex set $$$S_2$$$, if $$$S_1$$$ is a subset of $$$S_2$$$, $$$S_1 \neq S_2$$$, and for all pairs of vertices $$$i$$$ and $$$j$$$ in $$$S_1$$$, $$$\operatorname{LCA}(i,j)$$$ is in $$$S_1$$$, where $$$\operatorname{LCA}(x,y)$$$ denotes the lowest common ancestor of vertices $$$x$$$ and $$$y$$$ on the tree. Note that a vertex set can have many different partial virtual trees.Kawashiro Nitori wants to know for each possible $$$k$$$, if she performs the operation exactly $$$k$$$ times, in how many ways she can make $$$U=\{1\}$$$ in the end? Two ways are considered different if there exists an integer $$$z$$$ ($$$1 \le z \le k$$$) such that after $$$z$$$ operations the sets $$$U$$$ are different.Since the answer could be very large, you need to find it modulo $$$p$$$. It's guaranteed that $$$p$$$ is a prime number. | 256 megabytes | //package com.example.practice.codeforces.sc3000.prac;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Solution1 {
final static int M = 2000;
final static long[][] comb = new long[M+1][M+1];
static {
comb[0][0] = comb[1][0] = comb[1][1] = 1;
}
public static void main (String [] args) throws IOException {
// Use BufferedReader rather than RandomAccessFile; it's much faster
final BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
// input file name goes above
StringTokenizer st = new StringTokenizer(input.readLine());
final int n = Integer.parseInt(st.nextToken()), P = Integer.parseInt(st.nextToken());
final int[][] es = readArray2DInt(n-1, n-1, input);
long[] res = calc(n, P, es);
printArray(res, out);
out.close(); // close the output file
}
private static long[] calc(final int n, final int P, final int[][] es) {
ArrayList<Integer>[] als = new ArrayList[n+1];
for (int i=1;i<=n;++i){
als[i] = new ArrayList<>();
}
for (int[] e : es){
als[e[0]].add(e[1]);
als[e[1]].add(e[0]);
}
for (int i=2;i<=n;++i){
comb[i][0] = comb[i][i] = 1;
for (int j=1;j<i;++j){
comb[i][j] = (comb[i-1][j-1] + comb[i-1][j]) % P;
}
}
final Tree root = build(als, 1, -1, n-1);
todo(root, P);
long[] res = new long[n-1];
res[0] = 1;
for (int i=1;i<res.length;++i){
long t = 1;
for (int j=0;j<root.cd.size();++j){
t = (t * root.cd.get(j).dp2[i]) % P;
}
for (int j=1;j<=i;++j){
t -= (res[i-j] * comb[i+1][j]) % P;
if (t<0)t+=P;
}
res[i] = t;
}
return res;
}
private static void todo(Tree rt, final int P){
for (Tree tr : rt.cd){
todo(tr, P);
}
rt.dp2[0] = 1;
if (rt.cd.size()==0){
for (int i=1;i<rt.dp2.length;++i){
rt.dp2[i] = rt.dp2[i-1] + 1;
}
return;
}
if (rt.cd.size()==1){
Tree tr = rt.cd.get(0);
for (int i=1;i<rt.dp2.length;++i){
long t = tr.dp2[i] - tr.dp2[i-1];
if (t<0)t+=P;
rt.dp2[i] = (t * i) % P;
rt.dp2[i] += tr.dp2[i];
rt.dp2[i] = rt.dp2[i] % P;
rt.dp2[i] = (rt.dp2[i] + rt.dp2[i-1]) % P;
}
return;
}
long[][] rd1 = new long[rt.cd.size()-1][rt.dp2.length], rd2 = new long[rt.cd.size()][rt.dp2.length];
for (int i=0;i<rd2[0].length;++i){
long pre = 1;
for (int j=0;j<rd1.length;++j){
pre = (pre * rt.cd.get(j).dp2[i]) % P;
rd1[j][i] = pre;
}
pre = 1;
for (int j=rt.cd.size()-1; j>=0 ;--j){
pre = (pre * rt.cd.get(j).dp2[i]) % P;
rd2[j][i] = pre;
}
}
long[] dp3 = new long[rt.cd.size()];
Arrays.fill(dp3, 1L);
for (int i=1;i<rt.dp2.length;++i){
rt.dp2[i] = rd2[0][i];
for (int j=0;j<rt.cd.size();++j){
long t = rt.cd.get(j).dp2[i] - rt.cd.get(j).dp2[i-1];
if (t < 0)t+=P;
rt.dp2[i] += (dp3[j] * t) % P;
rt.dp2[i] = rt.dp2[i] % P;
long l = j>0 ? rd1[j-1][i] : 1;
long r = j+1<rt.cd.size() ? rd2[j+1][i] : 1;
dp3[j] += (l * r) % P;
dp3[j] = dp3[j] % P;
}
rt.dp2[i] = (rt.dp2[i] + rt.dp2[i-1]) % P;
}
}
private static Tree build(ArrayList<Integer>[] als, int idx, int pre, final int n){
Tree res = new Tree(n);
for (int a : als[idx]){
if (a != pre){
Tree tr = build(als, a, idx, n);
res.cd.add(tr);
}
}
return res;
}
private static void printArray(long[] ns, final PrintWriter out){
for (int i=0;i<ns.length;++i){
out.print(ns[i]);
if (i+1<ns.length)out.print(" ");
else out.println();
}
}
private static void printArrayInt(int[] ns, final PrintWriter out){
for (int i=0;i<ns.length;++i){
out.print(ns[i]);
if (i+1<ns.length)out.print(" ");
else out.println();
}
}
private static void printArrayVertical(long[] ns, final PrintWriter out){
for (long a : ns){
out.println(a);
}
}
private static void printArrayVerticalInt(int[] ns, final PrintWriter out){
for (int a : ns){
out.println(a);
}
}
private static void printArray2D(long[][] ns, final int len, final PrintWriter out){
int cnt = 0;
for (long[] kk : ns){
cnt++;
if (cnt > len)break;
for (int i=0;i<kk.length;++i){
out.print(kk[i]);
if (i+1<kk.length)out.print(" ");
else out.println();
}
}
}
private static void printArray2DInt(int[][] ns, final int len, final PrintWriter out){
int cnt = 0;
for (int[] kk : ns){
cnt++;
if (cnt > len)break;
for (int i=0;i<kk.length;++i){
out.print(kk[i]);
if (i+1<kk.length)out.print(" ");
else out.println();
}
}
}
private static long[] readArray(final int n, final BufferedReader input) throws IOException{
long[] ns = new long[n];
StringTokenizer st = new StringTokenizer(input.readLine());
for (int i=0;i<n;++i){
ns[i] = Long.parseLong(st.nextToken());
}
return ns;
}
private static int[] readArrayInt(final int n, final BufferedReader input) throws IOException{
int[] ns = new int[n];
StringTokenizer st = new StringTokenizer(input.readLine());
for (int i=0;i<n;++i){
ns[i] = Integer.parseInt(st.nextToken());
}
return ns;
}
private static long[] readArrayVertical(final int n, final BufferedReader input) throws IOException{
long[] ns = new long[n];
for (int i=0;i<n;++i){
ns[i] = Long.parseLong(input.readLine());
}
return ns;
}
private static int[] readArrayVerticalInt(final int n, final BufferedReader input) throws IOException{
int[] ns = new int[n];
for (int i=0;i<n;++i){
ns[i] = Integer.parseInt(input.readLine());
}
return ns;
}
private static long[][] readArray2D(final int n, final int len, final BufferedReader input) throws IOException{
long[][] ns = new long[len][];
for (int i=0;i<n;++i){
StringTokenizer st = new StringTokenizer(input.readLine());
ArrayList<Long> al = new ArrayList<>();
while (st.hasMoreTokens()){
al.add(Long.parseLong(st.nextToken()));
}
long[] kk = new long[al.size()];
for (int j=0;j<kk.length;++j){
kk[j] = al.get(j);
}
ns[i] = kk;
}
return ns;
}
private static int[][] readArray2DInt(final int n, final int len, final BufferedReader input) throws IOException{
int[][] ns = new int[len][];
for (int i=0;i<n;++i){
StringTokenizer st = new StringTokenizer(input.readLine());
ArrayList<Integer> al = new ArrayList<>();
while (st.hasMoreTokens()){
al.add(Integer.parseInt(st.nextToken()));
}
int[] kk = new int[al.size()];
for (int j=0;j<kk.length;++j){
kk[j] = al.get(j);
}
ns[i] = kk;
}
return ns;
}
static class Tree{
ArrayList<Tree> cd;
long[] dp2;
public Tree(int n){
cd = new ArrayList<>();
dp2 = new long[n];
}
}
} | Java | ["4 998244353\n1 2\n2 3\n1 4", "7 100000007\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7", "8 1000000007\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8"] | 2 seconds | ["1 6 6", "1 47 340 854 880 320", "1 126 1806 8400 16800 15120 5040"] | NoteIn the first test case, when $$$k=1$$$, the only possible way is: $$$\{1,2,3,4\} \to \{1\}$$$. When $$$k=2$$$, there are $$$6$$$ possible ways: $$$\{1,2,3,4\} \to \{1,2\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,4\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3,4\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,4\} \to \{1\}$$$. When $$$k=3$$$, there are $$$6$$$ possible ways: $$$\{1,2,3,4\} \to \{1,2,3\} \to \{1,2\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,3\} \to \{1,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,4\} \to \{1,2\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,4\} \to \{1,4\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3,4\} \to \{1,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3,4\} \to \{1,4\} \to \{1\}$$$. | Java 11 | standard input | [
"combinatorics",
"dp",
"math",
"trees"
] | 27e8cf87034934d067b0a8c4b9703271 | The first line contains two integers $$$n$$$ and $$$p$$$ ($$$2 \le n \le 2000$$$, $$$10^8 + 7 \le p \le 10^9+9$$$). It's guaranteed that $$$p$$$ is a prime number. Each of the next $$$n-1$$$ lines contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$), representing an edge between $$$u_i$$$ and $$$v_i$$$. It is guaranteed that the given edges form a tree. | 3,000 | The only line contains $$$n-1$$$ integers — the answer modulo $$$p$$$ for $$$k=1,2,\ldots,n-1$$$. | standard output | |
PASSED | 544dd7b8d4412788c488dd0599d9b802 | train_109.jsonl | 1657982100 | Kawashiro Nitori is a girl who loves competitive programming. One day she found a rooted tree consisting of $$$n$$$ vertices. The root is vertex $$$1$$$. As an advanced problem setter, she quickly thought of a problem.Kawashiro Nitori has a vertex set $$$U=\{1,2,\ldots,n\}$$$. She's going to play a game with the tree and the set. In each operation, she will choose a vertex set $$$T$$$, where $$$T$$$ is a partial virtual tree of $$$U$$$, and change $$$U$$$ into $$$T$$$.A vertex set $$$S_1$$$ is a partial virtual tree of a vertex set $$$S_2$$$, if $$$S_1$$$ is a subset of $$$S_2$$$, $$$S_1 \neq S_2$$$, and for all pairs of vertices $$$i$$$ and $$$j$$$ in $$$S_1$$$, $$$\operatorname{LCA}(i,j)$$$ is in $$$S_1$$$, where $$$\operatorname{LCA}(x,y)$$$ denotes the lowest common ancestor of vertices $$$x$$$ and $$$y$$$ on the tree. Note that a vertex set can have many different partial virtual trees.Kawashiro Nitori wants to know for each possible $$$k$$$, if she performs the operation exactly $$$k$$$ times, in how many ways she can make $$$U=\{1\}$$$ in the end? Two ways are considered different if there exists an integer $$$z$$$ ($$$1 \le z \le k$$$) such that after $$$z$$$ operations the sets $$$U$$$ are different.Since the answer could be very large, you need to find it modulo $$$p$$$. It's guaranteed that $$$p$$$ is a prime number. | 256 megabytes | //package com.example.practice.codeforces.sc3000.prac;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Solution1 {
final static int M = 2000;
final static long[][] comb = new long[M+1][M+1];
static {
comb[0][0] = comb[1][0] = comb[1][1] = 1;
}
public static void main (String [] args) throws IOException {
// Use BufferedReader rather than RandomAccessFile; it's much faster
final BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
// input file name goes above
StringTokenizer st = new StringTokenizer(input.readLine());
final int n = Integer.parseInt(st.nextToken()), P = Integer.parseInt(st.nextToken());
final int[][] es = readArray2DInt(n-1, n-1, input);
final long[] res = calc(n, P, es);
printArray(res, out);
out.close(); // close the output file
}
private static long[] calc(final int n, final int P, final int[][] es) {
ArrayList<Integer>[] als = new ArrayList[n+1];
for (int i=1;i<=n;++i){
als[i] = new ArrayList<>();
}
for (int[] e : es){
als[e[0]].add(e[1]);
als[e[1]].add(e[0]);
}
for (int i=2;i<=n;++i){
comb[i][0] = comb[i][i] = 1;
for (int j=1;j<i;++j){
comb[i][j] = (comb[i-1][j-1] + comb[i-1][j]) % P;
}
}
final Tree root = build(als, 1, -1, n-1);
todo(root, P);
long[] res = new long[n-1];
res[0] = 1;
for (int i=1;i<res.length;++i){
long t = 1;
for (int j=0;j<root.cd.size();++j){
t = (t * root.cd.get(j).dp2[i]) % P;
}
for (int j=1;j<=i;++j){
t -= (res[i-j] * comb[i+1][j]) % P;
if (t<0)t+=P;
}
res[i] = t;
}
return res;
}
private static void todo(Tree rt, final int P){
for (Tree tr : rt.cd){
todo(tr, P);
}
rt.dp2[0] = 1;
if (rt.cd.size()==0){
for (int i=1;i<rt.dp2.length;++i){
rt.dp2[i] = rt.dp2[i-1] + 1;
}
return;
}
if (rt.cd.size()==1){
Tree tr = rt.cd.get(0);
for (int i=1;i<rt.dp2.length;++i){
long t = tr.dp2[i] - tr.dp2[i-1];
if (t<0)t+=P;
rt.dp2[i] = (t * i) % P;
rt.dp2[i] += tr.dp2[i];
rt.dp2[i] = rt.dp2[i] % P;
rt.dp2[i] = (rt.dp2[i] + rt.dp2[i-1]) % P;
}
return;
}
long[][] rd1 = new long[rt.cd.size()-1][rt.dp2.length], rd2 = new long[rt.cd.size()][rt.dp2.length];
for (int i=0;i<rd2[0].length;++i){
long pre = 1;
for (int j=0;j<rd1.length;++j){
pre = (pre * rt.cd.get(j).dp2[i]) % P;
rd1[j][i] = pre;
}
pre = 1;
for (int j=rt.cd.size()-1; j>=0 ;--j){
pre = (pre * rt.cd.get(j).dp2[i]) % P;
rd2[j][i] = pre;
}
}
long[] dp3 = new long[rt.cd.size()];
Arrays.fill(dp3, 1L);
for (int i=1;i<rt.dp2.length;++i){
rt.dp2[i] = rd2[0][i];
for (int j=0;j<rt.cd.size();++j){
long t = rt.cd.get(j).dp2[i] - rt.cd.get(j).dp2[i-1];
if (t < 0)t+=P;
rt.dp2[i] += (dp3[j] * t) % P;
rt.dp2[i] = rt.dp2[i] % P;
long l = j>0 ? rd1[j-1][i] : 1;
long r = j+1<rt.cd.size() ? rd2[j+1][i] : 1;
dp3[j] += (l * r) % P;
dp3[j] = dp3[j] % P;
}
rt.dp2[i] = (rt.dp2[i] + rt.dp2[i-1]) % P;
}
}
private static Tree build(ArrayList<Integer>[] als, int idx, int pre, final int n){
Tree res = new Tree(n);
for (int a : als[idx]){
if (a != pre){
Tree tr = build(als, a, idx, n);
res.cd.add(tr);
}
}
return res;
}
private static void printArray(long[] ns, final PrintWriter out){
for (int i=0;i<ns.length;++i){
out.print(ns[i]);
if (i+1<ns.length)out.print(" ");
else out.println();
}
}
private static void printArrayInt(int[] ns, final PrintWriter out){
for (int i=0;i<ns.length;++i){
out.print(ns[i]);
if (i+1<ns.length)out.print(" ");
else out.println();
}
}
private static void printArrayVertical(long[] ns, final PrintWriter out){
for (long a : ns){
out.println(a);
}
}
private static void printArrayVerticalInt(int[] ns, final PrintWriter out){
for (int a : ns){
out.println(a);
}
}
private static void printArray2D(long[][] ns, final int len, final PrintWriter out){
int cnt = 0;
for (long[] kk : ns){
cnt++;
if (cnt > len)break;
for (int i=0;i<kk.length;++i){
out.print(kk[i]);
if (i+1<kk.length)out.print(" ");
else out.println();
}
}
}
private static void printArray2DInt(int[][] ns, final int len, final PrintWriter out){
int cnt = 0;
for (int[] kk : ns){
cnt++;
if (cnt > len)break;
for (int i=0;i<kk.length;++i){
out.print(kk[i]);
if (i+1<kk.length)out.print(" ");
else out.println();
}
}
}
private static long[] readArray(final int n, final BufferedReader input) throws IOException{
long[] ns = new long[n];
StringTokenizer st = new StringTokenizer(input.readLine());
for (int i=0;i<n;++i){
ns[i] = Long.parseLong(st.nextToken());
}
return ns;
}
private static int[] readArrayInt(final int n, final BufferedReader input) throws IOException{
int[] ns = new int[n];
StringTokenizer st = new StringTokenizer(input.readLine());
for (int i=0;i<n;++i){
ns[i] = Integer.parseInt(st.nextToken());
}
return ns;
}
private static long[] readArrayVertical(final int n, final BufferedReader input) throws IOException{
long[] ns = new long[n];
for (int i=0;i<n;++i){
ns[i] = Long.parseLong(input.readLine());
}
return ns;
}
private static int[] readArrayVerticalInt(final int n, final BufferedReader input) throws IOException{
int[] ns = new int[n];
for (int i=0;i<n;++i){
ns[i] = Integer.parseInt(input.readLine());
}
return ns;
}
private static long[][] readArray2D(final int n, final int len, final BufferedReader input) throws IOException{
long[][] ns = new long[len][];
for (int i=0;i<n;++i){
StringTokenizer st = new StringTokenizer(input.readLine());
ArrayList<Long> al = new ArrayList<>();
while (st.hasMoreTokens()){
al.add(Long.parseLong(st.nextToken()));
}
long[] kk = new long[al.size()];
for (int j=0;j<kk.length;++j){
kk[j] = al.get(j);
}
ns[i] = kk;
}
return ns;
}
private static int[][] readArray2DInt(final int n, final int len, final BufferedReader input) throws IOException{
int[][] ns = new int[len][];
for (int i=0;i<n;++i){
StringTokenizer st = new StringTokenizer(input.readLine());
ArrayList<Integer> al = new ArrayList<>();
while (st.hasMoreTokens()){
al.add(Integer.parseInt(st.nextToken()));
}
int[] kk = new int[al.size()];
for (int j=0;j<kk.length;++j){
kk[j] = al.get(j);
}
ns[i] = kk;
}
return ns;
}
static class Tree{
ArrayList<Tree> cd;
long[] dp2;
public Tree(int n){
cd = new ArrayList<>();
dp2 = new long[n];
}
}
} | Java | ["4 998244353\n1 2\n2 3\n1 4", "7 100000007\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7", "8 1000000007\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8"] | 2 seconds | ["1 6 6", "1 47 340 854 880 320", "1 126 1806 8400 16800 15120 5040"] | NoteIn the first test case, when $$$k=1$$$, the only possible way is: $$$\{1,2,3,4\} \to \{1\}$$$. When $$$k=2$$$, there are $$$6$$$ possible ways: $$$\{1,2,3,4\} \to \{1,2\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,4\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3,4\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,4\} \to \{1\}$$$. When $$$k=3$$$, there are $$$6$$$ possible ways: $$$\{1,2,3,4\} \to \{1,2,3\} \to \{1,2\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,3\} \to \{1,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,4\} \to \{1,2\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,2,4\} \to \{1,4\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3,4\} \to \{1,3\} \to \{1\}$$$; $$$\{1,2,3,4\} \to \{1,3,4\} \to \{1,4\} \to \{1\}$$$. | Java 11 | standard input | [
"combinatorics",
"dp",
"math",
"trees"
] | 27e8cf87034934d067b0a8c4b9703271 | The first line contains two integers $$$n$$$ and $$$p$$$ ($$$2 \le n \le 2000$$$, $$$10^8 + 7 \le p \le 10^9+9$$$). It's guaranteed that $$$p$$$ is a prime number. Each of the next $$$n-1$$$ lines contains two integers $$$u_i$$$, $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$), representing an edge between $$$u_i$$$ and $$$v_i$$$. It is guaranteed that the given edges form a tree. | 3,000 | The only line contains $$$n-1$$$ integers — the answer modulo $$$p$$$ for $$$k=1,2,\ldots,n-1$$$. | standard output | |
PASSED | 072f81c19170c9b2931853f98cd451b5 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Solution {
private static final FastScanner fs = new FastScanner();
public static void main(String[] args) {
int testCase = fs.nextInt();
for (int i = 1; i <= testCase; i++) {
solve();
}
}
public static void solve() {
int n = fs.nextInt();
int[] arr1 = new int[n];
int[] arr2 = new int[n];
for (int i = 0; i < n; i++) {
arr1[i] = fs.nextInt();
}
for (int i = 0; i < n; i++) {
arr2[i] = fs.nextInt();
}
sortInts(arr1);
sortInts(arr2);
for (int i = 0; i < arr1.length; i++) {
if (arr1[i] > arr2[i] || Math.abs(arr1[i] - arr2[i]) > 1) {
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
public static void sortInts(int[] arr) {
List<Integer> temp = new ArrayList<>();
for (int i : arr) temp.add(i);
Collections.sort(temp);
int start = 0;
for (int i : temp) arr[start++] = i;
}
}
class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | ab09e9e81ce015527862bbca55819064 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.Scanner;
import static java.util.Arrays.sort;
public class BE {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int t= in.nextInt();
while (t!=0){
t--;
final int n=in.nextInt();
int a[]=new int[n];
int b[]=new int[n];
for(int i=0;i<n;i++){
int x=in.nextInt();
a[i]=x;
}
for(int i=0;i<n;i++){
int x=in.nextInt();
b[i]=x;
}
sort(a);
sort(b);
boolean f=true;
for(int i=0;i<n;i++){
if(b[i]-a[i]==1||b[i]-a[i]==0)continue;
else {f=false;break;}
}
if(f) System.out.println("YES");
else System.out.println("NO");
}
}
private static int abs(int a,int b) {
if(a-b<0)return (a-b)*-1;
else return a-b;
}
private static int gcd(int f1,int f2) {
return f2==0?f1:gcd(f2,f1%f2);
}
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 16027d993b673b25ab46c81e584fe3dc | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class Q3 {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t-- > 0){
int n = scanner.nextInt();
int [] a =new int[n];
int [] b= new int[n];
for (int i =0 ;i<n;i++){
a[i] = scanner.nextInt();
}
for (int i =0 ;i<n;i++){
b[i] = scanner.nextInt();
}
Arrays.sort(a);
Arrays.sort(b);
if(Arrays.equals(a,b)){
System.out.println("YES");
continue;
}
boolean outerFlag = true;
outer:for (int i =0 ;i<a.length;i++){
boolean flag = false;
for (int j =0;j<b.length;j++){
if (a[i] ==b[j] || a[i] == b[j]-1){
flag = true;
b[j] = -9999;
continue outer;
}
}
if (!flag){
outerFlag = false;
break ;
}
}
if (outerFlag){
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 55343dc0a9da7cc8cd1699565941b423 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | // ceil using integer division: ceil(x/y) = (x+y-1)/y
import java.lang.reflect.Array;
import java.util.*;
import java.lang.*;
import java.io.*;
public class practice {
public static void main(String[] args) throws IOException {
Reader.init(System.in);
int t = Reader.nextInt();
while(t-->0){
int n = Reader.nextInt();
Integer[] arr1 = new Integer[n];
Integer[] arr2 = new Integer[n];
for(int i = 0;i<n;i++){
arr1[i] = Reader.nextInt();
}
for(int i = 0;i<n;i++){
arr2[i] = Reader.nextInt();
}
Arrays.parallelSort(arr1);
Arrays.parallelSort(arr2);
boolean flag = true;
for(int i = 0;i<n;i++){
if(arr2[i]-arr1[i]>1 || arr2[i]<arr1[i]){
flag = false;
break;
}
}
if(flag){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
}
public static void reverseArray(int[] arr,int start, int end) {
int temp;
while (start < end) {
temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
public static long gcd(long a, long b){
if(a==0){
return b;
}
return gcd(b%a,a);
}
public static long lcm(long a, long b){
if(a>b) return a/gcd(b,a) * b;
return b/gcd(a,b) * a;
}
public static long largeExponentMod(long x,long y,long mod){
// computing (x^y) % mod
x%=mod;
long ans = 1;
while(y>0){
if((y&1)==1){
ans = (ans*x)%mod;
}
x = (x*x)%mod;
y = y >> 1;
}
return ans;
}
public static boolean[] numOfPrimesInRange(long L, long R){
boolean[] isPrime = new boolean[(int) (R-L+1)];
Arrays.fill(isPrime,true);
long lim = (long) Math.sqrt(R);
for (long i = 2; i <= lim; ++i){
for (long j = Math.max(i * i, (L + i - 1) / i * i); j <= R; j += i){
isPrime[(int) (j - L)] = false;
}
}
if (L == 1) isPrime[0] = false;
return isPrime;
}
public static ArrayList<Long> primeFactors(long n){
ArrayList<Long> factorization = new ArrayList<>();
while(n%2==0){
factorization.add(2L);
n/=2;
}
while(n%3==0){
factorization.add(3L);
n/=3;
}
while(n%5==0){
factorization.add(5L);
n/=5;
}
int[] increments = {4, 2, 4, 2, 4, 6, 2, 6};
int i = 0;
for (long d = 7; d * d <= n; d += increments[i++]) {
while (n % d == 0) {
factorization.add(d);
n /= d;
}
if (i == 8)
i = 0;
}
if (n > 1)
factorization.add(n);
return factorization;
}
}
class Reader {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(
new InputStreamReader(input) );
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() throws IOException {
while ( ! tokenizer.hasMoreTokens() ) {
//TODO add check for eof if necessary
tokenizer = new StringTokenizer(
reader.readLine() );
}
return tokenizer.nextToken();
}
static String nextLine() throws IOException {
return reader.readLine();
}
static int nextInt() throws IOException {
return Integer.parseInt( next() );
}
static long nextLong() throws IOException{
return Long.parseLong(next() );
}
static double nextDouble() throws IOException {
return Double.parseDouble( next() );
}
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | ab7c35d3cd9243825aee903e2d815dbd | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | //package com.shroom;
import java.util.*;
public class dec {
private static final Scanner in = new Scanner(System.in);
static void re(char []s,int start,int end) {
char temp;
while(start<=end){
temp=s[start];
s[start] = s[end];
s[end] = temp;
start++;
end--;
}
}
static int knapsack(int []wt, int []val, int n,int cap) {
if(n==0 || cap==0){
return 0;
}
if(wt[n-1]>cap){
return knapsack(wt,val,n-1,cap);
}else{ //1.) including nth 2.) not including
return Math.max(val[n-1]+knapsack(wt, val, n-1, cap-wt[n-1]),knapsack(wt, val, n-1, cap));
}
}
public static void main(String[] args) throws Exception {
int t = in.nextInt();
in.nextLine();
while(t-->0){
int n = in.nextInt();
int []a = new int[n];
int []b = new int[n];
for (int i=0; i<n; i++) a[i] = in.nextInt();
for (int i=0; i<n; i++) b[i] = in.nextInt();
int ans = 0;
Arrays.sort(a);
Arrays.sort(b);
boolean check = true;
for(int i=0; i<n; i++){
ans = b[i]-a[i];
if(a[i]==b[i] || a[i]+1==b[i]) continue;
else check = false;
}
System.out.println(check?"YES":"NO");
}
}
static void printMatrix(int [][]arr,int n,int k) {
for(int i=0; i<n; i++){
for(int j=0; j<k; j++){
System.out.print(arr[i][j]+" ");
}
System.out.println();
}
}
// 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);
// }
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 76f092ecaff08829e63a7b4628b3c25a | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution
{
static PrintWriter out = new PrintWriter((System.out));
static Kioken sc = new Kioken();
public static void main(String[] args)
{
int t = 1;
t = sc.nextInt();
while (t-- > 0)
{
solve();
}
out.close();
}
public static boolean check(int[] a, int[] b){
Arrays.sort(a);
Arrays.sort(b);
for(int i = 0; i < a.length; i++){
if(a[i] != b[i]){
return false;
}
}
return true;
}
public static void solve()
{
int n = sc.nextInt();
int[] a = new int[n];
int[] b = new int[n];
for(int i = 0; i < n; i++){
a[i] = sc.nextInt();
}
for(int i = 0; i < n; i++){
b[i] = sc.nextInt();
}
int[] aaa = a.clone();
int[] bbb = b.clone();
if(check(aaa, bbb)) {
out.println("YES");
return;
}
Arrays.sort(a);
Arrays.sort(b);
for(int i = 0; i < n; i++){
if((a[i] > b[i]) || b[i] - a[i] > 1){
out.println("NO");
return;
}
}
out.println("YES");
return;
}
public static long leftShift(long a){
return (long)Math.pow(2, a);
}
public static int lower_bound(ArrayList<Integer> ar, int k)
{
int s = 0, e = ar.size();
while (s != e)
{
int mid = s + e >> 1;
if (ar.get(mid) <= k)
{
s = mid + 1;
}
else
{
e = mid;
}
}
return Math.abs(s) - 1;
}
public static int upper_bound(ArrayList<Integer> ar, int k)
{
int s = 0;
int e = ar.size();
while (s != e)
{
int mid = s + e >> 1;
if (ar.get(mid) < k)
{
s = mid + 1;
}
else
{
e = mid;
}
}
if (s == ar.size())
{
return -1;
}
return s;
}
static class Kioken
{
// FileInputStream br = new FileInputStream("input.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
public String next()
{
while (!st.hasMoreTokens())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (Exception e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt()
{
return Integer.parseInt(next());
}
public long nextLong()
{
return Long.parseLong(next());
}
public double nextDouble()
{
return Double.parseDouble(next());
}
public String nextLine()
{
try
{
return br.readLine();
}
catch (Exception e)
{
e.printStackTrace();
}
return null;
}
public boolean hasNext()
{
String next = null;
try
{
next = br.readLine();
}
catch (Exception e)
{
}
if (next == null || next.length() == 0)
{
return false;
}
st = new StringTokenizer(next);
return true;
}
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 883cf62ce092e6535b39706ea813c778 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
int i,j,k,l;
for(l=0;l<t;l++){
int n=sc.nextInt();
int[] a=new int[n];
int[] b=new int[n];
int[] c=new int[n];
ArrayList<Integer> arrayb=new ArrayList<Integer>();
for(i=0;i<n;i++){
a[i]=sc.nextInt();
}
for(i=0;i<n;i++){
b[i]=sc.nextInt();
}
boolean flag=true;
Arrays.sort(a);
Arrays.sort(b);
for(i=0;i<n;i++){
if(!((a[i]+1==b[i])||(a[i]==b[i]))){
System.out.println("NO");
flag=false;
break;
}
}
if(flag==true){
System.out.println("YES");
}
}
}
public static int frequency(int[] arr,int element){
int count=-1;
for(int p:arr){
if(p==element){
count++;
}
}
return count;
}
public static boolean check(int[] arr1,int[] arr2){
for(int i=0;i<arr1.length;i++){
if(frequency(arr1,arr1[i])!=frequency(arr2,arr1[i])){
return false;
}
}
return true;
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | ad883a6b4cb09dcceefa66529f2cb601 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.*;
public class aaa {
public static void main(String[] args) {
var in = new Scanner(System.in);
int n = in.nextInt();
for(int i = 0; i < n; i++) {
int m = in.nextInt();
var a1 = new int[m];
var a2 = new int[m];
for(int j = 0; j < m ; j++)
a1[j] = in.nextInt();
for(int j = 0; j < m ; j++)
a2[j] = in.nextInt();
Arrays.sort(a1);
Arrays.sort(a2);
boolean flag = true;
for(int j = 0; j < m; j++)
if(a1[j] != a2[j])
{
a1[j]++;
if(a1[j] != a2[j]) {
flag = false;
break;
}
}
System.out.println( flag ? "YES" : "NO");
}
}
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 6197a6e4311352e697e01e8f07fab297 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | /*LoudSilence*/
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Solution {
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static FastScanner s = new FastScanner();
static FastWriter out = new FastWriter();
final static int mod = (int)1e9 + 7;
final static int INT_MAX = Integer.MAX_VALUE;
final static int INT_MIN = Integer.MIN_VALUE;
final static long LONG_MAX = Long.MAX_VALUE;
final static long LONG_MIN = Long.MIN_VALUE;
final static double DOUBLE_MAX = Double.MAX_VALUE;
final static double DOUBLE_MIN = Double.MIN_VALUE;
final static float FLOAT_MAX = Float.MAX_VALUE;
final static float FLOAT_MIN = Float.MIN_VALUE;
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static class FastScanner{BufferedReader br;StringTokenizer st;
public FastScanner() {if(System.getProperty("ONLINE_JUDGE") == null){try {br = new BufferedReader(new FileReader("E:\\Competitive Coding\\input.txt"));}
catch (FileNotFoundException e) {br = new BufferedReader(new InputStreamReader(System.in));}}else{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());}
List<Integer> readIntList(int n){List<Integer> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextInt()); return arr;}
List<Long> readLongList(int n){List<Long> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextLong()); return arr;}
int[] readIntArr(int n){int[] arr = new int[n]; for (int i = 0; i < n; i++) arr[i] = s.nextInt(); return arr;}
long[] readLongArr(int n){long[] arr = new long[n]; for (int i = 0; i < n; i++) arr[i] = s.nextLong(); return arr;}
String nextLine(){String str = "";try{str = br.readLine();}catch (IOException e){e.printStackTrace();}return str;}}
static class FastWriter{private BufferedWriter bw;public FastWriter(){if(System.getProperty("ONLINE_JUDGE") == null){try {this.bw = new BufferedWriter(new FileWriter("E:\\Competitive Coding\\output.txt"));}
catch (IOException e) {this.bw = new BufferedWriter(new OutputStreamWriter(System.out));}}else{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 debug(int object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void debug(long object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void close() throws IOException{bw.close();}}
public static void println(Object str) throws IOException{out.println(""+str);}
public static void println(Object str, int nextLine) throws IOException{out.print(""+str);}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static ArrayList<Integer> seive(int n){ArrayList<Integer> list = new ArrayList<>();int arr[] = new int[n+1];for(long i = 2; i <= n; i++) {if(arr[(int)i] == 1) {continue;}else {list.add((int)i);for(long j = i*i; j <= n; j = j + i) {arr[(int)j] = 1;}}}return list;}
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 void swap(int[] arr, int i, int j) {arr[i] = arr[i] ^ arr[j]; arr[j] = arr[j] ^ arr[i]; arr[i] = arr[i] ^ arr[j];}
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 mod_add(long a, long b){ return (a%mod + b%mod)%mod;}
public static long mod_sub(long a, long b){ return (a%mod - b%mod + mod)%mod;}
public static long mod_mul(long a, long b){ return (a%mod * b%mod)%mod;}
public static long modInv(long a, long b){ return expo(a, b-2)%b;}
public static long mod_div(long a, long b){return mod_mul(a, modInv(b, mod));}
public static long expo (long a, long n){if(n == 0){return 1;}long recAns = expo(mod_mul(a,a), n/2);if(n % 2 == 0){return recAns;}else{return mod_mul(a, recAns);}}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Pair class
public static class Pair<X extends Comparable<X>,Y extends Comparable<Y>> implements Comparable<Pair<X, Y>>{
X first;
Y second;
public Pair(X first, Y second){
this.first = first;
this.second = second;
}
public String toString(){
return "( " + first+" , "+second+" )";
}
@Override
public int compareTo(Pair<X, Y> o) {
int t = first.compareTo(o.first);
if(t == 0) return second.compareTo(o.second);
return t;
}
}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Code begins
public static void solve() throws IOException {
int n = s.nextInt();
List<Integer> a = s.readIntList(n);
List<Integer> b = s.readIntList(n);
Collections.sort(a);
Collections.sort(b);
for(int i = 0; i < n; i++){
int diff = b.get(i) - a.get(i);
if(diff > 1 || diff < 0){
println("NO");
return;
}
}
println("YES");
}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static void main(String[] args) throws IOException {
int test = s.nextInt();
for(int t = 1; t <= test; t++) {
solve();
}
out.close();
}
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 99fadc3198b64499f947077686359adf | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.*;
public class practice {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
StringBuilder sb = new StringBuilder();
int t = scan.nextInt();
while (t --> 0) {
int n = scan.nextInt();
int a[] = new int[n];
int b[] = new int[n];
boolean flag = false;
for (int i = 0; i < n; i++) a[i] = scan.nextInt();
for (int i = 0; i < n; i++) b[i] = scan.nextInt();
Arrays.sort(a);
Arrays.sort(b);
for (int i = 0; i < n; i++) {
if (a[i] > b[i] || b[i] - a[i] > 1) {
flag = true;
break;
}
}
sb.append((!flag ? "Yes" : "No") + "\n");
}
System.out.println(sb.toString().trim());
scan.close();
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 5dcd49f817667f0895820f404ed20ce5 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes |
import java.util.*;
import java.math.*;
//import java.io.*;
public class Experiment {
static Scanner in=new Scanner(System.in);
// static BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
public static void sort(int ar[],int l,int r) {
if(l<r) {
int m=l+(r-l)/2;
sort(ar,l,m);
sort(ar,m+1,r);
merge(ar,l,m,r);
}
}
public static void merge(int ar[],int l,int m,int r) {
int n1=m-l+1;int n2=r-m;
int L[]=new int[n1];
int R[]=new int[n2];
for(int i=0;i<n1;i++)
L[i]=ar[l+i];
for(int i=0;i<n2;i++)
R[i]=ar[m+i+1];
int i=0;int j=0;int k=l;
while(i<n1 && j<n2) {
if(L[i]<=R[j]) {
ar[k++]=L[i++];
}else {
ar[k++]=R[j++];
}
}
while(i<n1)
ar[k++]=L[i++];
while(j<n2)
ar[k++]=R[j++];
}
public static int[] sort(int ar[]) {
sort(ar,0,ar.length-1);
//Array.sort uses O(n^2) in its worst case ,so better use merge sort
return ar;
}
public static void func() {
}
public static void main(String[] args) {
int t=in.nextInt();
while(t!=0) {
int n=in.nextInt();
ArrayList<Integer> list1=new ArrayList<Integer>();
ArrayList<Integer> list2=new ArrayList<Integer>();
for(int i=0;i<n;i++) {
list1.add(in.nextInt());
}
for(int i=0;i<n;i++) {
list2.add(in.nextInt());
}
Collections.sort(list1); Collections.sort(list2);int k=0;
for(int i=0;i<n;i++) {
if((list2.get(i)-list1.get(i))>1 || list1.get(i)>list2.get(i)) {
k=1;break;}
}
if(k==1) {
System.out.println("NO");
}else {
System.out.println("YES");
}
// if(list1.get(n-1)<=list2.get(n-1))
// System.out.println("YES");
// else
// System.out.println("NO");
//
// func(list,n,one,zero);
t--;
}
}
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 9bd44c98b2da97a72a0c45809a5c9575 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.io.*;
import java.util.*;
/**
* Provide prove of correctness before implementation. Implementation can cost a lot of time.
* Anti test that prove that it's wrong.
* <p>
* Do not confuse i j k g indexes, upTo and length. Do extra methods!!! Write more informative names to simulation
* <p>
* Will program ever exceed limit?
* Try all approaches with prove of correctness if task is not obvious.
* If you are given formula/rule: Try to play with it.
* Analytic solution/Hardcoded solution/Constructive/Greedy/DP/Math/Brute force/Symmetric data
* Number theory
* Game theory (optimal play) that consider local and global strategy.
*/
public class CF1584C {
private void solveOne() {
int n = nextInt();
int[] a = radixSort2(nextIntArr(n));
int[] b = radixSort2(nextIntArr(n));
// Arrays.sort(a);
// Arrays.sort(b);
// Map<Integer, Integer> multiset = new HashMap<>();
// for (int i = 0; i < n; i++) {
// multiset.merge(a[i], 1, Integer::sum);
// }
// List<Integer> toMatchB = new ArrayList<>();
// for (int i = 0; i < n; i++) {
// if (multiset.containsKey(b[i])) {
// if (multiset.get(b[i]) == 1) {
// multiset.remove(b[i]);
// } else {
// multiset.put(b[i], multiset.get(b[i]) - 1);
// }
// } else {
// toMatchB.add(b[i]);
// }
// }
// List<Integer> toMatchA = new ArrayList<>();
// for (Integer integer : multiset.keySet()) {
// for (int i = 0; i < multiset.get(integer); i++) {
// toMatchA.add(integer);
// }
// }
// toMatchB.sort(Integer::compareTo);
// toMatchA.sort(Integer::compareTo);
//
// if (toMatchB.size() != toMatchA.size()) {
// throw new RuntimeException("!!!");
// }
boolean ans = true;
for (int i = 0; i < n; i++) {
ans &= a[i] + 1 == b[i] || a[i] == b[i];
}
System.out.println(ans ? "Yes" : "No");
}
public static int[] radixSort2(int[] a) {
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for (int v : a) {
c0[(v & 0xff) + 1]++;
c1[(v >>> 8 & 0xff) + 1]++;
c2[(v >>> 16 & 0xff) + 1]++;
c3[(v >>> 24 ^ 0x80) + 1]++;
}
for (int i = 0; i < 0xff; i++) {
c0[i + 1] += c0[i];
c1[i + 1] += c1[i];
c2[i + 1] += c2[i];
c3[i + 1] += c3[i];
}
int[] t = new int[n];
for (int v : a) t[c0[v & 0xff]++] = v;
for (int v : t) a[c1[v >>> 8 & 0xff]++] = v;
for (int v : a) t[c2[v >>> 16 & 0xff]++] = v;
for (int v : t) a[c3[v >>> 24 ^ 0x80]++] = v;
return a;
}
private void solve() {
int t = System.in.readInt();
for (int tt = 0; tt < t; tt++) {
solveOne();
}
}
private int nextInt() {
return System.in.readInt();
}
private long nextLong() {
return System.in.readLong();
}
private String nextString() {
return System.in.readString();
}
private int[] nextIntArr(int n) {
return System.in.readIntArray(n);
}
private long[] nextLongArr(int n) {
return System.in.readLongArray(n);
}
public static void main(String[] args) {
new CF1584C().run();
}
static class System {
private static FastInputStream in;
private static FastPrintStream out;
}
private void run() {
System.in = new FastInputStream(java.lang.System.in);
System.out = new FastPrintStream(java.lang.System.out);
solve();
System.out.flush();
}
private static class FastPrintStream {
private static final int BUF_SIZE = 8192;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastPrintStream() {
this(java.lang.System.out);
}
public FastPrintStream(OutputStream os) {
this.out = os;
}
public FastPrintStream(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastPrintStream print(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE) innerflush();
return this;
}
public FastPrintStream print(char c) {
return print((byte) c);
}
public FastPrintStream print(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
}
return this;
}
public FastPrintStream print(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
});
return this;
}
//can be optimized
public FastPrintStream print0(char[] s) {
if (ptr + s.length < BUF_SIZE) {
for (char c : s) {
buf[ptr++] = (byte) c;
}
} else {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
}
}
return this;
}
//can be optimized
public FastPrintStream print0(String s) {
if (ptr + s.length() < BUF_SIZE) {
for (int i = 0; i < s.length(); i++) {
buf[ptr++] = (byte) s.charAt(i);
}
} else {
for (int i = 0; i < s.length(); i++) {
buf[ptr++] = (byte) s.charAt(i);
if (ptr == BUF_SIZE) innerflush();
}
}
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000) return 10;
if (l >= 100000000) return 9;
if (l >= 10000000) return 8;
if (l >= 1000000) return 7;
if (l >= 100000) return 6;
if (l >= 10000) return 5;
if (l >= 1000) return 4;
if (l >= 100) return 3;
if (l >= 10) return 2;
return 1;
}
public FastPrintStream print(int x) {
if (x == Integer.MIN_VALUE) {
return print((long) x);
}
if (ptr + 12 >= BUF_SIZE) innerflush();
if (x < 0) {
print((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L) return 19;
if (l >= 100000000000000000L) return 18;
if (l >= 10000000000000000L) return 17;
if (l >= 1000000000000000L) return 16;
if (l >= 100000000000000L) return 15;
if (l >= 10000000000000L) return 14;
if (l >= 1000000000000L) return 13;
if (l >= 100000000000L) return 12;
if (l >= 10000000000L) return 11;
if (l >= 1000000000L) return 10;
if (l >= 100000000L) return 9;
if (l >= 10000000L) return 8;
if (l >= 1000000L) return 7;
if (l >= 100000L) return 6;
if (l >= 10000L) return 5;
if (l >= 1000L) return 4;
if (l >= 100L) return 3;
if (l >= 10L) return 2;
return 1;
}
public FastPrintStream print(long x) {
if (x == Long.MIN_VALUE) {
return print("" + x);
}
if (ptr + 21 >= BUF_SIZE) innerflush();
if (x < 0) {
print((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastPrintStream print(double x, int precision) {
if (x < 0) {
print('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
// if(x < 0){ x = 0; }
print((long) x).print(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
print((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastPrintStream println(char c) {
return print(c).println();
}
public FastPrintStream println(int x) {
return print(x).println();
}
public FastPrintStream println(long x) {
return print(x).println();
}
public FastPrintStream println(String x) {
return print(x).println();
}
public FastPrintStream println(double x, int precision) {
return print(x, precision).println();
}
public FastPrintStream println() {
return print((byte) '\n');
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
}
private static class FastInputStream {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public FastInputStream(InputStream stream) {
this.stream = stream;
}
public double[] readDoubleArray(int size) {
double[] array = new double[size];
for (int i = 0; i < size; i++) {
array[i] = readDouble();
}
return array;
}
public String[] readStringArray(int size) {
String[] array = new String[size];
for (int i = 0; i < size; i++) {
array[i] = readString();
}
return array;
}
public char[] readCharArray(int size) {
char[] array = new char[size];
for (int i = 0; i < size; i++) {
array[i] = readCharacter();
}
return array;
}
public void readIntArrays(int[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readInt();
}
}
}
public void readLongArrays(long[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readLong();
}
}
}
public void readDoubleArrays(double[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readDouble();
}
}
}
public char[][] readTable(int rowCount, int columnCount) {
char[][] table = new char[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = this.readCharArray(columnCount);
}
return table;
}
public int[][] readIntTable(int rowCount, int columnCount) {
int[][] table = new int[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = readIntArray(columnCount);
}
return table;
}
public double[][] readDoubleTable(int rowCount, int columnCount) {
double[][] table = new double[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = this.readDoubleArray(columnCount);
}
return table;
}
public long[][] readLongTable(int rowCount, int columnCount) {
long[][] table = new long[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = readLongArray(columnCount);
}
return table;
}
public String[][] readStringTable(int rowCount, int columnCount) {
String[][] table = new String[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = this.readStringArray(columnCount);
}
return table;
}
public String readText() {
StringBuilder result = new StringBuilder();
while (true) {
int character = read();
if (character == '\r') {
continue;
}
if (character == -1) {
break;
}
result.append((char) character);
}
return result.toString();
}
public void readStringArrays(String[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readString();
}
}
}
public long[] readLongArray(int size) {
long[] array = new long[size];
for (int i = 0; i < size; i++) {
array[i] = readLong();
}
return array;
}
public int[] readIntArray(int size) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = readInt();
}
return array;
}
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 peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int peekNonWhitespace() {
while (isWhitespace(peek())) {
read();
}
return peek();
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public char readCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return readString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 1183c4afc3a6e07521c8f5b5692750b0 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.io.*;
import java.util.*;
/**
* Provide prove of correctness before implementation. Implementation can cost a lot of time.
* Anti test that prove that it's wrong.
* <p>
* Do not confuse i j k g indexes, upTo and length. Do extra methods!!! Write more informative names to simulation
* <p>
* Will program ever exceed limit?
* Try all approaches with prove of correctness if task is not obvious.
* If you are given formula/rule: Try to play with it.
* Analytic solution/Hardcoded solution/Constructive/Greedy/DP/Math/Brute force/Symmetric data
* Number theory
* Game theory (optimal play) that consider local and global strategy.
*/
public class CF1584C {
private void solveOne() {
int n = nextInt();
int[] a = nextIntArr(n);
int[] b = nextIntArr(n);
Arrays.sort(a);
Arrays.sort(b);
// Map<Integer, Integer> multiset = new HashMap<>();
// for (int i = 0; i < n; i++) {
// multiset.merge(a[i], 1, Integer::sum);
// }
// List<Integer> toMatchB = new ArrayList<>();
// for (int i = 0; i < n; i++) {
// if (multiset.containsKey(b[i])) {
// if (multiset.get(b[i]) == 1) {
// multiset.remove(b[i]);
// } else {
// multiset.put(b[i], multiset.get(b[i]) - 1);
// }
// } else {
// toMatchB.add(b[i]);
// }
// }
// List<Integer> toMatchA = new ArrayList<>();
// for (Integer integer : multiset.keySet()) {
// for (int i = 0; i < multiset.get(integer); i++) {
// toMatchA.add(integer);
// }
// }
// toMatchB.sort(Integer::compareTo);
// toMatchA.sort(Integer::compareTo);
//
// if (toMatchB.size() != toMatchA.size()) {
// throw new RuntimeException("!!!");
// }
boolean ans = true;
for (int i = 0; i < n; i++) {
ans &= a[i] + 1 == b[i] || a[i] == b[i];
}
System.out.println(ans ? "Yes" : "No");
}
private void solve() {
int t = System.in.readInt();
for (int tt = 0; tt < t; tt++) {
solveOne();
}
}
private int nextInt() {
return System.in.readInt();
}
private long nextLong() {
return System.in.readLong();
}
private String nextString() {
return System.in.readString();
}
private int[] nextIntArr(int n) {
return System.in.readIntArray(n);
}
private long[] nextLongArr(int n) {
return System.in.readLongArray(n);
}
public static void main(String[] args) {
new CF1584C().run();
}
static class System {
private static FastInputStream in;
private static FastPrintStream out;
}
private void run() {
System.in = new FastInputStream(java.lang.System.in);
System.out = new FastPrintStream(java.lang.System.out);
solve();
System.out.flush();
}
private static class FastPrintStream {
private static final int BUF_SIZE = 8192;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastPrintStream() {
this(java.lang.System.out);
}
public FastPrintStream(OutputStream os) {
this.out = os;
}
public FastPrintStream(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastPrintStream print(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE) innerflush();
return this;
}
public FastPrintStream print(char c) {
return print((byte) c);
}
public FastPrintStream print(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
}
return this;
}
public FastPrintStream print(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
});
return this;
}
//can be optimized
public FastPrintStream print0(char[] s) {
if (ptr + s.length < BUF_SIZE) {
for (char c : s) {
buf[ptr++] = (byte) c;
}
} else {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE) innerflush();
}
}
return this;
}
//can be optimized
public FastPrintStream print0(String s) {
if (ptr + s.length() < BUF_SIZE) {
for (int i = 0; i < s.length(); i++) {
buf[ptr++] = (byte) s.charAt(i);
}
} else {
for (int i = 0; i < s.length(); i++) {
buf[ptr++] = (byte) s.charAt(i);
if (ptr == BUF_SIZE) innerflush();
}
}
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000) return 10;
if (l >= 100000000) return 9;
if (l >= 10000000) return 8;
if (l >= 1000000) return 7;
if (l >= 100000) return 6;
if (l >= 10000) return 5;
if (l >= 1000) return 4;
if (l >= 100) return 3;
if (l >= 10) return 2;
return 1;
}
public FastPrintStream print(int x) {
if (x == Integer.MIN_VALUE) {
return print((long) x);
}
if (ptr + 12 >= BUF_SIZE) innerflush();
if (x < 0) {
print((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L) return 19;
if (l >= 100000000000000000L) return 18;
if (l >= 10000000000000000L) return 17;
if (l >= 1000000000000000L) return 16;
if (l >= 100000000000000L) return 15;
if (l >= 10000000000000L) return 14;
if (l >= 1000000000000L) return 13;
if (l >= 100000000000L) return 12;
if (l >= 10000000000L) return 11;
if (l >= 1000000000L) return 10;
if (l >= 100000000L) return 9;
if (l >= 10000000L) return 8;
if (l >= 1000000L) return 7;
if (l >= 100000L) return 6;
if (l >= 10000L) return 5;
if (l >= 1000L) return 4;
if (l >= 100L) return 3;
if (l >= 10L) return 2;
return 1;
}
public FastPrintStream print(long x) {
if (x == Long.MIN_VALUE) {
return print("" + x);
}
if (ptr + 21 >= BUF_SIZE) innerflush();
if (x < 0) {
print((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastPrintStream print(double x, int precision) {
if (x < 0) {
print('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
// if(x < 0){ x = 0; }
print((long) x).print(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
print((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastPrintStream println(char c) {
return print(c).println();
}
public FastPrintStream println(int x) {
return print(x).println();
}
public FastPrintStream println(long x) {
return print(x).println();
}
public FastPrintStream println(String x) {
return print(x).println();
}
public FastPrintStream println(double x, int precision) {
return print(x, precision).println();
}
public FastPrintStream println() {
return print((byte) '\n');
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
}
private static class FastInputStream {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public FastInputStream(InputStream stream) {
this.stream = stream;
}
public double[] readDoubleArray(int size) {
double[] array = new double[size];
for (int i = 0; i < size; i++) {
array[i] = readDouble();
}
return array;
}
public String[] readStringArray(int size) {
String[] array = new String[size];
for (int i = 0; i < size; i++) {
array[i] = readString();
}
return array;
}
public char[] readCharArray(int size) {
char[] array = new char[size];
for (int i = 0; i < size; i++) {
array[i] = readCharacter();
}
return array;
}
public void readIntArrays(int[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readInt();
}
}
}
public void readLongArrays(long[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readLong();
}
}
}
public void readDoubleArrays(double[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readDouble();
}
}
}
public char[][] readTable(int rowCount, int columnCount) {
char[][] table = new char[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = this.readCharArray(columnCount);
}
return table;
}
public int[][] readIntTable(int rowCount, int columnCount) {
int[][] table = new int[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = readIntArray(columnCount);
}
return table;
}
public double[][] readDoubleTable(int rowCount, int columnCount) {
double[][] table = new double[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = this.readDoubleArray(columnCount);
}
return table;
}
public long[][] readLongTable(int rowCount, int columnCount) {
long[][] table = new long[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = readLongArray(columnCount);
}
return table;
}
public String[][] readStringTable(int rowCount, int columnCount) {
String[][] table = new String[rowCount][];
for (int i = 0; i < rowCount; i++) {
table[i] = this.readStringArray(columnCount);
}
return table;
}
public String readText() {
StringBuilder result = new StringBuilder();
while (true) {
int character = read();
if (character == '\r') {
continue;
}
if (character == -1) {
break;
}
result.append((char) character);
}
return result.toString();
}
public void readStringArrays(String[]... arrays) {
for (int i = 0; i < arrays[0].length; i++) {
for (int j = 0; j < arrays.length; j++) {
arrays[j][i] = readString();
}
}
}
public long[] readLongArray(int size) {
long[] array = new long[size];
for (int i = 0; i < size; i++) {
array[i] = readLong();
}
return array;
}
public int[] readIntArray(int size) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = readInt();
}
return array;
}
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 peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int peekNonWhitespace() {
while (isWhitespace(peek())) {
read();
}
return peek();
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public char readCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return readString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 417b9345b6c2cd1eb6d4e5b67ff0da66 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class twoArrays {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
// -1 1 0
// 0 0 2
// -1 0 1
// 0 0 2
int t = sc.nextInt();
sc.nextLine();
for (int i = 0; i < t; i++) {
int N = sc.nextInt();
int[] aArr = new int[N];
int[] bArr = new int[N];
for (int j = 0; j < N; j++) {
aArr[j] = sc.nextInt();
}
for (int j = 0; j < N; j++) {
bArr[j] = sc.nextInt();
}
Arrays.sort(aArr);
Arrays.sort(bArr);
if (possible(aArr, bArr)) {
pw.println("YES");
} else {
pw.println("NO");
}
}
sc.close();
pw.close();
}
public static boolean possible(int[] aArr, int[] bArr) {
int count = 0;
for (int i = 0; i < aArr.length; i++) {
// return early when we reach a failed condition?
// 3 3, are they not equal , are not different by 1?
//
// ! (A || B)
// !A && !B
if (aArr[i] != bArr[i] && bArr[i] - aArr[i] != 1) {
return false;
}
}
return true;
}
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 2f96090eccd16d3c2f6d286784a48714 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class twoArrays {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
// -1 1 0
// 0 0 2
// -1 0 1
// 0 0 2
int t = sc.nextInt();
sc.nextLine();
for (int i = 0; i < t; i++) {
int N = sc.nextInt();
int[] aArr = new int[N];
int[] bArr = new int[N];
for (int j = 0; j < N; j++) {
aArr[j] = sc.nextInt();
}
for (int j = 0; j < N; j++) {
bArr[j] = sc.nextInt();
}
Arrays.sort(aArr);
Arrays.sort(bArr);
if (possible(aArr, bArr)) {
pw.println("YES");
} else {
pw.println("NO");
}
}
sc.close();
pw.close();
}
public static boolean possible(int[] aArr, int[] bArr) {
int count = 0;
for (int i = 0; i < aArr.length; i++) {
if (aArr[i] == bArr[i] || bArr[i] - aArr[i] == 1) {
count++;
}
}
if (count == aArr.length) {
return true;
}
return false;
}
}
| Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 4026b8dea2a4e520a2020f032d4cd2e4 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | /******************************************************************************
Welcome to GDB Online.
GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl,
C#, VB, Swift, Pascal, Fortran, Haskell, Objective-C, Assembly, HTML, CSS, JS, SQLite, Prolog.
Code, Compile, Run and Debug online from anywhere in world.
*******************************************************************************/
import java.util.*;
import java.io.*;
import java.lang.*;
public class sol{
static Scanner scan;
public static void main(String[] args){
scan = new Scanner(System.in);
int t = scan.nextInt();
for(int i = 0; i< t; i++){
res();
}
}
private static void res(){
int n = scan.nextInt();
int[] a = new int[n];
int[] b = new int[n];
for(int i = 0; i< n; i++){
a[i] = scan.nextInt();
}
for(int i = 0; i< n; i++){
b[i] = scan.nextInt();
}
Arrays.sort(a);
Arrays.sort(b);
boolean result = true;
for(int i = 0; i< n; i++){
int d = b[i]-a[i];
if(d<0 || d>1){
result = false;
break;
}
}
if(result){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | d1a8605b7bfa4b954e53bf2cd3c1f8eb | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes |
import java.io.PrintWriter;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// System.out.println(sc.nextByte());
PrintWriter pw = new PrintWriter(System.out);
twoarray(sc);
pw.close();
}
private static void twoarray(Scanner scanner){
int testcase=scanner.nextInt();
for(int i=0;i<testcase;i++){
int size=scanner.nextInt();
int[] arr=new int[size];
int[] arr2=new int[size];
for(int j=0;j<size;j++){
arr[j]=scanner.nextInt();
}
for(int j=0;j<size;j++){
arr2[j]=scanner.nextInt();
}
int start=0;
while (start<size && arr[start]==arr2[start])start++;
Arrays.sort(arr);
Arrays.sort(arr2);
boolean possible=true;
for(int j=0;j<size;j++){
int first=arr[j];
int second=arr2[j];
if(first!=second && (first+1)!=second){
System.out.println("NO");
possible=false;
break;
}
}
if(possible)System.out.println("YES");
}
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 169371babdfd4a19ce77b50e14ae4937 | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution{
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
static StringTokenizer st;
public static void main(String[] YDSV) throws IOException{
//int t=1;
int t=Integer.parseInt(br.readLine());
while(t-->0) Solve();
bw.flush();
}
public static void Solve() throws IOException{
st=new StringTokenizer(br.readLine());
int n=Integer.parseInt(st.nextToken());
int[] ar=getArrIn(n);
int[] brr=getArrIn(n);
Arrays.sort(ar);
Arrays.sort(brr);
for(int i=0;i<n;i++){
if(Math.abs(ar[i]-brr[i])>1 || ar[i]>brr[i]){
bw.write("NO\n");
return ;
}
}
bw.write("YES\n");
}
public static int[] getArrIn(int n) throws IOException{
st=new StringTokenizer(br.readLine());
int[] ar=new int[n];
for(int i=0;i<n;i++) ar[i]=Integer.parseInt(st.nextToken());
return ar;
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output | |
PASSED | 25f2a36edb6aa4912b9450b72dfdeddd | train_109.jsonl | 1636869900 | You are given two arrays of integers $$$a_1, a_2, \ldots, a_n$$$ and $$$b_1, b_2, \ldots, b_n$$$.Let's define a transformation of the array $$$a$$$: Choose any non-negative integer $$$k$$$ such that $$$0 \le k \le n$$$. Choose $$$k$$$ distinct array indices $$$1 \le i_1 < i_2 < \ldots < i_k \le n$$$. Add $$$1$$$ to each of $$$a_{i_1}, a_{i_2}, \ldots, a_{i_k}$$$, all other elements of array $$$a$$$ remain unchanged. Permute the elements of array $$$a$$$ in any order. Is it possible to perform some transformation of the array $$$a$$$ exactly once, so that the resulting array is equal to $$$b$$$? | 256 megabytes | import java.util.*;
public class Hello {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T = Integer.parseInt(sc.nextLine());
while (T-- > 0) {
int N = sc.nextInt();
int arr1[] = new int[N];
int arr2[] = new int[N];
for (int index = 0; index < N; index++) {
arr1[index] = sc.nextInt();
}
for (int index = 0; index < N; index++) {
arr2[index] = sc.nextInt();
}
Arrays.sort(arr1);
Arrays.sort(arr2);
String result = "YES";
for(int index = 0; index < N; index++) {
int val = arr2[index] - arr1[index];
if(val < 0 || val > 1) {
result = "NO";
break;
}
}
System.out.println(result);
}
}
} | Java | ["3\n3\n-1 1 0\n0 0 2\n1\n0\n2\n5\n1 2 3 4 5\n1 2 3 4 5"] | 1 second | ["YES\nNO\nYES"] | NoteIn the first test case, we can make the following transformation: Choose $$$k = 2$$$. Choose $$$i_1 = 1$$$, $$$i_2 = 2$$$. Add $$$1$$$ to $$$a_1$$$ and $$$a_2$$$. The resulting array is $$$[0, 2, 0]$$$. Swap the elements on the second and third positions. In the second test case there is no suitable transformation.In the third test case we choose $$$k = 0$$$ and do not change the order of elements. | Java 11 | standard input | [
"greedy",
"math",
"sortings"
] | 6ca98e655007bfb86f1039c9f096557e | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Descriptions of test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$) — the size of arrays $$$a$$$ and $$$b$$$. The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-100 \le a_i \le 100$$$). The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$-100 \le b_i \le 100$$$). | 900 | For each test case, print "YES" (without quotes) if it is possible to perform a transformation of the array $$$a$$$, so that the resulting array is equal to $$$b$$$. Print "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower). | standard output |
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