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PASSED
5d912aabea5813b8b6e2218f79494194
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.File; import java.io.FileReader; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.StringTokenizer; public class Solution { static PrintWriter pw; static FastScanner s; public static void main(String[] args) throws Exception { pw=new PrintWriter(System.out); s=new FastScanner(System.in); int t=s.nextInt(); while(t-->0) { int n=s.nextInt(); int a[]=new int[n]; for(int i=0;i<n;i++) { a[i]=s.nextInt(); } boolean flag=true; for(int i=1;i<n;i++) { if(a[i]%a[0]!=0) { flag=false; break; } } if(flag) pw.println("YES"); else pw.println("NO"); } pw.flush(); } } class FastScanner{ BufferedReader br; StringTokenizer st; public FastScanner(InputStream s) { br = new BufferedReader(new InputStreamReader(s)); } public FastScanner(String s) throws Exception { br = new BufferedReader(new FileReader(new File(s))); } public String next() throws Exception { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine()); return st.nextToken(); } public int nextInt() throws Exception { return Integer.parseInt(next()); } public long nextLong() throws Exception { return Long.parseLong(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 8
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
07e93ed846a8a07bc00c5a95138cc563
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 codeforcesProject; import java.util.ArrayList; import java.util.Arrays; import java.util.Scanner; public class DifferenceOperations { private int count; ArrayList<Element> elements = new ArrayList<>(); public static void main(String[] args) { DifferenceOperations differenceOperations = new DifferenceOperations(); differenceOperations.inputData(); differenceOperations.callingEachContainer(); } private void inputData() { Scanner in = new Scanner(System.in); String firstLine = in.nextLine(); count = Integer.parseInt(firstLine); for (int i = 0; i < count; i++) { String st1 = in.nextLine(); String st2 = in.nextLine(); elements.add(Element.createElement(st1, st2)); } } private void callingEachContainer() { for (int i = 0; i < count; i++) { logicsForAny(elements.get(i)); } } private void logicsForAny(Element element) { String result = "YES"; int firstElement = element.combination[0]; for (int i = 1; i < element.count; i++) { if (element.combination[i] % firstElement != 0) { result = "NO"; } } System.out.println(result); } } class Element { Integer count; int combination[]; public Element(int count, int[] combination) { this.count = count; this.combination = combination; } public static Element createElement(String st1, String st2) { int count = Integer.parseInt(st1); int[] numArr = Arrays.stream(st2.split(" ")).mapToInt(Integer::parseInt).toArray(); return new Element(count, numArr); } @Override public String toString() { return "Element{" + "count=" + count + ", combination=" + Arrays.toString(combination) + "}\n"; } } //https://codeforces.com/problemset/problem/1708/A
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 8
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
ed5797200f812eeb7b57d113c48c9f97
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 javax.naming.NamingEnumeration; import javax.swing.plaf.IconUIResource; import java.lang.reflect.Array; import java.util.*; import java.math.BigInteger; public class ques { 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]; arr[0] = sc.nextInt(); String ans="Yes"; for (int j = 1; j < n; j++) { arr[j] = sc.nextInt(); if (arr[j]%arr[0] !=0){ ans="No"; } } System.out.println(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 8
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
8e380f720a30087b3f2e02f4eb3c2f46
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 inMain = new Scanner(System.in); int t = inMain.nextInt(); while (t > 0) { differenceOperation(inMain); t--; } } public static void differenceOperation(Scanner input) { int n = input.nextInt(); int start = input.nextInt(); int next; String result = "YES"; for (int i=0; i < n-1; i++) { next = input.nextInt(); if (!(next % start == 0)) { result = "NO"; } } System.out.println(result); } }
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
fa82e23650bf0f5e5b78a576d1e6e1f4
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
// Working program with FastReader import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Scanner; import java.util.StringTokenizer; import java.util.*; import java.lang.*; import java.util.stream.*; import java.util.stream.Collectors; 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 { if(st.hasMoreTokens()){ str = st.nextToken("\n"); } else{ str = br.readLine(); } } catch (IOException e) { e.printStackTrace(); } return str; } } public static int calcdiv2(int n) { int counter = 0; while( ( 1 & n) != 1) { n = n>>1; counter++; } return counter; } public static int highestPowerOfTwo(int n) { if((n & (n-1)) == 0) return n; return n & (n-1); } public static boolean isPowerOfTwo(int n) { return (n & (n-1)) == 0; } public static int valueOfRightMostSetBit(int n) { return (n & (-n)); } // public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm) // { // HashMap<Integer, Integer> temp // = hm.entrySet() // .stream() // .sorted((i1, i2) // -> i1.getValue().compareTo( // i2.getValue())) // .collect(Collectors.toMap( // Map.Entry::getKey, // Map.Entry::getValue, // (e1, e2) -> e1, LinkedHashMap::new)); // return temp; // } // static int[] sortint(int a[]) { // ArrayList<Integer> al = new ArrayList<>(); // for (int i : a) { // al.add(i); // } // Collections.sort(al); // for (int i = 0; i < a.length; i++) { // a[i] = al.get(i); // } // return a; // } // static long[] sortlong(long a[]) { // ArrayList<Long> al = new ArrayList<>(); // for (long i : a) { // al.add(i); // } // Collections.sort(al); // for (int i = 0; i < al.size(); i++) { // a[i] = al.get(i); // } // return a; // } // static ArrayList<Integer> primesTilN(int n) { // boolean p[] = new boolean[n + 1]; // ArrayList<Integer> al = new ArrayList<>(); // Arrays.fill(p, true); // int i = 0; // for (i = 2; i * i <= n; i++) { // if (p[i] == true) { // al.add(i); // for (int j = i * i; j <= n; j += i) { // p[j] = false; // } // } // } // for (i = i; i <= n; i++) { // if (p[i] == true) { // al.add(i); // } // } // return al; // } // static int gcd(int a, int b) { // if (a == 0) { // return b; // } // return gcd(b % a, a); // } // static long gcd(long a, long b) { // if (a == 0) { // return b; // } // return gcd(b % a, a); // } // static long pow(long x, long y) { // long result = 1; // while (y > 0) { // if (y % 2 == 0) { // x = x * x; // y = y / 2; // } else { // result = result * x; // y = y - 1; // } // } // return result; // } // static boolean[] esieve(int n) { // boolean p[] = new boolean[n + 1]; // Arrays.fill(p, true); // for (int i = 2; i * i <= n; i++) { // if (p[i] == true) { // for (int j = i * i; j <= n; j += i) { // p[j] = false; // } // } // } // return p; // } // static int[] readarr(int n) { // int arr[] = new int[n]; // for (int i = 0; i < n; i++) { // arr[i] = x.nextInt(); // } // return arr; // } // static HashMap<Integer, Integer> mapCounter(int arr[]) { // for(int i : arr) { // if(map.containsKey(i)) map.put(arr[i], map.get(arr[i])+1); // else map.put(arr[i], 1); // } // } // static HashMap<Integer, Integer> mapCounter(int arr[]) { // for(int i : arr) { // if(map.containsKey(i)) map.put(arr[i], map.get(arr[i])+1); // else map.put(arr[i], 1); // } // } public static void main(String[] args) { FastReader sc = new FastReader(); int testcases = sc.nextInt(); int count = 0; while (count < testcases) { int n = sc.nextInt(); long arr[] = new long[n]; for(int i = 0; i < n; i++) { arr[i] = sc.nextInt(); } if(arr[0] == 1) { System.out.println("YES"); } else { boolean is = true; for(int i = 1; i < n; i++) { if((arr[i] % arr[i-1] != 0 && arr[i] % arr[0] != 0) && arr[i-1] != 0) { is = false; break; } } if(is) System.out.println("YES"); else System.out.println("NO"); } count++; } } }
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
ef4153f861ce32214fde5b3b59d2f76e
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.*; public class Substitute { //Global Declarations final static Random random = new Random(); final static FastReader in = new FastReader(); final static PrintWriter out = new PrintWriter(System.out); public static String solve(int a[]) { //Long range is -10^18 to 10^18 and is 64 bit (8 bytes) //Integer range is ~~ -2 * 10^9(-2^31) to 2 * 10^9(2^31 - 1) and is 32 bit (4 bytes) //out.println(Arrays.toString(a)); int i,n = a.length,temp=a[0]; if(a[0] == 1) return "YES"; for(i=1;i<n;++i){ if(a[i]%temp == 0){ continue; } else return "NO"; } return "YES"; } public static int solve2(int l, int r, int pre[]){ return (l ==0) ? pre[r] : pre[r] - pre[l-1]; } public static void reverseLong(long a[]) { int i,n=a.length,j=n-1; long temp; for(i=0;i<n/2;++i){ temp = a[i]; a[i] = a[j]; a[j] = temp; --j; } } public static void reverseInt(int a[]) { int i,n=a.length,j=n-1,temp; for(i=0;i<n/2;++i){ temp = a[i]; a[i] = a[j]; a[j] = temp; --j; } } public static void ruffleLong(long a[]) { //shuffle, then sort int i,j,n=a.length; long temp; for (i=0; i<n; i++) { j = random.nextInt(n); temp = a[i]; a[i] = a[j]; a[j] = temp; } Arrays.sort(a); } public static void ruffleInt(int a[]) { //shuffle, then sort int i,j,n=a.length,temp; for (i=0; i<n; i++) { j = random.nextInt(n); temp = a[i]; a[i] = a[j]; a[j] = temp; } Arrays.sort(a); } public static void main(String ... args) throws IOException { // << --> multiply // >> --> divide //variable_name <<= 1 --> (variable)multiply by 2 //variable_name >>= 1 --> (variable)divide by 2 //2<<3 --> (multiply)2 * 2^3 --> 16; //6>>1 --> (divide)6 / 2^1 --> 3; //out.println(solve(1,20000)); //out.println(1/solve(0,Math.abs(-2))); //out.println(solve(5,3)); //48782 //out.println(solve()); /* int n = in.nextInt(),q = in.nextInt(),i,l=n-1; //StringBuilder sb = new StringBuilder(in.next()); Pair pairs[] = new Pair[n]; String s = in.next(); for(i=0;i<n;++i) pairs[i] = new Pair(i,s.charAt(i)); while(q-->0){ int t = in.nextInt(); int x = in.nextInt(); i=0; int val = x-1,temp = l,temp2 = (l+1)%n; if(temp2 >= n) temp2 = 0; if(t == 1){ while(i++ < x){ pairs[temp] = new Pair(val, pairs[temp].getChar()); --temp; --val; if(temp == -1) temp = n-1; } i=0; while(i++ < n-x){ int l2 = (pairs[temp2].getIndex() + x)%n; pairs[temp2] = new Pair(l2, pairs[temp2].getChar()); if(i == n-x){ l = l2; //out.println(l); } ++temp2; if(temp2 == n) temp2 = 0; } } else{ //out.println(pairs[x-1].getChar()); for(i=0;i<n;++i){ if(pairs[i].getIndex() == x-1){ out.println(pairs[i].getChar()); break; } } } for(i=0;i<n;++i){ out.println(pairs[i].getIndex() + " " + pairs[i].getChar()); } }*/ int tt = in.nextInt(); while(tt-->0){ int n = in.nextInt(); int i; LinkedHashSet<Integer> st = new LinkedHashSet<>(); for(i=0;i<n;++i){ st.add(in.nextInt()); } if(st.size() == 1) out.println("YES"); else{ i=0; int a[] = new int[st.size()]; for(Integer val: st){ a[i++] = val; } out.println(solve(a)); } } out.close(); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader( new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } class Pair{ private int in; private char c; public Pair(int in, char c){ this.in = in; this.c = c; } public int getIndex(){ return this.in; } public char getChar(){ return this.c; } }
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
b36f660ee9b46a11195e161ebf3107af
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.*; public class Substitute { //Global Declarations final static Random random = new Random(); final static FastReader in = new FastReader(); final static PrintWriter out = new PrintWriter(System.out); public static String solve(int a[]) { //Long range is -10^18 to 10^18 and is 64 bit (8 bytes) //Integer range is ~~ -2 * 10^9(-2^31) to 2 * 10^9(2^31 - 1) and is 32 bit (4 bytes) //out.println(Arrays.toString(a)); int i,n = a.length,temp=a[0]; if(a[0] == 1) return "YES"; for(i=1;i<n;++i){ if(a[i]%temp == 0){ continue; } else return "NO"; } return "YES"; } public static int solve2(int l, int r, int pre[]){ return (l ==0) ? pre[r] : pre[r] - pre[l-1]; } public static void reverseLong(long a[]) { int i,n=a.length,j=n-1; long temp; for(i=0;i<n/2;++i){ temp = a[i]; a[i] = a[j]; a[j] = temp; --j; } } public static void reverseInt(int a[]) { int i,n=a.length,j=n-1,temp; for(i=0;i<n/2;++i){ temp = a[i]; a[i] = a[j]; a[j] = temp; --j; } } public static void ruffleLong(long a[]) { //shuffle, then sort int i,j,n=a.length; long temp; for (i=0; i<n; i++) { j = random.nextInt(n); temp = a[i]; a[i] = a[j]; a[j] = temp; } Arrays.sort(a); } public static void ruffleInt(int a[]) { //shuffle, then sort int i,j,n=a.length,temp; for (i=0; i<n; i++) { j = random.nextInt(n); temp = a[i]; a[i] = a[j]; a[j] = temp; } Arrays.sort(a); } public static void main(String ... args) throws IOException { // << --> multiply // >> --> divide //variable_name <<= 1 --> (variable)multiply by 2 //variable_name >>= 1 --> (variable)divide by 2 //2<<3 --> (multiply)2 * 2^3 --> 16; //6>>1 --> (divide)6 / 2^1 --> 3; //out.println(solve(1,20000)); //out.println(1/solve(0,Math.abs(-2))); //out.println(solve(5,3)); //48782 //out.println(solve()); /* int n = in.nextInt(),q = in.nextInt(),i,l=n-1; //StringBuilder sb = new StringBuilder(in.next()); Pair pairs[] = new Pair[n]; String s = in.next(); for(i=0;i<n;++i) pairs[i] = new Pair(i,s.charAt(i)); while(q-->0){ int t = in.nextInt(); int x = in.nextInt(); i=0; int val = x-1,temp = l,temp2 = (l+1)%n; if(temp2 >= n) temp2 = 0; if(t == 1){ while(i++ < x){ pairs[temp] = new Pair(val, pairs[temp].getChar()); --temp; --val; if(temp == -1) temp = n-1; } i=0; while(i++ < n-x){ int l2 = (pairs[temp2].getIndex() + x)%n; pairs[temp2] = new Pair(l2, pairs[temp2].getChar()); if(i == n-x){ l = l2; //out.println(l); } ++temp2; if(temp2 == n) temp2 = 0; } } else{ //out.println(pairs[x-1].getChar()); for(i=0;i<n;++i){ if(pairs[i].getIndex() == x-1){ out.println(pairs[i].getChar()); break; } } } for(i=0;i<n;++i){ out.println(pairs[i].getIndex() + " " + pairs[i].getChar()); } }*/ int tt = in.nextInt(); while(tt-->0){ int n = in.nextInt(); int i; LinkedHashSet<Integer> st = new LinkedHashSet<>(); for(i=0;i<n;++i){ st.add(in.nextInt()); } if(st.size() == 1) out.println("YES"); else{ i=0; int a[] = new int[st.size()]; for(Integer val: st){ a[i++] = Integer.valueOf(val); } out.println(solve(a)); } } out.close(); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader( new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } } class Pair{ private int in; private char c; public Pair(int in, char c){ this.in = in; this.c = c; } public int getIndex(){ return this.in; } public char getChar(){ return this.c; } }
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
c416ccc102011876c8bde7739e4cb0b4
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; public class CF1607{ static Scanner scan; public static void main(String args[]){ scan = new Scanner(System.in); int testCases = scan.nextInt(); while(testCases>0){ int arrayLength=scan.nextInt(); if(isPossible(arrayLength)){ System.out.println("YES"); }else{ System.out.println("NO"); } testCases--; } } public static boolean isPossible(int arrayLength){ int firstNumber= scan.nextInt(); for(int i=1; i<arrayLength; i++){ int num=scan.nextInt(); if(num%firstNumber!=0){ scan.nextLine(); return false; } } return true; } }
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
b16589ce456a8753b96c87cd31a71cfd
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 { 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[] a=new int[n]; for (int j = 0; j < n; j++) { a[j]=sc.nextInt(); } boolean flag=true; for (int j = 1; j < n; j++) { if(a[j]%a[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 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
ee1f1d384dae5e69f797e118152f5ec1
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{ public static void main(String[] args){ Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-->0){ int k=sc.nextInt(); long arr[]=new long[k]; boolean a=true; for(int i=0;i<k;i++){ arr[i]=sc.nextInt(); } for(int i=1;i<k;i++){ if(arr[i]%arr[0]!=0){ a=false; break; } } if(a==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 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
0a0bf3bb8451ac213d7c3b92b3e7d60c
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.ArrayList; import java.util.Arrays; import java.util.StringTokenizer; //DECD public class Main { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int t = Integer.parseInt(br.readLine()); for (int i = 0; i < t; i++) { int n = Integer.parseInt(br.readLine()); StringTokenizer st = new StringTokenizer(br.readLine()); long[] arr = new long[n]; for (int j = 0; j < n; j++) { arr[j]=Long.parseLong(st.nextToken()); } long result=findGCD(arr); boolean invalid=false; if(!((arr[1]-result)%arr[0]==0)){ invalid=true; } if(arr[1]%arr[0]!=0){ invalid=true; } if(!invalid) { System.out.println("YES"); } else { System.out.println("NO"); } } //7 8 2 4 4 3 5 3 //7 1 2 4 4 3 5 3 //7 1 1 1 1 1 1 1 // 5 10 15 // 4 10 36 42 // 2 8 36 42 //result + (x)*a[0]=a[1] //(a[1]-result)%a[0]==0 //2 4 12 30 //2 2 12 30 //2 2 2 2 } static long gcd(long a, long b) { if (a == 0) return b; return gcd(b % a, a); } // Function to find gcd of array of // numbers static long findGCD(long arr[]) { long result = arr[0]; for (long element: arr){ result = gcd(result, element); if(result == 1) { return 1; } } return result; } }
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
b5c1c4530386c79074e1d37ccc391185
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.math.*; public class A { public static void main(String[] args) throws java.lang.Exception { Reader scan = new Reader(); int tc = scan.nextInt(); for (int i = 0; i < tc; i++) { int n = scan.nextInt(); ArrayList <Long> arr = new ArrayList <Long>(); for (int j = 0; j < n; j++){ long l = scan.nextLong(); arr.add(l); } System.out.println(solve(arr)); } } public static String solve(ArrayList <Long> arr) { long first = arr.get(0); for (int i = 0; i < arr.size(); i++){ if (arr.get(i) % first != 0) return "NO"; } return "YES"; } public static void printarr(ArrayList<Long> arr) { for (int i = 0; i < arr.size(); i++) { System.out.print(arr.get(i) + " "); } System.out.println(); } static class Reader { final private int BUFFER_SIZE = 1 << 64; 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[1000000]; // line length int cnt = 0, c; while ((c = read()) != -1) { if (c == '\n') { if (cnt != 0) { break; } else { continue; } } buf[cnt++] = (byte) c; } return new String(buf, 0, cnt); } public int nextInt() throws IOException { int ret = 0; byte c = read(); while (c <= ' ') { c = read(); } boolean neg = (c == '-'); if (neg) c = read(); do { ret = ret * 10 + c - '0'; } while ((c = read()) >= '0' && c <= '9'); if (neg) return -ret; return ret; } public long nextLong() throws IOException { long ret = 0; byte c = read(); while (c <= ' ') c = read(); boolean neg = (c == '-'); if (neg) c = read(); do { ret = ret * 10 + c - '0'; } while ((c = read()) >= '0' && c <= '9'); if (neg) return -ret; return ret; } public double nextDouble() throws IOException { double ret = 0, div = 1; byte c = read(); while (c <= ' ') c = read(); boolean neg = (c == '-'); if (neg) c = read(); do { ret = ret * 10 + c - '0'; } while ((c = read()) >= '0' && c <= '9'); if (c == '.') { while ((c = read()) >= '0' && c <= '9') { ret += (c - '0') / (div *= 10); } } if (neg) return -ret; return ret; } private void fillBuffer() throws IOException { bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); if (bytesRead == -1) buffer[0] = -1; } private byte read() throws IOException { if (bufferPointer == bytesRead) fillBuffer(); return buffer[bufferPointer++]; } public void close() throws IOException { if (din == null) return; din.close(); } } }
Java
["4\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
7fbb5edb23c8fcf04066591e4ecfc3cb
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 C2 { 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()); while (t > 0) { int n = Integer.parseInt(br.readLine()); StringTokenizer st = new StringTokenizer(br.readLine()); long[] arr = new long[n]; boolean yes = true; for (int i = 0; i < n; i++) { arr[i] = Long.parseLong(st.nextToken()); if (i != 0) if (arr[i] % arr[0] != 0) { yes = false; break; } } if (yes) pw.println("YES"); else pw.println("NO"); --t; } 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
334714537fcdb42b2e7e871617b1cc1a
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.InputStreamReader; import java.util.*; public class notAcheapString { public static void main(String[] args)throws Exception { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); int t=Integer.parseInt(br.readLine()); while(t-->0) { boolean flag=true; int n=Integer.parseInt(br.readLine()); String [] input=br.readLine().split(" "); int [] arr=new int[n]; for(int i=0;i<n;i++) { arr[i]=Integer.parseInt(input[i]); } for(int i=1;i<n;i++) { if(arr[i]%arr[0]!=0 ) flag=false; } 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 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
ef6cb93bbe97bee4cb72cb87cba8829e
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.util.*; public class weird_algrithm { static BufferedWriter output = new BufferedWriter( new OutputStreamWriter(System.out)); static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in)); static int mod = 1000000007; static String toReturn = ""; static int steps = Integer.MAX_VALUE; static int maxlen = 1000005; /*MATHEMATICS FUNCTIONS START HERE MATHS MATHS MATHS MATHS*/ static long gcd(long a, long b) { if(b == 0) return a; else return gcd(b, a % b); } static long powerMod(long x, long y, int mod) { if(y == 0) return 1; long temp = powerMod(x, y / 2, mod); temp = ((temp % mod) * (temp % mod)) % mod; if(y % 2 == 0) return temp; else return ((x % mod) * (temp % mod)) % mod; } static long modInverse(long n, int p) { return powerMod(n, p - 2, p); } static long nCr(int n, int r, int mod, long [] fact, long [] ifact) { return ((fact[n] % mod) * ((ifact[r] * ifact[n - r]) % mod)) % mod; } static boolean isPrime(long a) { if(a == 1) return false; else if(a == 2 || a == 3 || a== 5) return true; else if(a % 2 == 0 || a % 3 == 0) return false; for(int i = 5; i * i <= a; i = i + 6) { if(a % i == 0 || a % (i + 2) == 0) return false; } return true; } static int [] seive(int a) { int [] toReturn = new int [a + 1]; for(int i = 0; i < a; i++) toReturn[i] = 1; toReturn[0] = 0; toReturn[1] = 0; toReturn[2] = 1; for(int i = 2; i * i <= a; i++) { if(toReturn[i] == 0) continue; for(int j = 2 * i; j <= a; j += i) toReturn[j] = 0; } return toReturn; } static long [] fact(int a) { long [] arr = new long[a + 1]; arr[0] = 1; for(int i = 1; i < a + 1; i++) { arr[i] = (arr[i - 1] * i) % mod; } return arr; } /*MATHS MATHS MATHS MATHS MATHEMATICS FUNCTIONS END HERE */ /*SWAP FUNCTION START HERE SWAP SWAP SWAP SWAP */ static void swap(int i, int j, long[] arr) { long temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } static void swap(int i, int j, int[] arr) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } static void swap(int i, int j, String [] arr) { String temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } static void swap(int i, int j, char [] arr) { char temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } /*SWAP SWAP SWAP SWAP SWAP FUNCTION END HERE*/ /*BINARY SEARCH METHODS START HERE * BINARY SEARCH * BINARY SEARCH * BINARY SEARCH * BINARY SEARCH */ static boolean BinaryCheck(long test, long [] arr, long health) { for(int i = 0; i <= arr.length - 1; i++) { if(i == arr.length - 1) health -= test; else if(arr[i + 1] - arr[i] > test) { health = health - test; }else { health = health - (arr[i + 1] - arr[i]); } if(health <= 0) return true; } return false; } static long binarySearchModified(long start1, long n, ArrayList<Long> arr, int a, long r) { long start = start1, end = n, ans = -1; while(start < end) { long mid = (start + end) / 2; //System.out.println(mid); if(arr.get((int)mid) + arr.get(a) <= r && mid != start1) { ans = mid; start = mid + 1; }else{ end = mid; } } //System.out.println(); return ans; } static int binarySearch(int start, int end, long [] arr, long val) { while(start < end) { int mid = (int)Math.ceil((start + end) / 2.0); //System.out.println(mid); if(arr[mid] > val) end = mid - 1; else start = mid; } return start; } /*BINARY SEARCH * BINARY SEARCH * BINARY SEARCH * BINARY SEARCH * BINARY SEARCH BINARY SEARCH METHODS END HERE*/ /*RECURSIVE FUNCTION START HERE * RECURSIVE * RECURSIVE * RECURSIVE * RECURSIVE */ static int recurse(int x, int y, int n, int steps1, Integer [][] dp) { if(x > n || y > n) return 0; if(dp[x][y] != null) { return dp[x][y]; } else if(x == n || y == n) { return steps1; } return dp[x][y] = Math.max(recurse(x + y, y, n, steps1 + 1, dp), recurse(x, x + y, n, steps1 + 1, dp)); } /*RECURSIVE * RECURSIVE * RECURSIVE * RECURSIVE * RECURSIVE RECURSIVE FUNCTION END HERE*/ /*GRAPH FUNCTIONS START HERE * GRAPH * GRAPH * GRAPH * GRAPH * */ static class edge{ int from, to; long weight; public edge(int x, int y, long weight2) { this.from = x; this.to = y; this.weight = weight2; } } static class sort implements Comparator<pair>{ @Override public int compare(pair o1, pair o2) { // TODO Auto-generated method stub return (int)o1.a - (int)o2.a; } } static void addEdge(ArrayList<ArrayList<edge>> graph, int from, int to, long weight) { edge temp = new edge(from, to, weight); edge temp1 = new edge(to, from, weight); graph.get(from).add(temp); //graph.get(to).add(temp1); } static int ans = 0; static int dfs(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, int [] childs) { //System.out.println(graph.get(vertex).size()); if(visited[vertex]) return 0; visited[vertex] = true; int ans = 0; for(int i = 0; i < graph.get(vertex).size(); i++) { int temp = graph.get(vertex).get(i); if(!visited[temp]) { ans += dfs(graph, temp, visited, childs) + 1; } } childs[vertex] = ans; return ans; } static void topoSort(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, ArrayList<Integer> toReturn) { if(visited[vertex]) return; visited[vertex] = true; for(int i = 0; i < graph.get(vertex).size(); i++) { if(!visited[graph.get(vertex).get(i)]) topoSort(graph, graph.get(vertex).get(i), visited, toReturn); } toReturn.add(vertex); } static boolean isCyclicDirected(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, boolean [] reStack) { if(reStack[vertex]) return true; if(visited[vertex]) return false; reStack[vertex] = true; visited[vertex] = true; for(int i = 0; i < graph.get(vertex).size(); i++) { if(isCyclicDirected(graph, graph.get(vertex).get(i), visited, reStack)) return true; } reStack[vertex] = false; return false; } static int e = 0; static long mst(PriorityQueue<edge> pq, int nodes) { long weight = 0; while(!pq.isEmpty()) { edge temp = pq.poll(); int x = parent(parent, temp.to); int y = parent(parent, temp.from); if(x != y) { //System.out.println(temp.weight); union(x, y, rank, parent); weight += temp.weight; e++; } } return weight; } static void floyd(long [][] dist) { // to find min distance between two nodes for(int k = 0; k < dist.length; k++) { for(int i = 0; i < dist.length; i++) { for(int j = 0; j < dist.length; j++) { if(dist[i][j] > dist[i][k] + dist[k][j]) { dist[i][j] = dist[i][k] + dist[k][j]; } } } } } static void dijkstra(ArrayList<ArrayList<edge>> graph, long [] dist, int src) { for(int i = 0; i < dist.length; i++) dist[i] = Long.MAX_VALUE / 2; dist[src] = 0; boolean visited[] = new boolean[dist.length]; PriorityQueue<pair> pq = new PriorityQueue<>(new sort()); pq.add(new pair(src, 0)); while(!pq.isEmpty()) { pair temp = pq.poll(); int index = (int)temp.a; for(int i = 0; i < graph.get(index).size(); i++) { if(dist[graph.get(index).get(i).to] > dist[index] + graph.get(index).get(i).weight) { dist[graph.get(index).get(i).to] = dist[index] + graph.get(index).get(i).weight; pq.add(new pair(graph.get(index).get(i).to, graph.get(index).get(i).weight)); } } } } static int parent1 = -1; static boolean ford(ArrayList<ArrayList<edge>> graph1, ArrayList<edge> graph, long [] dist, int src, int [] parent) { for(int i = 0; i < dist.length; i++) dist[i] = Long.MIN_VALUE / 2; dist[src] = 0; boolean hasNeg = false; for(int i = 0; i < dist.length - 1; i++) { for(int j = 0; j < graph.size(); j++) { int from = graph.get(j).from; int to = graph.get(j).to; long weight = graph.get(j).weight; if(dist[to] < dist[from] + weight) { dist[to] = dist[from] + weight; parent[to] = from; } } } for(int i = 0; i < graph.size(); i++) { int from = graph.get(i).from; int to = graph.get(i).to; long weight = graph.get(i).weight; if(dist[to] < dist[from] + weight) { parent1 = from; hasNeg = true; /* * dfs(graph1, parent1, new boolean[dist.length], dist.length - 1); * //System.out.println(ans); dfs(graph1, 0, new boolean[dist.length], parent1); */ //System.out.println(ans); if(ans == 2) break; else ans = 0; } } return hasNeg; } /*GRAPH FUNCTIONS END HERE * GRAPH * GRAPH * GRAPH * GRAPH */ /*disjoint Set START HERE * disjoint Set * disjoint Set * disjoint Set * disjoint Set */ static int [] rank; static int [] parent; static int parent(int [] parent, int x) { if(parent[x] == x) return x; else return parent[x] = parent(parent, parent[x]); } static boolean union(int x, int y, int [] rank, int [] parent) { if(parent(parent, x) == parent(parent, y)) { return true; } if(rank[x] > rank[y]) { swap(x, y, rank); } rank[x] += rank[y]; parent[y] = x; return false; } /*disjoint Set END HERE * disjoint Set * disjoint Set * disjoint Set * disjoint Set */ /*INPUT START HERE * INPUT * INPUT * INPUT * INPUT * INPUT */ static int nextInt() throws NumberFormatException, IOException { return Integer.parseInt(sc.readLine()); } static long nextLong() throws NumberFormatException, IOException { return Long.parseLong(sc.readLine()); } static long [] inputLongArr() throws NumberFormatException, IOException{ String [] s = sc.readLine().split(" "); long [] toReturn = new long[s.length]; for(int i = 0; i < s.length; i++) { toReturn[i] = Long.parseLong(s[i]); } return toReturn; } static int max = 0; static int [] inputIntArr() throws NumberFormatException, IOException{ String [] s = sc.readLine().split(" "); //System.out.println(s.length); int [] toReturn = new int[s.length]; for(int i = 0; i < s.length; i++) { toReturn[i] = Integer.parseInt(s[i]); } return toReturn; } static void print(int n) throws IOException { output.write(Integer.toString(n)); output.flush(); } static void print(String s) throws IOException { output.write(s); output.flush(); } static void print(long n) throws IOException { output.write(Long.toString(n) + "\n"); output.flush(); } static void print(char n) throws IOException { output.write(n); output.flush(); } /*INPUT * INPUT * INPUT * INPUT * INPUT * INPUT END HERE */ static long [] preCompute(int level) { long [] toReturn = new long[level]; toReturn[0] = 1; toReturn[1] = 16; for(int i = 2; i < level; i++) { toReturn[i] = ((toReturn[i - 1] % mod) * (toReturn[i - 1] % mod)) % mod; } return toReturn; } static class pair{ long a; long b; long d; public pair(long in, long y) { this.a = in; this.b = y; this.d = 0; } } static long smallestFactor(long a) { for(long i = 2; i * i <= a; i++) { if(a % i == 0) { return i; } } return a; } static int recurseRow(int [][] mat, int i, int j, int prev, int ans) { if(j >= mat[0].length) return ans; if(mat[i][j] == prev) { return recurseRow(mat, i, j + 1, prev, ans + 1); } else return ans; } static int recurseCol(int [][] mat, int i, int j, int prev, int ans) { if(i >= mat.length) return ans; if(mat[i][j] == prev) { return recurseCol(mat, i + 1, j, prev, ans + 1); } else return ans; } static int diff(char [] a, char [] b) { int sum = 0; for(int i = 0; i < a.length; i++) { sum += Math.abs((int)a[i] - (int)b[i]); } return sum; } static int [] nextGreaterBack(char [] s) { Stack<Integer> stack = new Stack<>(); int [] toReturn = new int[s.length]; for(int i = 0; i < s.length; i++) { if(!stack.isEmpty() && s[stack.peek()] >= s[i]) { stack.pop(); } if(stack.isEmpty()) { stack.push(i); toReturn[i] = -1; }else { toReturn[i] = stack.peek(); stack.push(i); } } return toReturn; } static int [] nextGreaterFront(char [] s) { Stack<Integer> stack = new Stack<>(); int [] toReturn = new int[s.length]; for(int i = s.length - 1; i >= 0; i--) { if(!stack.isEmpty() && s[stack.peek()] >= s[i]) { stack.pop(); } if(stack.isEmpty()) { stack.push(i); toReturn[i] = -1; }else { toReturn[i] = stack.peek(); stack.push(i); } } return toReturn; } static int lps(String s) { int max = 0; int [] lps = new int[s.length()]; lps[0] = 0; int j = 0; for(int i = 1; i < lps.length; i++) { j = lps[i - 1]; while(j > 0 && s.charAt(i) != s.charAt(j)) j = lps[j - 1]; if(s.charAt(i) == s.charAt(j)) { lps[i] = j + 1; max = Math.max(max, lps[i]); } } return max; } static int [][] vectors = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; static String dir = "DRUL"; static boolean check(int i, int j, boolean [][] visited) { if(i >= visited.length || j >= visited[0].length) return false; if(i < 0 || j < 0) return false; return true; } static int count = 0; static void recurse(boolean [][] visited, int i, int j, String p, int z, Integer [][][] dp) throws IOException { if(p.length() == z) { //System.out.println(i + " " + j); if(i == visited.length - 1 && j == 0) { count++; return; }else return; } if(i == visited.length - 1 && j == 0) { return; } if(visited[i][j]) return; visited[i][j] = true; for(int k = 0; k < vectors.length; k++) { int x = i + vectors[k][0], y = j + vectors[k][1]; char toAdd = dir.charAt(k); if(p.charAt(z) == '?' || p.charAt(z) == toAdd) { if(check(x, y, visited) && !visited[x][y]) { //recurse(visited, x, y, p, z + 1); } } } visited[i][j] = false; } static boolean check1(int k, long [] arr, long minSum) { int subArrays = 0; long sum = 0; for(int i = 0; i < arr.length; i++) { if(sum + arr[i] > minSum) { sum = arr[i]; subArrays++; if(arr[i] > minSum) return false; }else sum += arr[i]; } if(subArrays < k) return true; return false; } static long bin(long start, long end, long [] arr, int k) { long ans = 0; while(start < end) { long mid = (start + end) / 2; //System.out.println(mid); if(check1(k, arr, mid)) { ans = mid; end = mid; }else { start = mid + 1; } } return ans; } static void selectionSort(long arr[], long [] arr1, ArrayList<ArrayList<Integer>> ans) { int n = arr.length; for (int i = 0; i < n-1; i++) { int min_idx = i; for (int j = i+1; j < n; j++) if (arr[j] < arr[min_idx]) min_idx = j; else if(arr[j] == arr[min_idx]) { if(arr1[j] < arr1[min_idx]) min_idx = j; } if(i == min_idx) { continue; } ArrayList<Integer> p = new ArrayList<Integer>(); p.add(min_idx + 1); p.add(i + 1); ans.add(new ArrayList<Integer>(p)); swap(i, min_idx, arr); swap(i, min_idx, arr1); } } static void solve1() throws IOException { /* * for(int i = 0; i < dp.length; i++) { for(int j = 0; j < dp[0].length; j++) { * System.out.print(dp[i][j] + " "); } System.out.println(); } */ //print(recurse1(n[0], n[1], new Long[Math.max(n[0], n[1]) + 1][Math.max(n[0], n[1]) + 1])); } static int saved = Integer.MAX_VALUE; static String ans1 = ""; public static boolean isValid(int x, int y, String [] mat) { if(x >= mat.length || x < 0) return false; if(y >= mat[0].length() || y < 0) return false; return true; } static boolean recurse1(int i, int j, int [][] arr, int sum, Boolean [][][] dp) { if(i == arr.length - 1 && j == arr[0].length - 1) { if(sum + arr[i][j] == 0) return true; return false; } if(i == arr.length) return false; else if(j == arr[0].length) return false; if(sum < 0 && dp[i][j][arr.length + arr[0].length + 1 + sum] != null) return dp[i][j][arr.length + arr[0].length + 1 + sum]; if(sum > 0 && dp[i][j][sum] != null) return dp[i][j][sum]; if(sum < 0) dp[i][j][arr.length + arr[0].length + 1 + sum] = recurse1(i + 1, j, arr, sum + arr[i][j], dp) || recurse1(i, j + 1, arr, sum + arr[i][j], dp); return dp[i][j][sum] = recurse1(i + 1, j, arr, sum + arr[i][j], dp) || recurse1(i, j + 1, arr, sum + arr[i][j], dp); } static void bfs(String [] grid, int [][] distance, Queue<pair> q) { boolean [][] visited = new boolean[grid.length][grid[0].length()]; while(!q.isEmpty()) { pair head = q.poll(); for(int i = 0; i < vectors.length; i++) { int x = (int)head.a + vectors[i][0], y = (int)head.b + vectors[i][1]; if(isValid(x, y, grid) && grid[x].charAt(y) != '#' && distance[x][y] == -1) { distance[x][y] = (int)head.d + 1; pair toAdd = new pair(x, y); toAdd.d = distance[x][y]; q.add(toAdd); } } } } static pair bfs1(String [] grid, int [][] distance, Queue<pair> q, char [][] path) { boolean [][] visited = new boolean[grid.length][grid[0].length()]; while(!q.isEmpty()) { pair head = q.poll(); if(head.a == grid.length - 1 || head.b == grid[0].length() - 1 || head.a == 0 || head.b == 0) { return head; } for(int i = 0; i < vectors.length; i++) { int x = (int)head.a + vectors[i][0], y = (int)head.b + vectors[i][1]; int dist = (int)head.d + 1; if(isValid(x, y, grid) && grid[x].charAt(y) != '#' && !visited[x][y] && (dist < distance[x][y] || distance[x][y] == -1)) { path[x][y] = dir.charAt(i); visited[x][y] = true; pair toAdd = new pair(x, y); toAdd.d = dist; q.add(toAdd); } } } return null; } static int search(ArrayList<Integer> arr, int target) { int start = 0; int end = arr.size() - 1; while(start <= end) { int mid = (start + end) / 2; if(arr.get(mid) < target) start = mid + 1; else if(arr.get(mid) > target) end = mid - 1; else if(arr.get(mid) == target) { if(mid + 1 < arr.size() && arr.get(mid + 1) == target) { start = mid + 1; }else return start; } } return start; } static int end = -1; static boolean dfs1(int src, ArrayList<ArrayList<Integer>> graph, boolean [] visited, int k, HashMap<Integer, Integer> map) { visited[src] = true; if(k == 0) return true; //System.out.println(src + " " + k); for(int x : graph.get(src)) { if(!visited[x]) { if(map.containsKey(x)) { end = x; if(dfs1(x, graph, visited, k - 1, map)) return true; }else { if(dfs1(x, graph, visited, k, map)) return true; } } } return false; } public static void recurse3(ArrayList<Character> arr, int index, String s, int max, ArrayList<String> toReturn) { if(s.length() == max) { toReturn.add(s); return; } if(index == arr.size()) return; recurse3(arr, index + 1, s + arr.get(index), max, toReturn); recurse3(arr, index + 1, s, max, toReturn); } static void solve() throws IOException { int n = nextInt(); int [] arr = inputIntArr(); if(arr[0] == 1) { print("YES\n"); return; } int count = 0; int same = 1; for(int i = 1; i < arr.length; i++) { if(arr[i] % arr[0] == 0) continue; else { print("NO\n"); return; } } print("YES\n"); } public static void main(String[] args) throws IOException { // TODO Auto-generated method stub int t = Integer.parseInt(sc.readLine()); for(int i = 0; i < t; i++) solve(); } } class TreeNode { int val; TreeNode next; public TreeNode(int x, TreeNode y) { this.val = x; this.next = y; } } /* 1 10 6 10 7 9 11 99 45 20 88 31 */
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
9e22be138c79b2e7b3ea053a8e26665b
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.max; import static java.lang.Math.min; import static java.lang.Math.abs; import static java.lang.System.out; import java.util.*; import java.io.*; import java.math.*; public class CF808A { public static void main(String[] args) { FastScanner sc=new FastScanner(); int T=sc.nextInt(); StringBuilder sb = new StringBuilder(); for (int tt=0; tt<T; tt++) { boolean ans = true; int n = sc.nextInt(); int arr[] = sc.readArray(n); for(int i=1;i<n;i++) { if(arr[i]%arr[0]!=0) { ans = false; break; } } if(ans) sb.append("YES"); else sb.append("NO"); sb.append('\n'); } out.print(sb); out.flush(); } static void sort(int[] a) { ArrayList<Integer> l=new ArrayList<>(); for (int i:a) l.add(i); Collections.sort(l); for (int i=0; i<a.length; i++) a[i]=l.get(i); } static class FastScanner { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st=new StringTokenizer(""); String next() { while (!st.hasMoreTokens()) try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } int[] readArray(int n) { int[] a=new int[n]; for (int i=0; i<n; i++) a[i]=nextInt(); return a; } long nextLong() { return Long.parseLong(next()); } } }
Java
["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
77379865cee86e56de172a296e83d6ed
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 Practise_problems; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Arrays; import java.util.Random; import java.util.StringTokenizer; public class DifferenceOperations { public static void main(String []args) { FastScanner fs=new FastScanner(); int t=fs.nextInt(); while(t-->0) { int n=fs.nextInt(); int []arr=new int[n]; arr=fs.readArray(n); int gcd=arr[0]; //System.out.println(calc_gcd(3,4)); if(arr[0]==1) { System.out.println("YES"); continue; } for(int i=1;i<n;i++) { gcd=calc_gcd(gcd,arr[i]); } if(gcd==arr[0]) System.out.println("YES"); else System.out.println("NO"); } } public static int calc_gcd(int a,int b) { if(b==0) return a; else return calc_gcd(b,a%b); } static final Random random=new Random(); static void ruffleSort(int[] a) { int n=a.length;//shuffle, then sort for (int i=0; i<n; i++) { int oi=random.nextInt(n), temp=a[oi]; a[oi]=a[i]; a[i]=temp; } Arrays.sort(a); } static class FastScanner { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st=new StringTokenizer(""); String next() { while (!st.hasMoreTokens()) try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } int[] readArray(int n) { int[] a=new int[n]; for (int i=0; i<n; i++) a[i]=nextInt(); return a; } long nextLong() { return Long.parseLong(next()); } } }
Java
["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
97b882448be98289c08676dcffd169c9
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.IOException; import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.StringTokenizer; //import java.util.Arrays; //import java.util.Collections; //import java.util.ArrayList; public class CodeForces { public static void main(String args[])throws IOException { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); int t=Integer.valueOf(br.readLine()); while(t-->0) { int n=Integer.valueOf(br.readLine()); StringTokenizer st=new StringTokenizer(br.readLine()); long arr[]=new long[n]; for(int i=0;i<n;i++) arr[i]=Long.valueOf(st.nextToken()); //System.out.println((arr[1]%arr[0]==0)?"YES":"NO"); boolean flag=true; for(int i=1;i<n;i++) { if(arr[i]%arr[0]!=0) flag=false; } System.out.println(flag?"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 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
04ca38b4319e652bcdf1338da18966d1
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 practice { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); PrintWriter pw = new PrintWriter(new BufferedWriter(new PrintWriter(System.out))); StringBuilder sb = new StringBuilder(); int t = Integer.parseInt(br.readLine()); while (t --> 0) { int n = Integer.parseInt(br.readLine()); StringTokenizer st = new StringTokenizer(br.readLine()); int arr[] = new int[n]; for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(st.nextToken()); int num = arr[0]; boolean flag = true; for (int i = 1; i < n && flag; i++) { if (arr[i] % num != 0) flag = false; } sb.append(flag ? "Yes" : "No").append("\n"); } pw.println(sb.toString().trim()); 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 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
7df0c873c55daf87876612df2a0a09b2
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 { static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } public static void main(String[] args) { Scanner s = new Scanner(System.in); int t = s.nextInt(); while (t-- > 0) { int n = s.nextInt(); int arr[] = new int[n]; int gcd = 0; for (int i = 0; i < n; i++) { arr[i]=s.nextInt(); gcd = gcd(gcd, arr[i]); } if(gcd==arr[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 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
b17d34dba2e2dce5f06874d5cad05ff5
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 Solution { static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } static void sort(int a[]){ // int -> long ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long for(int i=0;i<a.length;i++) arr.add(a[i]); Collections.sort(arr); for(int i=0;i<a.length;i++) a[i]=arr.get(i); } private static long gcd(long a, long b){ if(b==0)return a; return gcd(b,a%b); } private static long pow(long x,long y){ if(y==0)return 1; long temp = pow(x, y/2); if(y%2==1) return x*temp*temp; else return temp*temp; } static int log(long n){ if(n<=1)return 0; long temp = n; int res = 0; while(n>1){ res++; n/=2; } return (1<<res)==temp?res:res+1; } static int mod = (int)1e9+7; static int INF = Integer.MAX_VALUE; static PrintWriter out; static FastReader sc ; public static void main(String[] args) throws IOException { sc = new FastReader(); out = new PrintWriter(System.out); // primes(); // ================================ // int test = sc.nextInt(); while (test-- > 0) { int n=sc.nextInt(); boolean flag = false; int k = sc.nextInt(); for(int i=0;i<n-1;i++){ int a = sc.nextInt(); if(a%k!=0){ flag = true; } } out.println( flag?"NO":"YES"); } out.flush(); } public static void solver() { } }
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
93c186dc0f9482d0d7df9698c78f1141
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 Practice { static boolean multipleTC = true; final static int mod = 1000000007; final static int mod2 = 998244353; final double E = 2.7182818284590452354; final double PI = 3.14159265358979323846; int MAX = 200005; int N = 40000; void pre() throws Exception { // dp = new int[N + 1]; // dp[0] = 1; // for (int a = 1; a <= N; a++) // if (reverse(a, 0) == a) { // for (int n = 0; n + a <= N; n++) { // dp[n + a] = (dp[n + a] + dp[n]) % mod; // } // } } int reverse(int a, int b) { return a == 0 ? b : reverse(a / 10, b * 10 + a % 10); } // All the best void solve(int TC) throws Exception { int n = ni(), arr[] = readArr(n); for (int i = 1; i < n; i++) { if(arr[i]%arr[0]!=0) { pn("NO"); return; } } pn("YES"); } double dist(int x1, int y1, int x2, int y2) { double a = x1 - x2, b = y1 - y2; return Math.sqrt((a * a) + (b * b)); } int[] readArr(int n) throws Exception { int arr[] = new int[n]; for (int i = 0; i < n; i++) { arr[i] = ni(); } return arr; } void sort(int arr[], int left, int right) { ArrayList<Integer> list = new ArrayList<>(); for (int i = left; i <= right; i++) list.add(arr[i]); Collections.sort(list); for (int i = left; i <= right; i++) arr[i] = list.get(i - left); } void sort(int arr[]) { ArrayList<Integer> list = new ArrayList<>(); for (int i = 0; i < arr.length; i++) list.add(arr[i]); Collections.sort(list); for (int i = 0; i < arr.length; i++) arr[i] = list.get(i); } public long max(long... arr) { long max = arr[0]; for (long itr : arr) max = Math.max(max, itr); return max; } public int max(int... arr) { int max = arr[0]; for (int itr : arr) max = Math.max(max, itr); return max; } public long min(long... arr) { long min = arr[0]; for (long itr : arr) min = Math.min(min, itr); return min; } public int min(int... arr) { int min = arr[0]; for (int itr : arr) min = Math.min(min, itr); return min; } public long sum(long... arr) { long sum = 0; for (long itr : arr) sum += itr; return sum; } public long sum(int... arr) { long sum = 0; for (int itr : arr) sum += itr; return sum; } String bin(long n) { return Long.toBinaryString(n); } String bin(int n) { return Integer.toBinaryString(n); } static int bitCount(int x) { return x == 0 ? 0 : (1 + bitCount(x & (x - 1))); } static void dbg(Object... o) { System.err.println(Arrays.deepToString(o)); } int bit(long n) { return (n == 0) ? 0 : (1 + bit(n & (n - 1))); } int abs(int a) { return (a < 0) ? -a : a; } long abs(long a) { return (a < 0) ? -a : a; } void p(Object o) { out.print(o); } void pn(Object o) { out.println(o); } void pni(Object o) { out.println(o); out.flush(); } void pn(int[] arr) { int n = arr.length; StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { sb.append(arr[i] + " "); } pn(sb); } void pn(long[] arr) { int n = arr.length; StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { sb.append(arr[i] + " "); } pn(sb); } String n() throws Exception { return in.next(); } String nln() throws Exception { return in.nextLine(); } int ni() throws Exception { return Integer.parseInt(in.next()); } long nl() throws Exception { return Long.parseLong(in.next()); } double nd() throws Exception { return Double.parseDouble(in.next()); } public static void main(String[] args) throws Exception { new Practice().run(); } FastReader in; PrintWriter out; void run() throws Exception { in = new FastReader(); out = new PrintWriter(System.out); int T = (multipleTC) ? ni() : 1; pre(); for (int t = 1; t <= T; t++) solve(t); out.flush(); out.close(); } class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } public FastReader(String s) throws Exception { br = new BufferedReader(new FileReader(s)); } String next() throws Exception { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { throw new Exception(e.toString()); } } return st.nextToken(); } String nextLine() throws Exception { String str = ""; try { str = br.readLine(); } catch (IOException e) { throw new Exception(e.toString()); } return str; } } }
Java
["4\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
50ceb73325818f93bf6f13ff12c053d1
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 java.lang.*; public class Main { static Scanner sc = new Scanner(System.in); public static void main(String[] args) { int tc = sc.nextInt(); while(tc-->0) { int n = sc.nextInt(); int d = sc.nextInt(); boolean isPossible = true; while(n-->1) { int val = sc.nextInt(); if(val % d != 0) { isPossible = false; } } if(isPossible) 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
94bd3ff9d38dcc52e08f14769d4f8ed6
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 java.lang.*; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int numC = sc.nextInt(); for (int c = 0; c < numC; c++) { int n = sc.nextInt(); int[] arr = new int[n]; boolean win = true; for (int i = 0; i < n; i++) { arr[i] = sc.nextInt(); if (arr[i] % arr[0] != 0) win = false; } System.out.println(win ? "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 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
2816e404ed6c6e776bf589cd8b7aa1f6
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(); po: while(t-->0){ int n = sc.nextInt(); int[] arr = new int[n]; int zeros = 0; for(int i=0;i<n;i++){ arr[i] = sc.nextInt(); if(arr[i]==0) zeros++; } if(zeros==n){ System.out.println("Yes"); continue po; } if(arr[0]==0){ System.out.println("No"); continue po; } for(int i=1;i<n;i++){ if(arr[i]%arr[0]!=0){ System.out.println("NO"); continue po; } } System.out.println("Yes"); } // your code goes here } }
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
5c011b2157d625e4dd0b9bc34f4869f6
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 java.util.stream.IntStream; public class Codeforces extends PrintWriter { Codeforces() { super(System.out); } Scanner sc = new Scanner(System.in); public static void main(String[] $) { Codeforces o = new Codeforces(); o.main(); o.flush(); o.close(); } final int mod = 998244353; // static int maxRepeat(int arr[], int n) // { // Map<Integer,Integer> map = new HashMap<>(); // for(int v: arr){ // map.put(v,map.getOrDefault(v,0)+1); // } // int max=0; // for(Integer val: map.values()){ // max=Math.max(max,val); // } // if(max>2) // return getKey(map,max); // else // return -1; // } String solve(int n){ boolean flag = true; int f = sc.nextInt(); for(int i=2;i<=n;i++){ if(sc.nextInt()%f!=0) flag = false; } if(flag) return "YES"; return "NO"; } void main() { int t = sc.nextInt(); while (t--> 0) { int n = sc.nextInt(); //Arrays.sort(arr); println(solve(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 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
644b403eb9480a56fffd5a27cb372617
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 in = new Scanner(System.in); int t = in.nextInt(); while(t--!=0) { int n = in.nextInt(); int arr[] = new int[n]; for(int i=0;i<n;i++) arr[i]=in.nextInt(); boolean flag = true; for(int i=1;i<n;i++) { if(arr[i]%arr[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 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
cf1f2d9a6b4fb72114f1f62f59a27b7f
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
// 20:18 ; 2'50" ; %%I02V2L4L // import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.*; public class A { static class pair implements Comparable<pair> { int a; int b; pair(int a, int b) { this.a = a; this.b = b; } public int compareTo(pair o) { return this.a - o.a; } } static ArrayList<Integer> g[]; static boolean vis[]; public static void main(String[] args) { FastScanner sc = new FastScanner(); int tt = sc.nextInt(); while (tt-- > 0) { int n = sc.nextInt(); int a[] = sc.readArray(n); int d = a[0]; boolean can = true; for (int i = 1; i < n; i++) { if (a[i] % d != 0 && a[i] % a[i - 1] != 0) { can = false; break; } } System.out.println(can ? "YES" : "NO"); } } static void sort(int[] a) { // ruffle int n = a.length; Random r = new Random(); for (int i = 0; i < a.length; i++) { int oi = r.nextInt(n), temp = a[i]; a[i] = a[oi]; a[oi] = temp; } // then sort Arrays.sort(a); } static 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()); } double nextDouble() { return Double.parseDouble(next()); } String str = ""; String nextLine() { 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[] readLongArray(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = nextLong(); return a; } long nextLong() { return Long.parseLong(next()); } } static long gcd(long n, long l) { if (l == 0) return n; return gcd(l, n % l); } static void sieveOfEratosthenes(int n, ArrayList<Integer> al) { boolean prime[] = new boolean[n + 1]; for (int i = 0; i <= n; i++) prime[i] = true; for (int p = 2; p * p <= n; p++) { if (prime[p] == true) { for (int i = p * p; i <= n; i += p) prime[i] = false; } } for (int i = 2; i <= n; i++) { if (prime[i] == true) al.get(i); } } static final int mod = 1_000_000_000 + 7; static final int max_val = 2147483647; static final int min_val = max_val + 1; 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 long mul(long a, long b) { return a * b % mod; } static int nCr(int n, int r) { return fact(n) / (fact(r) * fact(n - r)); } static int fact(int n) { int res = 1; for (int i = 2; i <= n; i++) res = res * i; return res; } // a -> z == 97 -> 122 // String.format("%.9f", ans) ,--> to get upto 9 decimal places , (ans is // double) }
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
a43d5b07b60a16029e59d0122f85cd7a
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; // static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out)); public static void main (String[] args) throws java.lang.Exception { FastReader sc =new FastReader(); int t=sc.nextInt(); // int t=1; outer : while(t-->0) { int n = sc.nextInt(); int arr[] = sc.readArray(n); for(int i=1;i<n;i++) { if(arr[i] % arr[0] != 0) { System.out.println("NO"); continue outer; } } 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; } } static class FenwickTree { //Binary Indexed Tree //1 indexed public int[] tree; public int size; public FenwickTree(int size) { this.size = size; tree = new int[size+5]; } public void add(int i, int v) { while(i <= size) { tree[i] += v; i += i&-i; } } public int find(int i) { int res = 0; while(i >= 1) { res += tree[i]; i -= i&-i; } return res; } public int find(int l, int r) { return find(r)-find(l-1); } } 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 static long mergeAndCount(int[] arr, int l, int m, int r) { // Left subarray int[] left = Arrays.copyOfRange(arr, l, m + 1); // Right subarray int[] right = Arrays.copyOfRange(arr, m + 1, r + 1); int i = 0, j = 0, k = l;long swaps = 0; while (i < left.length && j < right.length) { if (left[i] < right[j]) arr[k++] = left[i++]; else { arr[k++] = right[j++]; swaps += (m + 1) - (l + i); } } while (i < left.length) arr[k++] = left[i++]; while (j < right.length) arr[k++] = right[j++]; return swaps; } // Merge sort function private static long mergeSortAndCount(int[] arr, int l, int r) { // Keeps track of the inversion count at a // particular node of the recursion tree long count = 0; if (l < r) { int m = (l + r) / 2; // Total inversion count = left subarray count // + right subarray count + merge count // Left subarray count count += mergeSortAndCount(arr, l, m); // Right subarray count count += mergeSortAndCount(arr, m + 1, r); // Merge count count += mergeAndCount(arr, l, m, r); } return count; } static void my_sort(long[] arr) { ArrayList<Long> list = new ArrayList<>(); for(int i=0;i<arr.length;i++) { list.add(arr[i]); } Collections.sort(list); for(int i=0;i<arr.length;i++) { arr[i] = list.get(i); } } static void reverse_sorted(int[] arr) { ArrayList<Integer> list = new ArrayList<>(); for(int i=0;i<arr.length;i++) { list.add(arr[i]); } Collections.sort(list , Collections.reverseOrder()); for(int i=0;i<arr.length;i++) { arr[i] = list.get(i); } } static int LowerBound(int a[], int x) { // x is the target value or key int l=-1,r=a.length; while(l+1<r) { int m=(l+r)>>>1; if(a[m]>=x) r=m; else l=m; } return r; } static int UpperBound(ArrayList<Integer> list, int x) {// x is the key or target value int l=-1,r=list.size(); while(l+1<r) { int m=(l+r)>>>1; if(list.get(m)<=x) l=m; else r=m; } return l+1; } 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 class Queue_Pair implements Comparable<Queue_Pair> { int first , second; public Queue_Pair(int first, int second) { this.first=first; this.second=second; } public int compareTo(Queue_Pair o) { return Integer.compare(o.first, first); } } 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 long getSum(long n) { long 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(int n) { return (int)Math.floor(Math.log10(n) + 1); } } class Pair { int first; int second; Pair(int first , int second) { this.first = first; this.second = second; } }
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
da30b5478d1470da5146038ca1095893
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 static java.lang.Math.*; public class Main extends PrintWriter { Main() { super(System.out); } static boolean cases = true; // Solution void solve(int t) { int n = sc.nextInt(); int a[] = sc.readIntArray(n); for (int i = 1; i < n; i++) { if (a[i] % a[0] != 0) { System.out.println("NO"); return; } } System.out.println("YES"); } int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } void sort(int[] a) { ArrayList<Integer> l = new ArrayList<>(); for (int i : a) l.add(i); Collections.sort(l); for (int i = 0; i < a.length; i++) a[i] = l.get(i); } public static void main(String[] args) { Main obj = new Main(); int c = 1; for (int t = (cases ? sc.nextInt() : 0); t > 1; t--, c++) obj.solve(c); obj.solve(c); obj.flush(); } 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[] readIntArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt(); return a; } char[] readCharArray(int n) { char a[] = new char[n]; String s = sc.next(); for (int i = 0; i < n; i++) { a[i] = s.charAt(i); } 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()); } } final int ima = Integer.MAX_VALUE; final int imi = Integer.MIN_VALUE; final long lma = Long.MAX_VALUE; final long lmi = Long.MIN_VALUE; static final long mod = (long) 1e9 + 7; private static final FastScanner sc = new FastScanner(); private PrintWriter out = new PrintWriter(System.out); }
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
88bb3c57b58c2b9056631aaa7cc829e2
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 cf { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int testcase = scan.nextInt(); // int testcase = 1; for(int v = 0; v < testcase; v++){ int n = scan.nextInt(); int []arr = new int[n]; arr[0] = scan.nextInt(); boolean bool = true; for(int i = 1; i < n; i++){ arr[i] = scan.nextInt(); if(arr[i] % arr[0] != 0){ 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 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
ca2a2d9dd80622693e5e4e25f0460eff
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 DifferencesOp { 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 check=false; for(int i=1;i<n;i++){ if(arr[i]%arr[0]!=0){ check=true; break; } } if(check) 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
bcb1cc083fe10dc6cd450d87222a3a5d
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(); int m=0; for(int i=1;i<n;i++) { if(arr[i]%arr[0]!=0) m=1; } System.out.println(m==1?"NO":"YES"); } // your code goes here } }
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
55cf20bb723e4330f9c3a26d66e127b4
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.*; /* 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 a[]=new int[n]; for(int i=0;i<n;i++) a[i]=sc.nextInt(); int m=0; for(int i=1;i<n;i++){ if(a[i]%a[0]!=0||(a[i]==0&&i!=n-1)) m=1; } if(m==1) System.out.println("NO"); else System.out.println("YES"); } // your code goes here } }
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
fffc18a823423e74f3763bd65804ac6a
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 jin { public static int n; static Scanner source = new Scanner(System.in); static int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } public static void make(int n) { while (n > 0) { int x = source.nextInt(); int[] arr = new int[x]; boolean b = true; for(int i=0;i<x;i++){ arr[i] = source.nextInt(); if(i>0 && b && arr[i]%arr[0]!=0){ b = false; } } if(b){ System.out.println("yes"); } else{ System.out.println("no"); } n--; } } public static void main(String[] args) { n = source.nextInt(); make(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 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
abb42009cdfac1e38e96fc407c820e1b
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.ArrayList; import java.util.Collections; import java.util.StringTokenizer; public class A { static FastScanner fs=new FastScanner(); public static void main(String[] args) { int tc= fs.nextInt(); for (int i=0;i<tc;i++){ int n= fs.nextInt(); long[] arr= fs.readArray(n); boolean isPoss=true; for (int j=1;j<n;j++){ if (arr[j]%arr[0]!=0) { isPoss=false; break; } } System.out.println(isPoss?"YES":"NO"); } } static void sort(int[] a) { ArrayList<Integer> l=new ArrayList<>(); for (int i:a) l.add(i); Collections.sort(l); for (int i=0; i<a.length; i++) a[i]=l.get(i); } static class FastScanner { public 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[] readArray(int n) { long[] a=new long[n]; for (int i=0; i<n; i++) a[i]=nextLong(); return a; } long nextLong() { return Long.parseLong(next()); } } }
Java
["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
3bd61b32577e618698994bea8550153e
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[] strgs) { Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-- >0) { int n=sc.nextInt(); ArrayList<Long> list=new ArrayList<>(); //gcd for(int i=0;i<n;i++) { list.add(sc.nextLong()); } long gcdsum=list.get(0); for(int i=1;i<n;i++) { gcdsum=gcd(gcdsum,list.get(i)); } if(list.indexOf(gcdsum)!=-1 && list.indexOf(gcdsum)==0) { System.out.println("YES"); } else { System.out.println("NO"); } } } public static long gcd(long x,long y) { long a=(x>y)?x:y; long b=(x<y)?x:y; long r=b; while(a%b!=0) { r=a%b; a=b; b=r; } return r; } }
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
7643590caf370b9988649a7336b7d8c0
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.System.out; import static java.lang.Math.*; import java.util.*; import javax.naming.ContextNotEmptyException; import javax.swing.ImageIcon; import javax.swing.table.TableStringConverter; import java.io.*; import java.math.*; import java.rmi.ConnectIOException; public class Main { static FastReader sc; static long mod = ((long) 1e9) + 7; static Tree tree; static PrintWriter out, err; static int imax = Integer.MAX_VALUE; static int imin = Integer.MIN_VALUE; static boolean testcaseBool = false; static long lmax = Long.MAX_VALUE; static long lmin = Long.MIN_VALUE; static double dmax = Double.MAX_VALUE; static double dmin = Double.MIN_VALUE; static int max = 0; private static void solve() throws IOException{ int n=sc.nextInt(); int[] arr=readIntArray(n); boolean done=true; boolean equal=true; boolean z=true; boolean modequal=true; int prev=arr[1]%arr[0]; for(int i=1;i<n;i++){ if(arr[i]%arr[0]!=0){ done=false; } } if(done)print("yes"); else print("No"); } public static void main(String hi[]) throws IOException { initializeIO(); out = new PrintWriter(System.out); err = new PrintWriter(System.err); sc = new FastReader(); long startTimeProg = System.currentTimeMillis(); long endTimeProg; int testCase = 1; // Collections.sort(li); // debug(li+""); // debug(li+""); testCase = sc.nextInt(); while (testCase-- != 0) { solve(); } endTimeProg = System.currentTimeMillis(); debug("[finished : " + (endTimeProg - startTimeProg) + ".ms ]"); out.flush(); err.flush(); // System.out.println(String.format("%.9f", max)); } private static class Pair { int first = 0; int sec = 0; int[] arr; char ch; String s; Map<Integer, Integer> map; Pair(int first, int sec) { this.first = first; this.sec = sec; } Pair(int[] arr) { this.map = new HashMap<>(); for (int x : arr) this.map.put(x, map.getOrDefault(x, 0) + 1); this.arr = arr; } Pair(char ch, int first) { this.ch = ch; this.first = first; } Pair(String s, int first) { this.s = s; this.first = first; } } private static Set<Long> factors(long n) { Set<Long> res = new HashSet<>(); // res.add(n); for (long i = 1; i * i <= (n); i++) { if (n % i == 0) { res.add(i); if (n / i != i) { res.add(n / i); } } } return res; } // geometrics private static double areaOftriangle(double x1, double y1, double x2, double y2, double x3, double y3) { double[] mid_point = midOfaLine(x1, y1, x2, y2); // debug(Arrays.toString(mid_point)+" "+x1+" "+y1+" "+x2+" "+y2+" "+x3+" "+" // "+y3); double height = distanceBetweenPoints(mid_point[0], mid_point[1], x3, y3); double wight = distanceBetweenPoints(x1, y1, x2, y2); // debug(height+" "+wight); return (height * wight) / 2; } private static double distanceBetweenPoints(double x1, double y1, double x2, double y2) { double x = x2 - x1; double y = y2 - y1; return (Math.pow(x, 2) + Math.pow(y, 2)); } private static double[] midOfaLine(double x1, double y1, double x2, double y2) { double[] mid = new double[2]; mid[0] = (x1 + x2) / 2; mid[1] = (y1 + y2) / 2; return mid; } /* Function to calculate x raised to the power y in O(logn) */ static long power(long x, long y) { long temp; if (y == 0) return 1l; temp = power(x, y / 2); if (y % 2 == 0) return (temp * temp); else return (x * temp * temp); } private static StringBuilder reverseString(String s) { StringBuilder sb = new StringBuilder(s); int l = 0, r = sb.length() - 1; while (l <= r) { char ch = sb.charAt(l); sb.setCharAt(l, sb.charAt(r)); sb.setCharAt(r, ch); l++; r--; } return sb; } private static void swap(List<Integer> li, int i, int j) { int t = li.get(i); li.set(i, li.get(j)); li.set(j, t); } private static void swap(int[] arr, int i, int j) { int t = arr[i]; arr[i] = arr[j]; arr[j] = t; } static int lcm(int a, int b) { return (a / gcd(a, b)) * b; } static long lcm(long a, long b) { return (a / gcd(a, b)) * b; } private static boolean isPallindrome(String s, int l, int r) { while (l < r) { if (s.charAt(l) != s.charAt(r)) return false; l++; r--; } return true; } private static StringBuilder removeLeadingZero(StringBuilder sb) { int i = 0; while (i < sb.length() && sb.charAt(i) == '0') i++; // debug("remove "+i); if (i == sb.length()) return new StringBuilder(); return new StringBuilder(sb.substring(i, sb.length())); } private static void print(int[][] arr) { int n = arr.length; int m = arr[0].length; int i, j; for (i = 0; i < n; i++) { for (j = 0; j < m; j++) { out.print(arr[i][j] + " "); } out.println(); } } private static StringBuilder removeEndingZero(StringBuilder sb) { int i = sb.length() - 1; while (i >= 0 && sb.charAt(i) == '0') i--; if (i < 0) return new StringBuilder(); return new StringBuilder(sb.substring(0, i + 1)); } private static void debug(int[][] arr) { for (int i = 0; i < arr.length; i++) { err.println(Arrays.toString(arr[i])); } } private static void debug(long[][] arr) { for (int i = 0; i < arr.length; i++) { err.println(Arrays.toString(arr[i])); } } private static void debug(List<int[]> arr) { for (int[] a : arr) { err.println(Arrays.toString(a)); } } private static void debug(float[][] arr) { for (int i = 0; i < arr.length; i++) { err.println(Arrays.toString(arr[i])); } } private static void debug(double[][] arr) { for (int i = 0; i < arr.length; i++) { err.println(Arrays.toString(arr[i])); } } private static void debug(boolean[][] arr) { for (int i = 0; i < arr.length; i++) { err.println(Arrays.toString(arr[i])); } } // private static void print() throws IOException { // out.println(s); // } private static void debug(String s) throws IOException { err.println(s); } private static void print(double s) throws IOException { out.println(s); } private static void print(float s) throws IOException { out.println(s); } private static void print(long s) throws IOException { out.println(s); } private static void print(int s) throws IOException { out.println(s); } private static void debug(double s) throws IOException { err.println(s); } private static void debug(float s) throws IOException { err.println(s); } private static void debug(long s) { err.println(s); } private static void debug(int s) { err.println(s); } private 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 (int i = 3; i * i <= n; i += 2) { if (n % i == 0) return false; } return true; } private static List<List<Integer>> readUndirectedGraph(int n) { List<List<Integer>> graph = new ArrayList<>(); for (int i = 0; i <= n; i++) { graph.add(new ArrayList<>()); } for (int i = 0; i < n; i++) { int x = sc.nextInt(); int y = sc.nextInt(); graph.get(x).add(y); graph.get(y).add(x); } return graph; } private static List<List<Integer>> readUndirectedGraph(int[][] intervals, int n) { List<List<Integer>> graph = new ArrayList<>(); for (int i = 0; i <= n; i++) { graph.add(new ArrayList<>()); } for (int i = 0; i < intervals.length; i++) { int x = intervals[i][0]; int y = intervals[i][1]; graph.get(x).add(y); graph.get(y).add(x); } return graph; } private static List<List<Integer>> readDirectedGraph(int[][] intervals, int n) { List<List<Integer>> graph = new ArrayList<>(); for (int i = 0; i <= n; i++) { graph.add(new ArrayList<>()); } for (int i = 0; i < intervals.length; i++) { int x = intervals[i][0]; int y = intervals[i][1]; graph.get(x).add(y); // graph.get(y).add(x); } return graph; } private static List<List<Integer>> readDirectedGraph(int n) { List<List<Integer>> graph = new ArrayList<>(); for (int i = 0; i <= n; i++) { graph.add(new ArrayList<>()); } for (int i = 0; i < n; i++) { int x = sc.nextInt(); int y = sc.nextInt(); graph.get(x).add(y); // graph.get(y).add(x); } return graph; } static String[] readStringArray(int n) { String[] arr = new String[n]; for (int i = 0; i < n; i++) { arr[i] = sc.next(); } return arr; } private static Map<Character, Integer> freq(String s) { Map<Character, Integer> map = new HashMap<>(); for (char c : s.toCharArray()) { map.put(c, map.getOrDefault(c, 0) + 1); } return map; } private static Map<Long, Integer> freq(long[] arr) { Map<Long, Integer> map = new HashMap<>(); for (long x : arr) { map.put(x, map.getOrDefault(x, 0) + 1); } return map; } private static Map<Integer, Integer> freq(int[] arr) { Map<Integer, Integer> map = new HashMap<>(); for (int x : arr) { map.put(x, map.getOrDefault(x, 0) + 1); } return map; } static boolean[] sieveOfEratosthenes(long n) { boolean prime[] = new boolean[(int) n + 1]; for (int i = 2; i <= n; i++) prime[i] = true; for (int p = 2; p * p <= n; p++) { // If prime[p] is not changed, then it is a // prime if (prime[p] == true) { // Update all multiples of p for (int i = p * p; i <= n; i += p) prime[i] = false; } } return prime; } static int[] sieveOfEratosthenesInt(long n) { boolean prime[] = new boolean[(int) n + 1]; Set<Integer> li = new HashSet<>(); for (int i = 2; i <= n; i++) { prime[i] = true; } for (int p = 2; p * p <= n; p++) { if (prime[p] == true) { for (int i = p * p; i <= n; i += p) { prime[i] = false; } } } for (int i = 0; i <= n; i++) { if (prime[i]) li.add(i); } int[] arr = new int[li.size()]; int i = 0; for (int x : li) { arr[i++] = x; } return arr; } static boolean isMemberAC(int a, int d, int x) { // If difference is 0, then x must // be same as a. if (d == 0) return (x == a); // Else difference between x and a // must be divisible by d. return ((x - a) % d == 0 && (x - a) / d >= 0); } static boolean isMemberAC(long a, long d, long x) { // If difference is 0, then x must // be same as a. if (d == 0) return (x == a); // Else difference between x and a // must be divisible by d. return ((x - a) % d == 0 && (x - a) / d >= 0); } private static void sort(int[] arr) { int n = arr.length; List<Integer> li = new ArrayList<>(); for (int x : arr) { li.add(x); } Collections.sort(li); for (int i = 0; i < n; i++) { arr[i] = li.get(i); } } private static void sortReverse(int[] arr) { int n = arr.length; List<Integer> li = new ArrayList<>(); for (int x : arr) { li.add(x); } Collections.sort(li, Collections.reverseOrder()); for (int i = 0; i < n; i++) { arr[i] = li.get(i); } } private static void sort(double[] arr) { int n = arr.length; List<Double> li = new ArrayList<>(); for (double x : arr) { li.add(x); } Collections.sort(li); for (int i = 0; i < n; i++) { arr[i] = li.get(i); } } private static void sortReverse(double[] arr) { int n = arr.length; List<Double> li = new ArrayList<>(); for (double x : arr) { li.add(x); } Collections.sort(li, Collections.reverseOrder()); for (int i = 0; i < n; i++) { arr[i] = li.get(i); } } private static void sortReverse(long[] arr) { int n = arr.length; List<Long> li = new ArrayList<>(); for (long x : arr) { li.add(x); } Collections.sort(li, Collections.reverseOrder()); for (int i = 0; i < n; i++) { arr[i] = li.get(i); } } private static void sort(long[] arr) { int n = arr.length; List<Long> li = new ArrayList<>(); for (long x : arr) { li.add(x); } Collections.sort(li); for (int i = 0; i < n; i++) { arr[i] = li.get(i); } } private static long sum(int[] arr) { long sum = 0; for (int x : arr) { sum += x; } return sum; } private static long sum(long[] arr) { long sum = 0; for (long x : arr) { sum += x; } return sum; } private static void initializeIO() { try { System.setIn(new FileInputStream("input.txt")); System.setOut(new PrintStream(new FileOutputStream("output.txt"))); System.setErr(new PrintStream(new FileOutputStream("error.txt"))); } catch (Exception e) { System.err.println(e.getMessage()); } } private static int maxOfArray(int[] arr) { int max = Integer.MIN_VALUE; for (int x : arr) { max = Math.max(max, x); } return max; } private static long maxOfArray(long[] arr) { long max = Long.MIN_VALUE; for (long x : arr) { max = Math.max(max, x); } return max; } private static int[][] readIntIntervals(int n, int m) { int[][] arr = new int[n][m]; for (int j = 0; j < n; j++) { for (int i = 0; i < m; i++) { arr[j][i] = sc.nextInt(); } } return arr; } private static long gcd(long a, long b) { if (b == 0) return a; return gcd(b, a % b); } private static int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } private static int[] readIntArray(int n) { int[] arr = new int[n]; for (int i = 0; i < n; i++) { arr[i] = sc.nextInt(); } return arr; } private static double[] readDoubleArray(int n) { double[] arr = new double[n]; for (int i = 0; i < n; i++) { arr[i] = sc.nextDouble(); } return arr; } private static long[] readLongArray(int n) { long[] arr = new long[n]; for (int i = 0; i < n; i++) { arr[i] = sc.nextLong(); } return arr; } private static void print(int[] arr) throws IOException { out.println(Arrays.toString(arr)); } private static void print(long[] arr) throws IOException { out.println(Arrays.toString(arr)); } private static void print(String[] arr) throws IOException { out.println(Arrays.toString(arr)); } private static void print(double[] arr) throws IOException { out.println(Arrays.toString(arr)); } private static void debug(String[] arr) { err.println(Arrays.toString(arr)); } private static void debug(Boolean[][] arr) { for (int i = 0; i < arr.length; i++) err.println(Arrays.toString(arr[i])); } private static void debug(int[] arr) { err.println(Arrays.toString(arr)); } private static void debug(long[] arr) { err.println(Arrays.toString(arr)); } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine().trim(); } catch (Exception e) { e.printStackTrace(); } return str; } } static class Tree { List<List<Integer>> adj = new ArrayList<>(); int parent = 1; int[] arr; int n; Map<Integer, Integer> node_parent; List<Integer> leaf; public Tree(int n) { this.n = n; leaf = new ArrayList<>(); node_parent = new HashMap<>(); arr = new int[n + 2]; for (int i = 0; i <= n + 1; i++) { adj.add(new ArrayList<>()); } for (int i = 0; i < n; i++) { arr[i] = sc.nextInt(); node_parent.put(i + 1, arr[i]); if ((i + 1) == arr[i]) parent = arr[i]; else adj.get(arr[i]).add(i + 1); } } private List<List<Integer>> getTree() { return adj; } private void formLeaf(int v) { boolean leaf = true; for (int x : adj.get(v)) { formLeaf(x); leaf = false; } if (leaf) this.leaf.add(v); } private List<Integer> getLeaf() { return leaf; } private Map<Integer, Integer> getParents() { return node_parent; } int[] getArr() { return arr; } } static class Dsu { int[] parent, size; Dsu(int n) { parent = new int[n + 1]; size = new int[n + 1]; for (int i = 0; i <= n; i++) { parent[i] = i; size[i] = 1; } } private int findParent(int u) { if (parent[u] == u) return u; return parent[u] = findParent(parent[u]); } private boolean union(int u, int v) { // System.out.println("uf "+u+" "+v); int pu = findParent(u); // System.out.println("uf2 "+pu+" "+v); int pv = findParent(v); // System.out.println("uf3 " + u + " " + pv); if (pu == pv) return false; if (size[pu] <= size[pv]) { parent[pu] = pv; size[pv] += size[pu]; } else { parent[pv] = pu; size[pu] += size[pv]; } return true; } } // =================================================================================== public static int log2(int N) { return (int) (Math.log(N) / Math.log(2)); } private static long fact(long n) { if (n <= 2) return n; return n * fact(n - 1); } private static void print(String s) throws IOException { out.println(s); } public static boolean isNumeric(String strNum) { if (strNum == null) { return false; } try { double d = Double.parseDouble(strNum); } catch (NumberFormatException nfe) { return false; } return true; } private static boolean isSorted(List<Integer> li) { int n = li.size(); if (n <= 1) return true; for (int i = 0; i < n - 1; i++) { if (li.get(i) > li.get(i + 1)) return false; } return true; } private static boolean isSorted(int[] arr) { int n = arr.length; if (n <= 1) return true; for (int i = 0; i < n - 1; i++) { if (arr[i] > arr[i + 1]) return false; } return true; } private static boolean ispallindromeList(List<Integer> res) { int l = 0, r = res.size() - 1; while (l < r) { if (res.get(l) != res.get(r)) return false; l++; r--; } return true; } private static long ncr(long n, long r) { return fact(n) / (fact(r) * fact(n - r)); } private static void swap(char[] s, int i, int j) { char t = s[i]; s[i] = s[j]; s[j] = t; } static boolean isPowerOfTwo(long x) { return x != 0 && ((x & (x - 1)) == 0); } private static long gcd(long[] arr) { long ans = 0; for (long x : arr) { ans = gcd(x, ans); } return ans; } private static int gcd(int[] arr) { int ans = 0; for (int x : arr) { ans = gcd(x, ans); } return ans; } private static long power(long x, long y, long p) { long res = 1; x = x % p; if (x == 0) return 0; while (y > 0) { if ((y & 1) != 0) res = (res * x) % p; y = y >> 1; x = (x * x) % p; } 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
dbfb43b9c16e2e62e9831038eadaa446
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
/* _oo0oo_ o8888888o 88" . "88 (| -_- |) 0\ = /0 ___/`---'\___ .' \\| |// '. / \\||| : |||// \ / _||||| -:- |||||- \ | | \\\ - /// | | | \_| ''\---/'' |_/ | \ .-\__ '-' ___/-. / ___'. .' /--.--\ `. .'___ ."" '< `.___\_<|>_/___.' >' "". | | : `- \`.;`\ _ /`;.`/ - ` : | | \ \ `_. \_ __\ /__ _/ .-` / / =====`-.____`.___ \_____/___.-`___.-'===== `=---=' */ import java.util.*; import java.math.*; import java.io.*; import java.lang.Math.*; public class KickStart2020 { static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } float nextFloat() { return Float.parseFloat(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } } static long gcd(long a, long b) { if (b == 0) return a; return gcd(b, a % b); } static long lcm(long a, long b) { return a / gcd(a, b) * b; } public static class Pair implements Comparable<Pair> { public int index; public long value; public Pair(int index, long value) { this.index = index; this.value = value; } @Override public int compareTo(Pair other) { // multiplied to -1 as the author need descending sort order if(other.index < this.index) return 1; if(other.index > this.index) return -1; return 0; } @Override public String toString() { return this.index + " " + this.value; } } static boolean isPrime(long d) { if(d == 0) return true; if (d == 1) return false; for (int i = 2; i <= (long) Math.sqrt(d); i++) { if (d % i == 0) return false; } return true; } static boolean isPali(String n) { String s = n; int l = 0; int r = s.length() - 1; while(l < r) if(s.charAt(l++) != s.charAt(r--)) return false; return true; } static void decimalTob(long n, int k , int arr[], int i) { arr[i] += (n % k); n /= k; if(n > 0) { decimalTob(n, k, arr, i + 1); } } static long powermod(long x, long y, long mod) { if(y == 0) return 1; long value = powermod(x, y / 2, mod); if(y % 2 == 0) return (value * value) % mod; return (value * (value * x) % mod) % mod; } static long power(long x, long y) { if(y == 0) return 1; long value = power(x, y / 2); if(y % 2 == 0) return (value * value); return value * value * x; } static int bS(int l, int r, int find, Integer arr[]) { int ans = -1; while(l <= r) { int mid = (l + r) / 2; if(arr[mid] >= find) { ans = mid; r = mid - 1; } else l = mid + 1; } return ans; } static void build(int index, int l, int r, int seqtree[][], int i, int arr[][]) { if(l == r) { seqtree[i][index] = arr[i][l]; return; } int mid = (l + r) / 2; build(2 * index + 1, l, mid, seqtree, i, arr); build(2 * index + 2, mid + 1, r, seqtree, i, arr); seqtree[i][index] = Math.max(seqtree[i][2 * index + 1], seqtree[i][2 * index + 2]); } static int query(int index, int low, int high, int l, int r, int seqtree[][], int i) { if(high < l || low > r) return Integer.MIN_VALUE; if(l <= low && r >= high) { return seqtree[i][index]; } int mid = (low + high) / 2; int left = query(2 * index + 1, low , mid, l, r, seqtree, i); int right = query(2 * index + 2, mid + 1, high, l, r, seqtree,i); return Math.max(left, right); } static int brr[] = new int[100000 + 1]; static int dp(int i, int j, int arr[], int dp[][], int n) { if(i == n) return 0; if(j == 0) return 0; if(dp[i][j] != -1) return dp[i][j]; if(arr[i] <= j) { return dp[i][j] = 1 + dp(i + 1, j, arr, dp, n); } else { int max = 1 + dp(i + 1, j - 1, arr, dp, n); max = Math.max(max, dp(i + 2, j , arr, dp, n)); return dp[i][j] = max; } } public static void main(String[] args) throws Exception { FastReader sc = new FastReader(); PrintWriter out = new PrintWriter(System.out); int t = sc.nextInt(); outerloop: while(t-- > 0) { int n = sc.nextInt(); int arr[] = new int[n]; for(int i = 0; i < n; i++) arr[i] = sc.nextInt(); for(int i = 1; i < n; i++) { arr[i] = arr[i] % arr[i - 1]; if(arr[i] == 0) arr[i] = arr[i - 1]; } for(int i = 1; i < n; i++) { if(arr[i] % arr[i - 1] != 0) { out.println("NO"); continue outerloop; } } out.println("YES"); } 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
6014c187c285ef58bc26596d196d9815
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 java.math.BigInteger; import java.util.stream.Collectors; public class Main { InputStream is; PrintWriter out; String INPUT = ""; void run() throws Exception { boolean oj = false; if(System.getProperty("ONLINE_JUDGE") == null) oj = true; if(!oj) { is = System.in; out = new PrintWriter(System.out); solve(); out.flush(); out.close(); } else { PrintStream ps = new PrintStream(new File("output2.txt")); InputStream iss = new FileInputStream("input.txt"); System.setIn(iss); System.setOut(ps); is = System.in; out = new PrintWriter(System.out); solve(); out.flush(); out.close(); } } public static void main(String[] args) throws Exception { new Main().run(); } public byte[] inbuf = new byte[1 << 16]; public int lenbuf = 0, ptrbuf = 0; public int readByte() { if (lenbuf == -1) { throw new InputMismatchException(); } if (ptrbuf >= lenbuf) { ptrbuf = 0; try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); } if (lenbuf <= 0) return -1; } return inbuf[ptrbuf++]; } public boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); } public int skip() { int b; while ((b = readByte()) != -1 && isSpaceChar(b)) ; return b; } public double nd() { return Double.parseDouble(ns()); } public char nc() { return (char) skip(); } private String ns() { int b = skip(); StringBuilder sb = new StringBuilder(); while (!(isSpaceChar(b))) { sb.appendCodePoint(b); b = readByte(); } return sb.toString(); } private int ni() { return (int) nl(); } private long nl() { long num = 0; int b; boolean minus = false; while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ; if (b == '-') { minus = true; b = readByte(); } while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else { return minus ? -num : num; } b = readByte(); } } class Pair { int first; int second; Pair(int a, int b) { first = a; second = b; } } long[] nal(int n) { long[] arr = new long[n]; for (int i = 0; i < n; i++) { arr[i] = nl(); } return arr; } // Sorting an array (O(nlogn)):- int partition(long[] array, int begin, int end) { int pivot = end; int counter = begin; for (int i = begin; i < end; i++) { if (array[i] < array[pivot]) { long temp = array[counter]; array[counter] = array[i]; array[i] = temp; counter++; } } long temp = array[pivot]; array[pivot] = array[counter]; array[counter] = temp; return counter; } void sort(long[] array, int begin, int end) { if (end <= begin) return; int pivot = partition(array, begin, end); sort(array, begin, pivot-1); sort(array, pivot+1, end); } void printArr(long[] arr) { for(int i=0; i<arr.length; i++) { out.print(arr[i] + " "); } out.println(); } long[] arr; // ********************************************************************************** class Segment_Tree { long[] t = new long[4 * arr.length]; void build(int index, int left, int right) { if(left == right) { t[index] = arr[left]; return; } int mid = (left + right) / 2; build(index * 2, left, mid); build(index * 2 + 1, mid + 1, right); t[index] = t[2 * index] + t[2 * index + 1]; } // indexPos -> For what index we have to update the value of an array. void update(int index, int left, int right, int indexPos, int value) { if(indexPos < left || indexPos > right) return; if(left == right) { t[index] = value; arr[left] = value; return; } int mid = (left + right) / 2; update(2 * index, left, mid, indexPos, value); update(2 * index + 1, mid + 1, right, indexPos, value); t[index] = t[2 * index] + t[2 * index + 1]; } long query(int index, int left, int right, int leftRange, int rightRange) { if(left > rightRange || leftRange > right) return 0; if(leftRange <= left && right <= rightRange) { return t[index]; } int mid = (left + right) / 2; long a = query(index * 2, left, mid, leftRange, rightRange); long b = query(index * 2 + 1, mid + 1, right, leftRange, rightRange); return a + b; } } // ********************************************************************************** void solve() { int test_case = 1; test_case = ni(); while (test_case-- > 0) { go(); } } // WRITE CODE FROM HERE :- void go() { int n = ni(); arr = nal(n); long p = arr[0]; for(int i=0; i<n; i++) { if(arr[i] % p != 0) { out.println("NO"); return; } } 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
5c71ef566331c08de762c4019ab52eb3
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 cp { //start public static void main(String[] args) { int t = in(); while(t-->0) { int n = in(); var a = in(n); var ans = true; if(a[1]%a[0]!=0) ans=false; if(ans) { a[1]=a[0]; for(int i = 2; i < n; ++i) { if(a[i]%a[i-1]!=0) ans=false; a[i]=a[i-1]; } } System.out.println(ans?"YES":"NO"); } } //end public static int m=(int)1e9+7; public static Scanner sc = new Scanner(System.in); public static int in() { int n = sc.hasNextInt()?sc.nextInt():0; sc.nextLine(); return n; } public static int[] in(int n) { int[] a = new int[n]; for(int i=0;i<n;++i) a[i]=sc.hasNextInt()?sc.nextInt():0; sc.nextLine(); return a; } public static String in(boolean s) { return sc.hasNextLine()?sc.nextLine():""; } public static void out(String s) { System.out.println(s); } public static void out(int[] a) { for(int i=0;i<a.length;++i) System.out.print(a[i]+" "); System.out.print("\n"); } public static void out(long[] a) { for(int i=0;i<a.length;++i) System.out.print(a[i]+" "); System.out.print("\n"); } public static void out(int n) { System.out.println(n); } public static void out(long n) { System.out.println(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 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
a96ed60563b9f694b7a04d0f6ab4224f
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.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; import java.io.Writer; import java.io.OutputStreamWriter; import java.io.BufferedReader; 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[] arr = in.nextIntArray(n); for (int i = 1; i < n; i++) { if (arr[i] % arr[0] != 0) { out.println("NO"); return; } } out.println("YES"); } } 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(); } } static class InputReader { BufferedReader reader; 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[] nextIntArray(int n) { int[] array = new int[n]; for (int i = 0; i < n; ++i) array[i] = nextInt(); return array; } public int nextInt() { return Integer.parseInt(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 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
02ae9ab8f7436bc5db43d01a994c09b3
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.*; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /* Name of the class has to be "Main" only if the class is public. */ public class C { static short m_short = Short.MAX_VALUE; static Scanner sc= new Scanner(System.in); public static void main (String[] args) throws java.lang.Exception { // your code goes here int t=sc.nextInt(); while(t-->0) solve(); } static final long modulo = 1000000007; static int res=0; static int mod = (int)Math.pow(10,9)+7; // static int c[]; public static void solve() { // DataInputStream ds= new DataInputStream(System.in); // BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); // long supreme = 100000; int n=sc.nextInt(); int[] a = new int[n]; //String a =sc.next(); //String b=sc.next(); // char [][] a = new char[n][n]; // int[] b = new int[n+1]; // int[] b = new int[n+11]; HashMap<String,Integer> map = new HashMap<>(); int ans=-1; for(int i=0;i<10;i++) ans&=i; int rs=1; // List<Long> nn = new AanswertaskayList<>(n); for(int x=-1;x>10;x+=5) ans++; // int ptr=n-1; int z=3; int sum=0; for(int i=0;i<n;i++){ a[i]=sc.nextInt(); sum+=a[i]; } // Arrays.sort(a); int currops=0; boolean f=false; for(int i=1;i<n;i++){ if(a[i]%a[0]!=0){ f=true; break; } } System.out.println((f)?"NO":"YES"); } public static void swap(int[] a , int i ,int j){ int temp = a[i]; a[i]=a[j]; a[j]=temp; } public static int reverseN(int n){ int rev=0; while(n>0){ rev = rev*10+n%10; n/=10; } return rev; } public static boolean palin(char[] ch , int i ,int j){ while(i<=j){ if(ch[i]!=ch[j]) return false; } return true; } // public static void dfs(StringBuilder s,int[][] que,int i){ // if(i==que.length) return; // s.append(s.substring(que[i][0]-1,que[i][1])); // dfs(s,que,i+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
068e2fd99131f9e18d974ad21f5266c9
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 Tworivalstudents{ public static void main(String[] args){ Scanner sc= new Scanner(System.in); int t =sc.nextInt(); while(t>0){ int n=sc.nextInt(); int flag=0; int[] a= new int[n]; for(int i=0;i<n;i++) a[i]=sc.nextInt(); for(int i=1;i<n;i++){ if(a[i]%a[0]!=0){ flag=1; break; } } if(flag==1) System.out.println("NO"); else System.out.println("YES"); 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 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
732f0e394320a9df4e21720ff6486040
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 Codeforces { public static void main(String[] args) { try { Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while (t-- > 0) { int n = sc.nextInt(); int[] a = new int[n]; for(int i = 0;i<n;i++){ a[i] = sc.nextInt(); } int flag=0,init=a[0]; for(int i = 1;i<n;i++){ if(a[i]%init!=0){ flag=1; break; } } System.out.println(flag==1? "NO":"YES"); } } catch (Exception e) { System.out.println(e); } } }
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
1352d275e72b99faa88c9f5ca1e3439a
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 Codeforces { public static void main(String[] args) { try { Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while (t-- > 0) { int n = sc.nextInt(); int[] a = new int[n]; for(int i = 0;i<n;i++){ a[i] = sc.nextInt(); } int flag=0; for(int i = 1;i<n;i++){ if(a[i]%a[0]!=0){ flag=1; break; } } System.out.println(flag==0? "YES":"NO"); } } catch (Exception e) { System.out.println(e); } } }
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
446b80cc34f5ef05f6bbf3327bb2c967
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
/****************************************************************************** Welcome to GDB Online. GDB online is an online compiler and debugger tool for C, C++, Python, Java, PHP, Ruby, Perl, C#, OCaml, 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.*; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int i,j,k; for(j=0;j<n;j++) { int len = in.nextInt(); int[] arr = new int[len+1]; for(i=1;i<=len;i++) { arr[i] = in.nextInt(); } for(i=2;i<=len;i++) { if(arr[i]%arr[1]!=0) { System.out.println("NO"); break; } } if(i==len+1) 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
f0d9fe1e01985c42fae870a64e70559d
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.StringTokenizer; public class DifferenceOperations { static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader( new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { if (st.hasMoreTokens()) { str = st.nextToken("\n"); } else { str = br.readLine(); } } catch (IOException e) { e.printStackTrace(); } return str; } } public static void main(String[] args) { FastReader scanner = new FastReader(); int t = scanner.nextInt(); while(t-- > 0){ int n = scanner.nextInt(); int[] arr = new int[n]; boolean flag = true; for(int i = 0; i < n; i++){ arr[i] = scanner.nextInt(); } for(int i = 1; i < n; i++){ if(arr[i]%arr[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 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
1ce68241000f80445b6cbd66fe38389a
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 codeforces; // don't place package name! */ // algo_messiah23 , NIT RKL ... import java.util.*; import java.lang.*; import java.io.*; import java.math.*; import static java.lang.System.*; import java.util.stream.IntStream; import java.util.Map.Entry; /* Name of the class has to be "Main" only if the class is public. */ public class CodeForces { static PrintWriter out = new PrintWriter((System.out)); static FastReader in = new FastReader(); static int INF = Integer.MAX_VALUE; static int NINF = Integer.MIN_VALUE; public static void main(String[] args) throws java.lang.Exception { // your code goes here int t = i(); while (t-- > 0) { solve(); } out.close(); } public static void solve() { int n = i(); int a[] = input(n); if (n == 2) { if (a[1] % a[0] == 0) out.println("YES"); else out.println("NO"); } else { int f = 0; for (int i = 1; i < n; i++) { if (a[i] % a[0] == 0) continue; else { f = 1; break; } } if (f == 0) out.println("YES"); else out.println("NO"); } } static int[] input(int N) { int[] A = new int[N]; for (int i = 0; i < N; i++) A[i] = in.nextInt(); return A; } public static void print(int[] arr) { int n = arr.length; for (int i = 0; i < n; i++) { out.print(arr[i] + " "); } out.println(); } public static void sort(int[] arr) { ArrayList<Integer> ls = new ArrayList<>(); for (int x : arr) { ls.add(x); } Collections.sort(ls); for (int i = 0; i < arr.length; i++) { arr[i] = ls.get(i); } } public static void reverse(int[] arr) { int n = arr.length; for (int i = 0; i < n / 2; i++) { int temp = arr[i]; arr[i] = arr[n - 1 - i]; arr[n - 1 - i] = temp; } } public static void reverse(long[] arr) { int n = arr.length; for (int i = 0; i < n / 2; i++) { long temp = arr[i]; arr[i] = arr[n - 1 - i]; arr[n - 1 - i] = temp; } } public static void print(ArrayList<Integer> arr) { int n = arr.size(); for (int i = 0; i < n; i++) { out.print(arr.get(i) + " "); } out.println(); } static int i() { return in.nextInt(); } static long l() { return in.nextLong(); } static char ich() { return in.next().charAt(0); } static String is() { return in.next(); } static String isl() { return in.nextLine(); } public static int max(int[] arr) { int max = -1; int n = arr.length; for (int i = 0; i < n; i++) max = Math.max(max, arr[i]); return max; } public static int min(int[] arr) { int min = INF; int n = arr.length; for (int i = 0; i < n; i++) min = Math.min(min, arr[i]); return min; } public static int gcd(int x, int y) { if (y == 0) { return x; } return gcd(y, x % y); } } 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 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
71a08bd125dfd435a43ff3348a3fc8ce
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(); } int flag =1; int fir=arr[0]; for(int i=1;i<n;i++){ if((arr[i]%fir)!=0){ flag=0; break; } } if(flag==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 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
56b497b0d5361c6204983aed4ab12bfb
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.*; public class Competitive { public static void main(String[] args) throws IOException{ // TODO Auto-generated method stub BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int t = Integer.parseInt(br.readLine()); while(t-- > 0) { int n = Integer.parseInt(br.readLine()); String[] in = br.readLine().split(" "); boolean flag = true; for(int i = 1; i < n; i++) { if(Integer.parseInt(in[i])%Integer.parseInt(in[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 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
8955d57d2b526d3387f0f93cbf5c18cd
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 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 flag = 0; for(int i = 1;i<=n-1;i++){ if(arr[i]%arr[0]!=0){ flag = 1; break; } } if(flag == 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 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
4f76e175b0eec76eb302bb21b8c6c738
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 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
cad017fe3545bb4d128634f2adf218aa
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.*; import static java.lang.Math.*; import static java.util.Arrays.sort; /* * 11223456 * * */ public class Codeforces { public static void main(String[] args) { FastReader fastReader = new FastReader(); PrintWriter out = new PrintWriter(System.out); int t = fastReader.nextInt(); while (t-- > 0) { int n = fastReader.nextInt(); int a[] = fastReader.ria(n); boolean ans = true; for (int i = 1; i < n; i++) { if (a[i] % a[0] != 0) { ans = false; break; } } out.println(ans ? "YES" : "NO"); } out.close(); } // constants static final int IBIG = 1000000007; static final int IMAX = 2147483647; static final long LMAX = 9223372036854775807L; static Random __r = new Random(); // math util static int minof(int a, int b, int c) { return min(a, min(b, c)); } static int minof(int... x) { if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min; } static long minof(long a, long b, long c) { return min(a, min(b, c)); } static long minof(long... x) { if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min; } static int maxof(int a, int b, int c) { return max(a, max(b, c)); } static int maxof(int... x) { if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max; } static long maxof(long a, long b, long c) { return max(a, max(b, c)); } static long maxof(long... x) { if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max; } static int powi(int a, int b) { if (a == 0) return 0; int ans = 1; while (b > 0) { if ((b & 1) > 0) ans *= a; a *= a; b >>= 1; } return ans; } static long powl(long a, int b) { if (a == 0) return 0; long ans = 1; while (b > 0) { if ((b & 1) > 0) ans *= a; a *= a; b >>= 1; } return ans; } static int fli(double d) { return (int) d; } static int cei(double d) { return (int) ceil(d); } static long fll(double d) { return (long) d; } static long cel(double d) { return (long) ceil(d); } static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } static int lcm(int a, int b) { return (a / gcd(a, b)) * b; } static long gcd(long a, long b) { return b == 0 ? a : gcd(b, a % b); } static long lcm(long a, long b) { return (a / gcd(a, b)) * b; } static int[] exgcd(int a, int b) { if (b == 0) return new int[] { 1, 0 }; int[] y = exgcd(b, a % b); return new int[] { y[1], y[0] - y[1] * (a / b) }; } static long[] exgcd(long a, long b) { if (b == 0) return new long[] { 1, 0 }; long[] y = exgcd(b, a % b); return new long[] { y[1], y[0] - y[1] * (a / b) }; } static int randInt(int min, int max) { return __r.nextInt(max - min + 1) + min; } static long mix(long x) { x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31); } public static boolean[] findPrimes(int limit) { assert limit >= 2; final boolean[] nonPrimes = new boolean[limit]; nonPrimes[0] = true; nonPrimes[1] = true; int sqrt = (int) Math.sqrt(limit); for (int i = 2; i <= sqrt; i++) { if (nonPrimes[i]) continue; for (int j = i; j < limit; j += i) { if (!nonPrimes[j] && i != j) nonPrimes[j] = true; } } return nonPrimes; } // array util static void reverse(int[] a) { for (int i = 0, n = a.length, half = n / 2; i < half; ++i) { int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap; } } static void reverse(long[] a) { for (int i = 0, n = a.length, half = n / 2; i < half; ++i) { long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap; } } static void reverse(double[] a) { for (int i = 0, n = a.length, half = n / 2; i < half; ++i) { double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap; } } static void reverse(char[] a) { for (int i = 0, n = a.length, half = n / 2; i < half; ++i) { char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap; } } static void shuffle(int[] a) { int n = a.length - 1; for (int i = 0; i < n; ++i) { int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap; } } static void shuffle(long[] a) { int n = a.length - 1; for (int i = 0; i < n; ++i) { int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap; } } static void shuffle(double[] a) { int n = a.length - 1; for (int i = 0; i < n; ++i) { int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap; } } static void rsort(int[] a) { shuffle(a); sort(a); } static void rsort(long[] a) { shuffle(a); sort(a); } static void rsort(double[] a) { shuffle(a); sort(a); } static int[] copy(int[] a) { int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans; } static long[] copy(long[] a) { long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans; } static double[] copy(double[] a) { double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans; } static char[] copy(char[] a) { char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans; } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader( new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } int[] ria(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = Integer.parseInt(next()); return a; } long nextLong() { return Long.parseLong(next()); } long[] rla(int n) { long[] a = new long[n]; for (int i = 0; i < n; i++) a[i] = Long.parseLong(next()); return a; } 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 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
91ee540be246f0e907788d40455e3058
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 scanner = new Scanner(System.in); int t = scanner.nextInt(); for (int i = 0; i < t; i++) { int n = scanner.nextInt(); int[] nums = new int[n]; for (int j = 0; j < n; j++) { nums[j] = scanner.nextInt(); } if (satisfied(nums)) { System.out.println("YES"); } else { System.out.println("NO"); } } } public static boolean satisfied(int[] nums) { int n = nums.length, base = nums[0], pos = 0; while (pos < n && base == 0) { base = nums[pos++]; } for (int i = pos; i < n; i++) { if (nums[i] % base > 0) { return false; } } return true; } }
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
52d015066f16040212a1b00b3e864914
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 scanner = new Scanner(System.in); int t = scanner.nextInt(); while(t-- > 0){ int n = scanner.nextInt(); String check = "YES"; int[] arr = new int[n]; for(int i = 0 ; i < n ; i++){ arr[i] = scanner.nextInt(); if (arr[i] % arr[0] != 0) { check = "NO"; } } System.out.println(check); } } }
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
c930cb7bd0e57c486732fb3a7e1e92b3
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
/* IF I HAD CHOICE, I WOULD HAVE BEEN A PIRATE, LOL XD ; _____________############# _____________##___________## ______________#____________# _______________#____________#_## _______________#__############### _____##############______________# _____##____________#_____################ ______#______##############______________# ______##_____##___________#______________## _______#______#____PARTH___#______________# ______##_______#___________##____________# ______###########__________##___________## ___________#_#_##________###_____########_____### ___________#_#_###########_#######_______#####__######### ######_____#_#______##_#_______#_#########___######## ##___#########______##_#____#######________##### __##________##################____##______### ____#____________________________##_#____## _____#_____###_____##_____###_____###___## ______#___##_##___##_#____#_##__________# ______##____##_____###_____##__________## ________##___________________________## ___________######################### */ import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.*; public class Main { static PrintWriter out = new PrintWriter(System.out); static FastReader sc = new FastReader(); 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) { int tes = sc.nextInt(); while(tes-->0){ solve(); } out.close(); } private static void solve() { int n = sc.nextInt(); int [] arr = new int[n]; for (int i = 0; i < n; i++) { arr[i]=sc.nextInt(); } for (int i = 1; i < n; i++) { if (arr[i]%arr[0]!=0){ out.println("NO"); return; } } out.println("YES"); } } /* Test case 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 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
2cd844704f8420acc5e2974e0e412c71
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.FileNotFoundException; import java.io.FileReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; import java.io.OutputStream; import java.io.BufferedOutputStream; import java.util.*; public class CodeForces_808 { public static void main(String[] args)throws Exception{ FastScanner in = new FastScanner();OutputStream out = new BufferedOutputStream ( System.out ); int t=in.nextInt(); while (t-->0){ int n=in.nextInt();int[] a=new int[n]; for(int i=0;i<n;i++){ a[i]=in.nextInt(); } boolean flag=true; for(int i=1;i<n;i++){ if((a[i]%a[0])!=0){ flag=false; break; } } if(flag) System.out.println("YES"); else System.out.println("NO"); } } public static class FastScanner { BufferedReader br; StringTokenizer st; public FastScanner(String s) { try { br = new BufferedReader(new FileReader(s)); } catch (FileNotFoundException e) { // TODO Auto-generated catch block e.printStackTrace(); } } public FastScanner() { br = new BufferedReader(new InputStreamReader(System.in)); } String nextToken() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(nextToken()); } long nextLong() { return Long.parseLong(nextToken()); } double nextDouble() { return Double.parseDouble(nextToken()); } String nextLine() { String str = ""; try { 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 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
00968eea9566809ff8a5e5ae9163138e
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.TreeSet; public class Solution { static class FastScanner { private BufferedReader in; private PrintWriter out; private StringTokenizer tok = new StringTokenizer(""); public FastScanner() { in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); } String readString() throws IOException { while (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); } return tok.nextToken(); } int readInt() throws IOException { return Integer.parseInt(readString()); } long readLong() throws IOException { return Long.parseLong(readString()); } double readDouble() throws IOException { return Double.parseDouble(readString()); } public void print(Object object) { out.println(object); } public void print(Object[] arr) { out.println(Arrays.toString(arr)); } public void printMatrix(Object[][] matrix) { int row = matrix.length, col = matrix[0].length; for (int i = 0; i < row; ++i) { for (int j = 0; j < col; ++j) { out.print(matrix[i][j] + " "); } out.println(); } } public void preDestroy() { try { in.close(); out.close(); } catch (Exception e) { e.printStackTrace(); } } } public static void main(String[] args) throws IOException { Solution obj = new Solution(); FastScanner fastScanner = new FastScanner(); obj.solve(fastScanner, fastScanner.out); fastScanner.preDestroy(); } public void solve(FastScanner in, PrintWriter out) throws IOException { int test = in.readInt(); while(test > 0) { int n = in.readInt(); int[] arr = new int[n]; boolean flag = true; for(int i = 0; i < n; ++ i) { arr[i] = in.readInt(); } if (arr[0] != 1) { int end = 0; for(int i = 1; i < n; ++ i) { if ((arr[i] - arr[i - 1] == 1) || arr[i] == 1) { end = i; break; } } end = end == 0 ? n - 1 : end; for(int i = 1; i <= end; ++ i) { if(arr[i] % arr[0] != 0) { flag = false; break; } } } out.println(flag ? "YES" : "NO"); --test; } } }
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
9e7332eec9bc79e9f744988bf6205059
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.awt.image.AreaAveragingScaleFilter; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.*; public class div8072 { static class Pair implements Comparable<Pair>{ int val; char ch; int freq; Pair(int val, char ch, int freq){ this.val = val; this.ch = ch; this.freq = freq; } @Override public int compareTo(Pair o) { return this.val - o.val; } } public static void main(String[] args) { FastScanner scn = new FastScanner(); PrintWriter out = new PrintWriter(System.out); try{ 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(); } boolean flag = true; HashSet<Integer> hs = new HashSet<>(); int min = arr[0]; int smin = Integer.MAX_VALUE; int sum = 0; for(int i = 1;i<arr.length;i++){ // System.out.print(arr[i] + " " + arr[i-1]); if(arr[i]%arr[0]!=0){ flag = false; break; } } // System.out.println(); if(flag){ System.out.println("YES"); } else{ System.out.println("NO"); } } } catch(Exception e){ return; } out.close(); } static final Random random=new Random(); static final int mod=1_000_000_007; static void ruffleSort(int[] a) { int n=a.length;//shuffle, then sort for (int i=0; i<n; i++) { int oi=random.nextInt(n), temp=a[oi]; a[oi]=a[i]; a[i]=temp; } Arrays.sort(a); } static long add(long a, long b) { return (a+b)%mod; } static long sub(long a, long b) { return ((a-b)%mod+mod)%mod; } static long mul(long a, long b) { return (a*b)%mod; } static long exp(long base, long exp) { if (exp==0) return 1; long half=exp(base, exp/2); if (exp%2==0) return mul(half, half); return mul(half, mul(half, base)); } static long[] factorials=new long[2_000_001]; static long[] invFactorials=new long[2_000_001]; static void precompFacts() { factorials[0]=invFactorials[0]=1; for (int i=1; i<factorials.length; i++) factorials[i]=mul(factorials[i-1], i); invFactorials[factorials.length-1]=exp(factorials[factorials.length-1], mod-2); for (int i=invFactorials.length-2; i>=0; i--) invFactorials[i]=mul(invFactorials[i+1], i+1); } static long nCk(int n, int k) { return mul(factorials[n], mul(invFactorials[k], invFactorials[n-k])); } static void sort(int[] a) { ArrayList<Integer> l=new ArrayList<>(); for (int i:a) l.add(i); Collections.sort(l); for (int i=0; i<a.length; i++) a[i]=l.get(i); } static class FastScanner { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st=new StringTokenizer(""); String next() { while (!st.hasMoreTokens()) try { st=new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } int[] readArray(int n) { int[] a=new int[n]; for (int i=0; i<n; i++) a[i]=nextInt(); return a; } long nextLong() { return Long.parseLong(next()); } } }
Java
["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
60ba57ec0473826246a885fd5b67eec1
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 scn = new Scanner(System.in); int A = scn.nextInt(); StringBuilder sb = new StringBuilder(""); while(A-- > 0) { int n = scn.nextInt(); int[] arr = new int[n]; for(int i=0 ; i<n ; i++) { arr[i] = scn.nextInt(); } String ans = "YES"; for(int i=1 ; i<n ; i++) { if(arr[i] % arr[i - 1] == 0) { arr[i] = arr[i-1]; }else { ans = "NO"; break; } } sb.append(ans + "\n"); } System.out.println(sb); } // public static int gcd(int a, int b) { // if (b == 0) // return a; // return gcd(b, a % b); // } }
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
e7b6196456acb276633967939fce74ae
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 _cf { public static void main (String[] args) { new _cf(); } public _cf() { Scanner s = new Scanner(); PrintWriter out = new PrintWriter(System.out); int T = s.nextInt(); while(T-- > 0) { int n = s.nextInt(); int[] a = new int[n]; boolean possible = true; for(int i = 0; i < n; i++) { a[i] = s.nextInt(); if (a[i] % a[0] != 0) possible = false; } if(possible) out.println("YES"); else out.println("NO"); } out.close(); } class Scanner { public int BS = 1<<16; public char NC = (char)0; byte[] buf = new byte[BS]; int bId = 0, size = 0; char c = NC; double num = 1; BufferedInputStream in; public Scanner() { in = new BufferedInputStream(System.in, BS); } public Scanner(String s) { try { in = new BufferedInputStream(new FileInputStream(new File(s)), BS); } catch (Exception e) { in = new BufferedInputStream(System.in, BS); } } public char nextChar(){ while(bId==size) { try { size = in.read(buf); }catch(Exception e) { return NC; } if(size==-1)return NC; bId=0; } return (char)buf[bId++]; } public int nextInt() { return (int)nextLong(); } public long nextLong() { num=1; boolean neg = false; if(c==NC)c=nextChar(); for(;(c<'0' || c>'9'); c = nextChar()) { if(c=='-')neg=true; } long res = 0; for(; c>='0' && c <='9'; c=nextChar()) { res = (res<<3)+(res<<1)+c-'0'; num*=10; } return neg?-res:res; } public double nextDouble() { double cur = nextLong(); return c!='.' ? cur:cur+nextLong()/num; } public String next() { StringBuilder res = new StringBuilder(); while(c<=32)c=nextChar(); while(c>32) { res.append(c); c=nextChar(); } return res.toString(); } public String nextLine() { StringBuilder res = new StringBuilder(); while(c<=32)c=nextChar(); while(c!='\n') { res.append(c); c=nextChar(); } return res.toString(); } public boolean hasNext() { if(c>32)return true; while(true) { c=nextChar(); if(c==NC)return false; else if(c>32)return true; } } public int[] nextIntArray(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
723d1f50f08cb03559df5546a74723a6
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 CODE { public static void main(String[] args) { try { System.setIn(new FileInputStream("input.txt")); System.setOut(new PrintStream(new FileOutputStream("output.txt"))); } catch (Exception e) { System.err.println("Error"); } FastReader scn = new FastReader(); int np = scn.nextInt(); for (int l = 0; l < np; l++) { int n = scn.nextInt(); long[] ar = new long[n]; boolean f = true; for (int i = 0; i < n; i++) { ar[i] = scn.nextLong(); } for (int i = 1; i < n; i++) { if (ar[i] % ar[0] != 0) { f = false; break; } } if (f) { System.out.println("YES"); } else { System.out.println("NO"); } } } 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 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
b12c7987e7058911a4359c0406a54744
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 { static FastScanner scan = new FastScanner(); static Scanner scanner = new Scanner(System.in); static PrintWriter printWriter = new PrintWriter(System.out); static long MOD = 998244353; static long MOD2 = MOD * MOD; static int positiveInf = Integer.MAX_VALUE; static int negativeInf = Integer.MIN_VALUE; static long ded = (long) (1e17) + 9; static long end = Long.MAX_VALUE; public static void main(String[] args) { int t = scan.nextInt(); while (t-- > 0) { solve(); } printWriter.close(); scanner.close(); } public static void solve(){ int n = scan.nextInt(); int[] arr = new int[n]; int max = 0; Set<Integer> set = new HashSet<>(); for(int i =0;i<arr.length;i++){ arr[i]= scan.nextInt(); max = Math.max(arr[i],max); set.add(arr[i]); } if(set.size()==1){ printWriter.println("YES"); return; } boolean div = true; int a =arr[0]; for(int i =1;i<arr.length;i++){ if(arr[i] % a !=0){ div=false; break; } } if(div){ printWriter.println("YES"); } else{ printWriter.println("NO"); } } 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 static void yes(){ printWriter.println("YES"); } public static void no(){ printWriter.println("NO"); } static class Pair implements Comparable<Pair> { int x; int y; public Pair(int x, int y) { this.x = x; this.y = y; } @Override public int compareTo(Pair o){ if(this.y==o.y){ return this.x-o.x; } return this.y-o.y; } } public static long mul(long a, long b) { return ((a % MOD) * (b % MOD)) % MOD; } public static long add(long a, long b) { return ((a % MOD) + (b % MOD)) % MOD; } public static long c2(long n) { if ((n & 1) == 0) { return mul(n / 2, n - 1); } else { return mul(n, (n - 1) / 2); } } //Shuffle Sort static final Random random = new Random(); static void ruffleSort(int[] a) { int n = a.length;//shuffle, then sort for (int i = 0; i < n; i++) { int oi = random.nextInt(n); int temp= a[oi]; a[oi] = a[i]; a[i] = temp; } Arrays.sort(a); } //Brian Kernighan's Algorithm static long countSetBits(long n) { if (n == 0) return 0; return 1 + countSetBits(n & (n - 1)); } //Euclidean Algorithm static long gcd(long A, long B) { if (B == 0) return A; return gcd(B, A % B); } //Modular Exponentiation static long fastExpo(long x, long n) { if (n == 0) return 1; if ((n & 1) == 0) return fastExpo((x * x) % MOD, n / 2) % MOD; return ((x % MOD) * fastExpo((x * x) % MOD, (n - 1) / 2)) % MOD; } //AKS Algorithm 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 <= Math.sqrt(n); i += 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } public static long modinv(long x) { return modpow(x, MOD - 2); } public static long modpow(long a, long b) { if (b == 0) { return 1; } long x = modpow(a, b / 2); x = (x * x) % MOD; if (b % 2 == 1) { return (x * a) % MOD; } return x; } public static boolean isInteger(int n) { return Math.sqrt(n) % 1 == 0; } static class FastScanner { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st = new StringTokenizer(""); String next() { while (!st.hasMoreTokens()) try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } int[] readArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = nextInt(); return a; } long nextLong() { return Long.parseLong(next()); } } }
Java
["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
bda9b1fdb0c15a6803e621aa003a69d2
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 { public static void main (String[] args) throws java.lang.Exception { Scanner scn = new Scanner(System.in); int tc = scn.nextInt(); while (tc != 0) { int n = scn.nextInt(); int[] arr = new int[n]; for (int i = 0 ; i < n ; i++) { arr[i] = scn.nextInt(); } solve(n, arr); tc--; } } public static void solve(int n, int[] arr) { for (int i = 1 ; i < n ; i++) { if (arr[i] % arr[0] != 0) { System.out.println("NO"); return; } } System.out.println("YES"); 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 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
31d80885abc256a5b544f2d9b9ef5282
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 class pair{ int x; int y; pair(int x1,int y1){ x=x1; y=y1; } } static FastReader in=new FastReader(); static final Random random=new Random(); static long mod=1000000007L; static HashMap<String,Integer>map=new HashMap<>(); public static void main(String args[]) throws IOException { int t=in.nextInt(); while(t-->0){ int n=in.nextInt(); int arr[]=in.readintarray(n); int ind=arr[0]; boolean flag=false; for(int i=1;i<n;i++){ if(arr[i]%ind!=0){ flag=true; break; } } if(flag){ print("NO"); } else{ print("YES"); } } } static int max(int a, int b) { if(a<b) return b; return a; } static void ruffleSort(int[] a) { int n=a.length; for (int i=0; i<n; i++) { int oi=random.nextInt(n), temp=a[oi]; a[oi]=a[i]; a[i]=temp; } Arrays.sort(a); } static < E > void print(E res) { System.out.println(res); } static int gcd(int a,int b) { if(b==0) { return a; } return gcd(b,a%b); } static int lcm(int a, int b) { return (a / gcd(a, b)) * b; } static int abs(int a) { if(a<0) return -1*a; return a; } static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreElements()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } int [] readintarray(int n) { int res [] = new int [n]; for(int i = 0; i<n; i++)res[i] = nextInt(); return res; } long [] readlongarray(int n) { long res [] = new long [n]; for(int i = 0; i<n; i++)res[i] = nextLong(); return res; } } }
Java
["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
cdd83a8535cb01f378952abe79b08180
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 int mod = 998244353; static int mod = (int)1e9+7; static boolean[] prime = new boolean[1]; static int[][] dir1 = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; static int[][] dir2 = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}}; static int inf = 0x3f3f3f3f; static { for (int i = 2; i < prime.length; i++) prime[i] = true; for (int i = 2; i < prime.length; i++) { if (prime[i]) { for (int k = 2; i * k < prime.length; k++) { prime[i * k] = false; } } } } static class JoinSet { int[] fa; JoinSet(int n) { fa = new int[n]; for (int i = 0; i < n; i++) fa[i] = i; } int find(int t) { if (t != fa[t]) fa[t] = find(fa[t]); return fa[t]; } void join(int x, int y) { x = find(x); y = find(y); fa[x] = y; } } static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out)); static long gcd(long a, long b) { return b == 0 ? a : gcd(b, a % b); } static long lcm(long a, long b) { return a * b / gcd(a, b); } static int get() throws Exception { String ss = bf.readLine(); if (ss.contains(" ")) ss = ss.substring(0, ss.indexOf(" ")); return Integer.parseInt(ss); } static long getx() throws Exception { String ss = bf.readLine(); if (ss.contains(" ")) ss = ss.substring(0, ss.indexOf(" ")); return Long.parseLong(ss); } static int[] getint() throws Exception { String[] s = bf.readLine().split(" "); int[] a = new int[s.length]; for (int i = 0; i < a.length; i++) { a[i] = Integer.parseInt(s[i]); } return a; } static long[] getlong() throws Exception { String[] s = bf.readLine().split(" "); long[] a = new long[s.length]; for (int i = 0; i < a.length; i++) { a[i] = Long.parseLong(s[i]); } return a; } static char[] getstr() throws Exception { return bf.readLine().toCharArray(); } static void println() throws Exception { bw.write("\n"); } static void print(int a) throws Exception { bw.write(a + "\n"); } static void print(long a) throws Exception { bw.write(a + "\n"); } static void print(String a) throws Exception { bw.write(a + "\n"); } static void print(int[] a) throws Exception { for (int i : a) { bw.write(i + " "); } println(); } static void print(long[] a) throws Exception { for (long i : a) { bw.write(i + " "); } println(); } static void print(int[][] a) throws Exception { for (int i[] : a) print(i); } static void print(long[][] a) throws Exception { for (long[] i : a) print(i); } static void print(char[] a) throws Exception { for (char i : a) { bw.write(i + ""); } println(); } static long pow(long a, long b) { long ans = 1; while (b > 0) { if ((b & 1) == 1) { ans *= a; } a *= a; b >>= 1; } return ans; } static int powmod(long a, long b, int mod) { long ans = 1; while (b > 0) { if ((b & 1) == 1) { ans = ans * a % mod; } a = a * a % mod; b >>= 1; } return (int) ans; } static void sort(int[] a) { int n = a.length; Integer[] b = new Integer[n]; for (int i = 0; i < n; i++) b[i] = a[i]; Arrays.sort(b); for (int i = 0; i < n; i++) a[i] = b[i]; } static void sort(long[] a) { int n = a.length; Long[] b = new Long[n]; for (int i = 0; i < n; i++) b[i] = a[i]; Arrays.sort(b); for (int i = 0; i < n; i++) a[i] = b[i]; } static void resort(int[] a) { int n = a.length; Integer[] b = new Integer[n]; for (int i = 0; i < n; i++) b[i] = a[i]; Arrays.sort(b); for (int i = 0; i < n; i++) a[i] = b[n - 1 - i]; } static void resort(long[] a) { int n = a.length; Long[] b = new Long[n]; for (int i = 0; i < n; i++) b[i] = a[i]; Arrays.sort(b); for (int i = 0; i < n; i++) a[i] = b[n - 1 - i]; } static int max(int a, int b) { return Math.max(a, b); } static int min(int a, int b) { return Math.min(a, b); } static long max(long a, long b) { return Math.max(a, b); } static long min(long a, long b) { return Math.min(a, b); } static int max(int[] a) { int max = a[0]; for (int i : a) max = max(max, i); return max; } static int min(int[] a) { int min = a[0]; for (int i : a) min = min(min, i); return min; } static long max(long[] a) { long max = a[0]; for (long i : a) max = max(max, i); return max; } static long min(long[] a) { long min = a[0]; for (long i : a) min = min(min, i); return min; } static int abs(int a) { return Math.abs(a); } static long abs(long a) { return Math.abs(a); } static void yes() throws Exception { print("Yes"); } static void no() throws Exception { print("No"); } static int[] getarr(List<Integer> list) { int n = list.size(); int[] a = new int[n]; for (int i = 0; i < n; i++) a[i] = list.get(i); return a; } static List<Integer>[] lists; static class LCA{ int[] dep; int[][] fa; int[] log; boolean[] v; public LCA(int n){ dep = new int[n+5]; log = new int[n+5]; fa = new int[n+5][31]; v = new boolean[n+5]; for (int i = 2; i <= n; ++i) { log[i] = log[i/2] + 1; } dfs(1,0); } private void dfs(int cur, int pre){ if(v[cur]) return; v[cur] = true; dep[cur] = dep[pre]+1; fa[cur][0] = pre; for (int i = 1; i <= log[dep[cur]]; ++i) { fa[cur][i] = fa[fa[cur][i - 1]][i - 1]; } for(int i : lists[cur]){ dfs(i,cur); } } private int lca(int a, int b){ if(dep[a] > dep[b]){ int t = a; a = b; b = t; } while (dep[a] != dep[b]){ b = fa[b][log[dep[b] - dep[a]]]; } if(a == b) return a; for (int k = log[dep[a]]; k >= 0; k--) { if (fa[a][k] != fa[b][k]) { a = fa[a][k]; b = fa[b][k]; } } return fa[a][0]; } } public static void main(String[] args) throws Exception { int T = 1; T = get(); f:while (T-- > 0) { int n = get(); int a[] = getint(); for(int i = 1;i < n;i++){ if(a[i]%a[0] != 0) { no(); continue f; } } yes(); } bw.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 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
b266950423e3005d5fd45e28ed01ca47
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
// LARGEST SQUARE // package com.company /* * @author :: Yuvraj Singh * CS UNDERGRAD AT IT GGV BILASPUR */ import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.*; import static java.lang.Math.*; import static java.lang.System.out; public class Main { // Solution here ArrayList<Pair> arr; void solve(int t) { int n= in.nextInt(); long ans = 0; long[] arr= longArr(n); boolean f= true; for(int i=1; i<n; i++){ if(arr[i] % arr[0] != 0){ f= false; } } if(f){ sb.append("YES"); }else sb.append("NO"); sb.append("\n"); } void start(){ sb= new StringBuffer(); int t= in.nextInt(); for(int i=1;i<=t;i++) { solve(i); } out.print(sb); } // Starter Code FastReader in; StringBuffer sb; public static void main(String[] args) { new Main().run(); } void run(){ in= new FastReader(); start(); } void bubbleSort(int[] arr){ int n= arr.length; for(int i=n-1; i>0; i--){ for(int j=0; j<i; j++){ if(arr[i] < arr[j]){ swap(arr, i, j); } } } } void swap(int[] arr, int i, int j){ int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } long numberOfWays(long n, long k) { // Base Cases if (n == 0) return 1; if (k == 0) return 1; if (n >= (int)Math.pow(2, k)) { int curr_val = (int)Math.pow(2, k); return numberOfWays(n - curr_val, k) + numberOfWays(n, k - 1); } else return numberOfWays(n, k - 1); } boolean palindrome(String s){ int i=0, j= s.length()-1; while(i<j){ if(s.charAt(i)!=s.charAt(j)) return false; i++; j--; } return true; } int call(int[] A, int N, int K) { int i = 0, j = 0, sum = 0; int maxLen = Integer.MIN_VALUE; while (j < N) { sum += A[j]; if (sum < K) { j++; } else if (sum == K) { maxLen = Math.max(maxLen, j-i+1); j++; } else if (sum > K) { while (sum > K) { sum -= A[i]; i++; } if(sum == K){ maxLen = Math.max(maxLen, j-i+1); } j++; } } return maxLen; } int largestNum(int n) { int num = 0; // Iterate through all possible values for (int i = 0; i <= 32; i++) { // Multiply the number by 2 i times int x = (1 << i); if ((x - 1) >= n) num = (1 << i) - 1; else break; } // Return the final result return num; } static boolean isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to n-1 for (int i = 2; i <= sqrt(n); i++) if (n % i == 0) return false; return true; } // Useful Functions void swap( int i , int j) { int tmp = i; i = j; j = tmp; } int call(int i,int j, int[][] mat, int[][] dp){ int m= mat.length; int n= mat[0].length; if(i>=m || j>=n){ return Integer.MIN_VALUE; } if(i==m-1 && j==n-1){ return mat[i][j]; } if(dp[i][j] != -1){ return dp[i][j]; } return dp[i][j] = max(call(i+1, j, mat, dp), call(i, j+1, mat, dp)) + mat[i][j]; } int util(int i,int j, int[][] mat, int[][] dp){ int m= mat.length; int n= mat[0].length; if(i>=m || j>=n){ return Integer.MAX_VALUE; } if(i==m-1 && j==n-1){ return mat[i][j]; } if(dp[i][j] != -1){ return dp[i][j]; } return dp[i][j] = min(util(i+1, j, mat, dp), util(i, j+1, mat, dp)) + mat[i][j]; } long power(long x, long y, long p) { long res = 1; x = x % p; while (y > 0) { if (y % 2 == 1) res = (res * x) % p; y = y >> 1; x = (x * x) % p; } return res; } int lower_bound(long[] a, long x) { // x is the target value or key int l=-1,r=a.length; while(l+1<r) { int m=(l+r)>>>1; if(a[m]>=x) r=m; else l=m; } return r; } int upper_bound(long[] arr, int key) { int i=0, j=arr.length-1; if (arr[j]<=key) return j+1; if(arr[i]>key) return i; while (i<j){ int mid= (i+j)/2; if(arr[mid]<=key){ i= mid+1; }else{ j=mid; } } return i; } void sort(long[] A){ int n = A.length; Random rnd = new Random(); for(int i=0; i<n; ++i){ long tmp = A[i]; int randomPos = i + rnd.nextInt(n-i); A[i] = A[randomPos]; A[randomPos] = tmp; } Arrays.sort(A); } int[] intArr(int n){ int[] res= new int[n]; for(int i=0;i<n;i++){ res[i]= in.nextInt(); } return res; } long[] longArr(int n){ long[] res= new long[n]; for(int i=0;i<n;i++){ res[i]= in.nextLong(); } return res; } // // sieve of eratosthenes code for precomputing whether numbers are prime or not up to MAX_VALUE long MAX= 100000000; int[] precomp; void sieve(){ long n= MAX; precomp = new int[(int) (n+1)]; boolean[] prime = new boolean[(int) (n+1)]; for(int i=0;i<=n;i++) prime[i] = true; for(long p = 2; p*p <=n; p++) { // If prime[p] is not changed, then it is a prime if(prime[(int) p]) { // Update all multiples of p for(long i = p*p; i <= n; i += p) prime[(int) i] = false; } } // Print all prime numbers for(long i = 2; i <= n; i++) { if(prime[(int) i]) precomp[(int) i]= 1; } } long REVERSE(long N) { // code here long rev=0; long org= N; while (N!=0){ long d= N%10; rev = rev*10 +d; N /= 10; } return rev; } long sumOfDigits(String n){ long sum= 0; for (char c: n.toCharArray()){ sum += Integer.parseInt(String.valueOf(c)); } return sum; } long[] revArray(long[] arr) { int n= arr.length; int i=0, j=n-1; while (i<j){ long temp= arr[i]; arr[i]= arr[j]; arr[j]= temp; i++; j--; } return arr; } long gcd(long a, long b){ if (b==0) return a; return gcd(b, a%b); } long lcm(long a,long b){ return (a*b)/gcd(a,b); } static class Pair implements Comparable<Pair>{ long first; long second; Pair(long x, long y){ this.first=x; this.second=y; } @Override public int compareTo(Pair o) { return 0; } } // static class Compare { // static void compare(ArrayList<Pair> arr, int n) { // arr.sort(new Comparator<Pair>() { // @Override // public int compare(Pair p1, Pair p2) { // return (int) (p2.first - p1.first); // } // }); // } // } public static class FastReader{ BufferedReader br; StringTokenizer st; public FastReader(){ br=new BufferedReader(new InputStreamReader(System.in)); } String next(){ while (st==null || !st.hasMoreElements()){ try{ st=new StringTokenizer(br.readLine()); }catch (Exception 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 (Exception 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 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
bcec05c7d801dd08bb879a0cf6f2b88b
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 CodeForces { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-->0){ int n=sc.nextInt(); int[] a=new int[n]; boolean flag=true; for (int i = 0; i < n; i++){ a[i]= sc.nextInt(); } for (int i = 1; i < n; i++) { if(a[i]%a[0]!=0){ flag=false; break; } } if (!flag) 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
b5f3e83a7b302f2ae70c9f1a30e4ebaa
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.awt.image.AreaAveragingScaleFilter; import java.io.*; import java.math.BigInteger; import java.text.DecimalFormat; import java.util.*; public class Pratice { static final long mod = 1000000007; static StringBuilder sb = new StringBuilder(); static int xn = (int) (2e5 + 10); static long ans; static boolean prime[] = new boolean[1000001]; // calculate sqrt and cuberoot // static Set<Long> set=new TreeSet<>(); // static // { // long n=1000000001; // // // for(int i=1;i*i<=n;i++) // { // long x=i*i; // set.add(x); // } // for(int i=1;i*i*i<=n;i++) // { // long x=i*i*i; // set.add(x); // } // } static void sieveOfEratosthenes() { for (int i = 0; i <= 1000000; i++) prime[i] = true; prime[0] = false; prime[1] = false; for (int p = 2; p * p <= 1000000; p++) { if (prime[p] == true) { for (int i = p * p; i <= 1000000; i += p) prime[i] = false; } } } public static void main(String[] args) throws IOException { Reader reader = new Reader(); int t = reader.nextInt(); while (t-- > 0) { int n=reader.nextInt(); long arr[]=new long[n]; for (int i = 0; i <n; i++) { arr[i] = reader.nextLong(); } boolean flag=false; for(int i=1;i<n;i++) { if(arr[i]%arr[0]!=0) { flag=true; } } if (flag) System.out.println("No"); else System.out.println("Yes"); } } // } // static void SieveOfEratosthenes(int n, boolean prime[], // boolean primesquare[], int a[]) // { // // Create a boolean array "prime[0..n]" and // // initialize all entries it as true. A value // // in prime[i] will finally be false if i is // // Not a prime, else true. // for (int i = 2; i <= n; i++) // prime[i] = true; // // /* Create a boolean array "primesquare[0..n*n+1]" // and initialize all entries it as false. // A value in squareprime[i] will finally // be true if i is square of prime, // else false.*/ // for (int i = 0; i < ((n * n) + 1); i++) // primesquare[i] = false; // // // 1 is not a prime number // prime[1] = false; // // for (int p = 2; p * p <= n; p++) { // // If prime[p] is not changed, // // then it is a prime // if (prime[p] == true) { // // Update all multiples of p // for (int i = p * 2; i <= n; i += p) // prime[i] = false; // } // } // // int j = 0; // for (int p = 2; p <= n; p++) { // if (prime[p]) { // // a[j] = p; // // // primesquare[p * p] = true; // j++; // } // } // } // // // static int countDivisors(int n) // { // // if (n == 1) // return 1; // // boolean prime[] = new boolean[n + 1]; // boolean primesquare[] = new boolean[(n * n) + 1]; // // int a[] = new int[n]; // // // SieveOfEratosthenes(n, prime, primesquare, a); // // // int ans = 1; // // // Loop for counting factors of n // for (int i = 0;; i++) { // // a[i] is not less than cube root n // if (a[i] * a[i] * a[i] > n) // break; // // int cnt = 1; // // // if a[i] is a factor of n // while (n % a[i] == 0) { // n = n / a[i]; // // // incrementing power // cnt = cnt + 1; // } // // // ans = ans * cnt; // } // // // if (prime[n]) // ans = ans * 2; // // // Second case // else if (primesquare[n]) // ans = ans * 3; // // // Third case // else if (n != 1) // ans = ans * 4; // // return ans; // Total divisors // } public static long[] inarr(long n) throws IOException { Reader reader = new Reader(); long arr[]=new long[(int) n]; for (long i = 0; i < n; i++) { arr[(int) i]=reader.nextLong(); } return arr; } public static boolean checkPerfectSquare(int number) { int x=number % 10; if (x==2 || x==3 || x==7 || x==8) { return false; } for (int i=0; i<=number/2 + 1; i++) { if (i*i==number) { return true; } } return false; } // check number is prime or not public static boolean isPrime(int n) { return BigInteger.valueOf(n).isProbablePrime(1); } // return the gcd of two numbers public static long gcd(long a, long b) { BigInteger b1 = BigInteger.valueOf(a); BigInteger b2 = BigInteger.valueOf(b); BigInteger gcd = b1.gcd(b2); return gcd.longValue(); } // return lcm of number static long lcm(int a, int b) { return (a / gcd(a, b)) * b; } // number of digits in the given number public static long numberOfDigits(long n) { long ans= (long) (Math.floor((Math.log10(n)))+1); return ans; } // return most significant bit in the number public static long mostSignificantNumber(long n) { double k=Math.log10(n); k=k-Math.floor(k); int ans=(int)Math.pow(10,k); return ans; } static class Reader { final private int BUFFER_SIZE = 1 << 16; private DataInputStream din; private byte[] buffer; private int bufferPointer, bytesRead; public Reader() { din = new DataInputStream(System.in); buffer = new byte[BUFFER_SIZE]; bufferPointer = bytesRead = 0; } public Reader(String file_name) throws IOException { din = new DataInputStream(new FileInputStream(file_name)); buffer = new byte[BUFFER_SIZE]; bufferPointer = bytesRead = 0; } public String readLine() throws IOException { byte[] buf = new byte[64]; // line length int cnt = 0, c; while ((c = read()) != -1) { if (c == '\n') break; buf[cnt++] = (byte) c; } return new String(buf, 0, cnt); } public int nextInt() throws IOException { int ret = 0; byte c = read(); while (c <= ' ') c = read(); boolean neg = (c == '-'); if (neg) c = read(); do { ret = ret * 10 + c - '0'; } while ((c = read()) >= '0' && c <= '9'); if (neg) return -ret; return ret; } public long nextLong() throws IOException { long ret = 0; byte c = read(); while (c <= ' ') c = read(); boolean neg = (c == '-'); if (neg) c = read(); do { ret = ret * 10 + c - '0'; } while ((c = read()) >= '0' && c <= '9'); if (neg) return -ret; return ret; } public double nextDouble() throws IOException { double ret = 0, div = 1; byte c = read(); while (c <= ' ') c = read(); boolean neg = (c == '-'); if (neg) c = read(); do { ret = ret * 10 + c - '0'; } while ((c = read()) >= '0' && c <= '9'); if (c == '.') { while ((c = read()) >= '0' && c <= '9') { ret += (c - '0') / (div *= 10); } } if (neg) return -ret; return ret; } private void fillBuffer() throws IOException { bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); if (bytesRead == -1) buffer[0] = -1; } private byte read() throws IOException { if (bufferPointer == bytesRead) fillBuffer(); return buffer[bufferPointer++]; } public void close() throws IOException { if (din == null) return; din.close(); } } }
Java
["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
c88d88405f46921eeb057bf7f7c7e2df
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 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
6184f44c6ec73ac39dfb75c74324cf76
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 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
b5990c40cc19202ecf4e54a94f0370b0
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 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
f4139b9d094edf0d1f3c38a27dec36cf
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 tc = sc.nextInt(); for(int i = 0; i < tc; i++) { int n = sc.nextInt(); int arr[] = new int[n]; for(int j = 0; j < n; j++) { arr[j] = sc.nextInt(); } int elem = arr[0]; boolean flag = false; for(int j = 1; j < n; j++) { if(arr[j] % elem != 0) { flag =true; break; } } if(flag) { 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
67a7fc0875074559e58aa9334ed184df
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.math.*; import java.util.*; public class A_Difference_Operations { public static void main(String[] args)throws Exception { new Solver().solve(); } } //* Success is not final, failure is not fatal: it is the courage to continue that counts. class Solver { void solve() throws Exception { int t = sc.nextInt(); while(t-->0){ int n =sc.nextInt(); long[] arr = sc.getLongArray(n); boolean isPos = true; for(int i = 1;i<n;i++){ long diff= sc.gcd(arr[i],arr[i-1]); if(diff<0 || diff%arr[0]!=0){ isPos = false; break; } } System.out.println(isPos?"YES":"NO"); } sc.flush(); } final Helper sc; final int MAXN = 1000_006; final long MOD = (long) 1e9 + 7; Solver() { sc = new Helper(MOD, MAXN); sc.initIO(System.in, System.out); } } class Helper { final long MOD; final int MAXN; final Random rnd; public Helper(long mod, int maxn) { MOD = mod; MAXN = maxn; rnd = new Random(); } public static int[] sieve; public static ArrayList<Integer> primes; public void setSieve() { primes = new ArrayList<>(); sieve = new int[MAXN]; int i, j; for (i = 2; i*i < MAXN; ++i) if (sieve[i] == 0) { primes.add(i); for (j = i*i; j < MAXN; j += i) { sieve[j] = i; } } } public static long[] factorial; public void setFactorial() { factorial = new long[MAXN]; factorial[0] = 1; for (int i = 1; i < MAXN; ++i) factorial[i] = factorial[i - 1] * i % MOD; } public long getFactorial(int n) { if (factorial == null) setFactorial(); return factorial[n]; } public long ncr(int n, int r) { if (r > n) return 0; if (factorial == null) setFactorial(); long numerator = factorial[n]; long denominator = factorial[r] * factorial[n - r] % MOD; return numerator * pow(denominator, MOD - 2, MOD) % MOD; } public long[] getLongArray(int size) throws Exception { long[] ar = new long[size]; for (int i = 0; i < size; ++i) ar[i] = nextLong(); return ar; } public int[] getIntArray(int size) throws Exception { int[] ar = new int[size]; for (int i = 0; i < size; ++i) ar[i] = nextInt(); return ar; } public long gcd(long a, long b) { return b == 0 ? a : gcd(b, a % b); } public int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } public long max(long[] ar) { long ret = ar[0]; for (long itr : ar) ret = Math.max(ret, itr); return ret; } public int max(int[] ar) { int ret = ar[0]; for (int itr : ar) ret = Math.max(ret, itr); return ret; } public long min(long[] ar) { long ret = ar[0]; for (long itr : ar) ret = Math.min(ret, itr); return ret; } public int min(int[] ar) { int ret = ar[0]; for (int itr : ar) ret = Math.min(ret, itr); return ret; } public long sum(long[] ar) { long sum = 0; for (long itr : ar) sum += itr; return sum; } public long sum(int[] ar) { long sum = 0; for (int itr : ar) sum += itr; return sum; } public void shuffle(int[] ar) { int r; for (int i = 0; i < ar.length; ++i) { r = rnd.nextInt(ar.length); if (r != i) { ar[i] ^= ar[r]; ar[r] ^= ar[i]; ar[i] ^= ar[r]; } } } public void shuffle(long[] ar) { int r; for (int i = 0; i < ar.length; ++i) { r = rnd.nextInt(ar.length); if (r != i) { ar[i] ^= ar[r]; ar[r] ^= ar[i]; ar[i] ^= ar[r]; } } } public long pow(long base, long exp, long MOD) { base %= MOD; long ret = 1; while (exp > 0) { if ((exp & 1) == 1) ret = ret * base % MOD; base = base * base % MOD; exp >>= 1; } return ret; } static byte[] buf = new byte[1000_006]; static int index, total; static InputStream in; static BufferedWriter bw; public void initIO(InputStream is, OutputStream os) { try { in = is; bw = new BufferedWriter(new OutputStreamWriter(os)); } catch (Exception e) { } } public void initIO(String inputFile, String outputFile) { try { in = new FileInputStream(inputFile); bw = new BufferedWriter(new OutputStreamWriter( new FileOutputStream(outputFile))); } catch (Exception e) { } } private int scan() throws Exception { if (index >= total) { index = 0; total = in.read(buf); if (total <= 0) return -1; } return buf[index++]; } public String nextLine() throws Exception { int c; for (c = scan(); c <= 32; c = scan()) ; StringBuilder sb = new StringBuilder(); for (; c >= 32; c = scan()) sb.append((char) c); return sb.toString(); } public String next() throws Exception { int c; for (c = scan(); c <= 32; c = scan()) ; StringBuilder sb = new StringBuilder(); for (; c > 32; c = scan()) sb.append((char) c); return sb.toString(); } public int nextInt() throws Exception { int c, val = 0; for (c = scan(); c <= 32; c = scan()) ; boolean neg = c == '-'; if (c == '-' || c == '+') c = scan(); for (; c >= '0' && c <= '9'; c = scan()) val = (val << 3) + (val << 1) + (c & 15); return neg ? -val : val; } public long nextLong() throws Exception { int c; long val = 0; for (c = scan(); c <= 32; c = scan()) ; boolean neg = c == '-'; if (c == '-' || c == '+') c = scan(); for (; c >= '0' && c <= '9'; c = scan()) val = (val << 3) + (val << 1) + (c & 15); return neg ? -val : val; } public void print(Object a) throws Exception { bw.write(a.toString()); } public void printsp(Object a) throws Exception { print(a); print(" "); } public void println() throws Exception { bw.write("\n"); } public void printArray(int[] arr) throws Exception{ for(int i = 0;i<arr.length;i++){ print(arr[i]+ " "); } println(); } public void println(Object a) throws Exception { print(a); println(); } public void flush() throws Exception { bw.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 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
37421001b510cb41f5da96f5f9a6d6c8
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; /** * * @author Lenovo */ public class Main { public static void main(String[] args) { Scanner input = new Scanner(System.in); int t = input.nextInt(); int flag = 0; for (int j = 0; j < t; j++) { int n = input.nextInt(); int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = input.nextInt(); } for (int i = n - 1; i > 0; i--) { if (a[i] % a[0] == 0) { flag = 1; } else { flag = 0; break; } } if (flag == 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 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
e99b4209bc8dd579ef9cc34e78d63db0
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.Integer.parseInt; import static java.lang.Long.parseLong; import static java.lang.Double.parseDouble; import static java.lang.Math.PI; import static java.lang.Math.min; import static java.lang.System.arraycopy; import static java.lang.System.exit; import static java.util.Arrays.copyOf; import java.util.LinkedList; import java.util.List; import java.util.Iterator; import java.io.FileReader; import java.io.FileWriter; import java.math.BigInteger; import java.util.ArrayList; import java.util.Arrays; import java.util.HashSet; import java.util.Set; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.HashMap; import java.util.Map; import java.util.Map.Entry; import java.util.NoSuchElementException; import java.util.PriorityQueue; import java.util.Queue; import java.util.Deque; import java.util.Stack; import java.util.StringTokenizer; import java.util.Comparator; import java.lang.StringBuilder; import java.util.Collections; import java.util.*; import java.text.DecimalFormat; public class Solution { // static class Edge{ // int u, v, w; // Edge(int u, int v, int w){ // this.u = u; // this.v = v; // this.w = w; // } // } static class Pair{ int first, second; Pair(int first, int second){ this.first = first; this.second = second; } @Override public String toString(){ return (this.first+" "+this.second); } } // static class Tuple{ // int first, second, third; // Tuple(int first, int second, int third){ // this.first = first; // this.second =second; // this.third = third; // } // @Override // public String toString(){ // return first+" "+second+" "+third; // } // } // static class Point{ // int x, y; // Point(int x, int y){ // this.x = x; // this.y = y; // } // @Override // public String toString(){ // return x+" "+y; // } // } static BufferedReader in; static PrintWriter out; static StringTokenizer tok; static int dx[] = {0, -1, 0, 1}; //left up right down static int dy[] = {-1, 0, 1, 0}; static int MOD = (int)1e9+7, INF = (int)1e9; static boolean vis[]; static ArrayList<Integer> adj[]; private static void solve() throws IOException{ int n = scanInt(); int arr[] = inputArray(n); if(arr[0] == 1){ out.println("YES"); return; } for(int i = 1; i<n; ++i){ if(arr[i]%arr[0]!=0){ out.println("NO"); return; } } out.println("YES"); } private static int[] inputArray(int n) throws IOException { int arr[] = new int[n]; for(int i=0; i<n; ++i) arr[i] = scanInt(); return arr; } private static void displayArray(int arr[]){ for(int i = 0; i<arr.length; ++i) out.print(arr[i]+" "); out.println(); } public static void main(String[] args) { try { long startTime = System.currentTimeMillis(); in = new BufferedReader(new InputStreamReader(System.in)); out = new PrintWriter(System.out); //in = new BufferedReader(new FileReader("input.txt")); //out = new PrintWriter(new FileWriter("output.txt")); int test = scanInt(); for(int t=1; t<=test; t++){ // out.print("Case #"+t+": "); solve(); } long endTime = System.currentTimeMillis(); long totalTime = endTime - startTime; //out.println(totalTime+"---------- "+System.currentTimeMillis() ); in.close(); out.close(); } catch (Throwable e) { e.printStackTrace(); exit(1); } } static int scanInt() throws IOException { return parseInt(scanString()); } static long scanLong() throws IOException { return parseLong(scanString()); } static double scanDouble() throws IOException { return parseDouble(scanString()); } static String scanString() throws IOException { if (tok == null || !tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); } return tok.nextToken(); } static String scanLine() throws IOException { return in.readLine(); } private static void sort(int arr[]){ mergeSort(arr, 0, arr.length-1); } private static void sort(int arr[], int start, int end){ mergeSort(arr, start, end-1); } private static void mergeSort(int arr[], int start, int end){ if(start >= end) return; int mid = (start+end)/2; mergeSort(arr, start, mid); mergeSort(arr, mid+1, end); merge(arr, start, mid, end); } private static void merge(int arr[], int start, int mid, int end){ int n1 = mid - start+1; int n2 = end - mid; int left[] = new int[n1]; int right[] = new int[n2]; for(int i = 0; i<n1; ++i){ left[i] = arr[i+start]; } for(int i = 0; i<n2; ++i){ right[i] = arr[mid+1+i]; } int i = 0, j = 0, curr = start; while(i <n1 && j <n2){ if(left[i] <= right[j]){ arr[curr] = left[i]; ++i; } else{ arr[curr] = right[j]; ++j; } ++curr; } while(i<n1){ arr[curr] = left[i]; ++i; ++curr; } while(j<n2){ arr[curr] = right[j]; ++j; ++curr; } } }
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
1027062cb8625fe3a0f8c1e1a19693e5
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 s=new Scanner(System.in); int t=s.nextInt(); while(t-->0) { int n=s.nextInt(); int a[]=new int[n]; for(int i=0;i<n;i++) { a[i]=s.nextInt(); } int diff=a[0]; int f=0; for(int i=1;i<n;i++) { if(a[i]%diff==0) { } else { System.out.println("NO"); f=1; break; } } if(f==0) { 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
a39f3aed931373536ceb8443263df159
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 CodeForces2 { public static void main(String[] args) { // TODO Auto-generated method stub Scanner sc = new Scanner(System.in); int t = sc.nextInt(); for (int i = 0; i < t; i++) { int len = sc.nextInt(); int[] A = new int[len]; for (int j = 0; j < len; j++) { A[j] = sc.nextInt(); } boolean ans = differenceOperations(A); if (ans) { System.out.println("YES"); } else { System.out.println("NO"); } } } private static boolean differenceOperations(int[] A) { int count=0; for(int i = A.length-1 ; i>=1 ; i--) { if((A[i] % A[0]) == 0) { count++; } } if(count == A.length-1) { return true; } else { return false; } } }
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
aef938fb1d1aeda952ec3bf628cc0d2d
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 Srhossain{ public static void main(String[] args){ new Srhossain(); } public Srhossain(){ Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while(t-- > 0){ int n = sc.nextInt(); int x = sc.nextInt(); for(int i = n; i > 1; i--){ int y = sc.nextInt(); if((y%x) != 0){ n = -1; } } if(n == -1){ System.out.println("NO"); } else{ System.out.println("YES"); } } sc.close(); } public void showarr(int[] arr){ for(int i=0; i<arr.length; i++) System.out.print(arr[i]+" "); showln(""); } public void show(String s){ System.out.print(s); } public void showln(String s){ System.out.println(s); } }
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
8107ed6033d4591c20cd6d76eabeef69
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(); long[] nums = new long[n]; for(int i=0;i<n;i++) { nums[i] = sc.nextLong(); } boolean flag = true; for(int i=1;i<n;i++) { if(nums[i]%nums[0]!=0) { flag = false; break; } } if(flag==false) 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
81b086ac45bb03ae8f62402ea102decb
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.StringTokenizer; public class App { public static Scanner scan = new Scanner(System.in); public static void main(String[] args) throws Exception { int n = Integer.parseInt(scan.nextLine()); while (n > 0) { System.out.println(solve() ? "YES" : "NO"); n--; } } public static boolean solve() { StringTokenizer st = new StringTokenizer("a b"); int length = Integer.parseInt(scan.nextLine()); int arr[] = new int[length]; st = new StringTokenizer(scan.nextLine()); for (int i = 0; i < length; i++) { arr[i] = Integer.parseInt(st.nextToken()); if(i >= 1) { if (arr[i] % arr[0] != 0) { return false; } } } return true; } }
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
39a6b2d0b1158f46b197f64c48c2ff42
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 { public static void main(String[] args) throws java.lang.Exception { PrintWriter out = new PrintWriter(System.out); FastScanner sc = new FastScanner(); int tater = sc.nextInt(); while (tater-- > 0) { int n = sc.nextInt(); int a[] = new int[n]; boolean isIncreasing = true; for (int i = 0; i < n; i++) { a[i] = sc.nextInt(); if (a[i] %a[0] != 0){ isIncreasing = false; } } if (isIncreasing) { System.out.println("YES"); } else { System.out.println("NO"); } } out.flush(); } static class FastScanner { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st = new StringTokenizer(""); String next() { while (!st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } int[] readArray(int n) { int a[] = new int[n]; for (int i = 0; i < n; i++) { a[i] = nextInt(); } return a; } long nextLong() { return Long.parseLong(next()); } } }
Java
["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
7e0aba290e213d94c7deac64826d4744
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.io.*; import java.util.*; public class Main{ static final Random random = new Random(); static boolean[] primecheck; static ArrayList<Integer>[] adj; static int[] vis; static int[] parent; static int[] rank; static int[] fact; static int[] invFact; static int[] ft; static int mod = (int)1e9 + 7; public static void main(String[] args) { OutputStream outputStream = System.out; FastReader in = new FastReader(); PrintWriter out = new PrintWriter(outputStream); PROBLEM solver = new PROBLEM(); int t = 1; t = in.ni(); for (int i = 0; i < t; i++) { //out.print("Case #" + (i+1) + ": "); solver.solve(in, out); } out.close(); } static class PROBLEM { public static void solve(FastReader in, PrintWriter out) { int n = in.ni(), a[] = in.ra(n); for (int i = 1; i < n; i++) { if((a[i]/a[i-1])*a[i-1] == a[i]){ a[i] = a[i-1]; continue; } out.println("NO"); return; } out.println("YES"); } } static int findLower(int[] a, int x){ int l = 0, r = a.length-1; while(l<=r){ int mid = l + (r-l)/2; if(a[mid]>=x) r = mid-1; else l = mid+1; } return l; } static int findHigher(int[] a, int x){ int l = 0, r = a.length-1; while(l<=r){ int mid = l + (r-l)/2; if(a[mid]<=x) l = mid+1; else r = mid-1; } return r; } static void graph(int n, int e, FastReader in){ adj = new ArrayList[n+1]; vis = new int[n+1]; for (int i = 0; i < n + 1; i++) adj[i] = new ArrayList<>(); for (int i = 0; i < e; i++) { int a = in.ni(), b = in.ni(); adj[a].add(b); adj[b].add(a); } } static int[] dx = {0, 0, 1, -1, 0, 0}, dy = {1, -1, 0, 0, 0, 0}, dz = {0, 0, 0, 0, 1, -1}; static int lower_bound(int array[], int key) { int index = Arrays.binarySearch(array, key); // If key is not present in the array if (index < 0) { // Index specify the position of the key // when inserted in the sorted array // so the element currently present at // this position will be the lower bound return Math.abs(index) - 1; } // If key is present in the array // we move leftwards to find its first occurrence else { // Decrement the index to find the first // occurrence of the key // while (index > 0) { // // // If previous value is same // if (array[index - 1] == key) // index--; // // // Previous value is different which means // // current index is the first occurrence of // // the key // else // return index; // } return index; } } static int upper_bound(int arr[], int key) { int index = Arrays.binarySearch(arr, key); int n = arr.length; // If key is not present in the array if (index < 0) { // Index specify the position of the key // when inserted in the sorted array // so the element currently present at // this position will be the upper bound return Math.abs(index) - 1; } // If key is present in the array // we move rightwards to find next greater value else { // Increment the index until value is equal to // key // while (index < n) { // // // If current value is same // if (arr[index] == key) // index++; // // // Current value is different which means // // it is the greater than the key // else { // return index; // } // } return index; } } static int getKthElement(int k){ int l = 0, r = ft.length-1, el = Integer.MAX_VALUE; while(l <= r){ int mid = l+r>>1; if(sumFenwick(ft, mid) >= k){ el = Math.min(el, mid); r = mid-1; }else l = mid+1; } return el; } static int rangeFenwick(int[] ft, int i, int j){ return sumFenwick(ft, j) - sumFenwick(ft, i); } static int sumFenwick(int[] ft, int i){ int sum = 0; for(; i>0; i&=~(i&-i)) sum+=ft[i]; return sum; } static void addFenwick(int[] ft, int i, int x){ int n = ft.length; for(; i<n; i+=i&-i) ft[i] += x; } static void combi(){ fact = new int[200001]; fact[0] = 1; invFact = new int[200001]; invFact[0] = 1; for (int i = 1; i < fact.length; i++) { fact[i] = (int)(((long)fact[i-1]%mod * i%mod) % mod); invFact[i] = (mod - ((mod/i)*invFact[mod%i])%mod)%mod; } } static int nCk(int n, int k){ return fact[n] * invFact[k]%mod * invFact[n-k]%mod; } static String reverse(String s){ char[] ch = s.toCharArray(); for (int j = 0; j < ch.length / 2; j++) { char temp = ch[j]; ch[j] = ch[ch.length-j-1]; ch[ch.length-j-1] = temp; } return String.valueOf(ch); } static int[][] matrixMul(int[][] a, int[][] m){ if(a[0].length == m.length) { int[][] b = new int[a.length][m.length]; for (int i = 0; i < m.length; i++) { for (int j = 0; j < m.length; j++) { int sum = 0; for (int k = 0; k < m.length; k++) { sum += m[i][k] * m[k][j]; } b[i][j] = sum; } } return b; } return null; } static void dfs(int i, int d){ vis[i] = 1; for(int j : adj[i]){ if (vis[j] == 0) dfs(j, d+1); } } static int find(int u){ if(u == parent[u]) return u; return parent[u] = find(parent[u]); } static void union(int u, int v){ int a = find(u), b = find(v); if(a == b) return; if(rank[a] > rank[b]) parent[b] = a; else if(rank[a] < rank[b]) parent[a] = b; else{ parent[b] = a; rank[a]++; } } static void dsu(int n){ parent = new int[n]; rank = new int[n]; for (int i = 0; i < parent.length; i++) { parent[i] = i; rank[i] = 1; } } static boolean isPalindrome(char[] s){ boolean b = true; for (int i = 0; i < s.length / 2; i++) { if(s[i] != s[s.length-1-i]){ b = false; break; } } return b; } static void yn(boolean b, PrintWriter out){ if(b) out.println("YES"); else out.println("NO"); } static void pa(int[] a, PrintWriter out){ for (int j : a) out.print(j + " "); out.println(); } static void pa(long[] a, PrintWriter out){ for (long j : a) out.print(j + " "); out.println(); } public static void sortByColumn(int arr[][], int col) { // Using built-in sort function Arrays.sort Arrays.sort(arr, new Comparator<>() { @Override // Compare values according to columns public int compare(final int[] entry1, final int[] entry2) { // To sort in descending order revert // the '>' Operator if (entry1[col] > entry2[col]) return 1; else return -1; } }); // End of function call sort(). } static boolean isPoT(long n){ return ((n&(n-1)) == 0); } static long sigmaK(long k){ return (k*(k+1))/2; } static void swap(int[] a, int l, int r) { int temp = a[l]; a[l] = a[r]; a[r] = temp; } static int binarySearch(int[] a, int l, int r, int x){ if(r>=l){ int mid = l + (r-l)/2; if(a[mid] == x) return mid; if(a[mid] > x) return binarySearch(a, l, mid-1, x); else return binarySearch(a,mid+1, r, x); } return -1; } static long gcd(long a, long b) { if (b == 0) return a; return gcd(b, a % b); } static long lcm(long a, long b){ return (a / gcd(a, b)) * b; } static int ceil(int a, int b){ return (a+b-1)/b; } static long ceil(long a, long b){ return (a+b-1)/b; } static boolean isSquare(double a) { boolean isSq = false; double b = Math.sqrt(a); double c = Math.sqrt(a) - Math.floor(b); if (c == 0) isSq = true; return isSq; } static long fast_pow(long a, long b) { //Jeel bhai OP if(b == 0) return 1L; long val = fast_pow(a, b / 2); if(b % 2 == 0) return val * val % mod; else return val * val % mod * a % mod; } // 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){ // // int ans = Integer.compare(x, o.x); // if(o.x == x) ans = Integer.compare(y, o.y); // // return ans; // //// int ans = Integer.compare(y, o.y); //// if(o.y == y) ans = Integer.compare(x, o.x); //// //// return ans; // } // } static class Tuple implements Comparable<Tuple>{ int x, y, id; Tuple(int x, int y, int id){ this.x = x; this.y = y; this.id = id; } public int compareTo(Tuple o){ int ans = Integer.compare(x, o.x); if(o.x == x) ans = Integer.compare(y, o.y); return ans; } } public static class Pair<U extends Comparable<U>, V extends Comparable<V>> implements Comparable<Pair<U, V>> { public U x; public V y; public Pair(U x, V y) { this.x = x; this.y = y; } public int hashCode() { return (x == null ? 0 : x.hashCode() * 31) + (y == null ? 0 : y.hashCode()); } public boolean equals(Object o) { if (this == o) return true; if (o == null || getClass() != o.getClass()) return false; Pair<U, V> p = (Pair<U, V>) o; return (x == null ? p.x == null : x.equals(p.x)) && (y == null ? p.y == null : y.equals(p.y)); } public int compareTo(Pair<U, V> b) { int cmpU = x.compareTo(b.x); return cmpU != 0 ? cmpU : y.compareTo(b.y); } public int compareToY(Pair<U, V> b) { int cmpU = y.compareTo(b.y); return cmpU != 0 ? cmpU : x.compareTo(b.x); } public String toString() { return String.format("(%s, %s)", x.toString(), y.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 ni() { return Integer.parseInt(next()); } long nl() { return Long.parseLong(next()); } double nd() { return Double.parseDouble(next()); } char nc() { return next().charAt(0); } boolean nb() { return !(ni() == 0); } // boolean nextBoolean(){return Boolean.parseBoolean(next());} String nline() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } public int[] ra(int size) { int[] array = new int[size]; for (int i = 0; i < size; i++) array[i] = ni(); return array; } public long[] ral(int size) { long[] array = new long[size]; for (int i = 0; i < size; i++) array[i] = nl(); return array; } } static void ruffleSort(int[] a) { int n=a.length;//shuffle, then sort for (int i=0; i<n; i++) { int oi=random.nextInt(n), temp=a[oi]; a[oi]=a[i]; a[i]=temp; } Arrays.sort(a); } public static void Sieve(int n) { Arrays.fill(primecheck, true); primecheck[0] = false; primecheck[1] = false; for (int i = 2; i * i < n + 1; i++) { if (primecheck[i]) { for (int j = i * 2; j < n + 1; j += i) { primecheck[j] = false; } } } } }
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
785adad9371a578b8dcf9c8244c1618f
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 = sc.nextInt(); while(t>0){ t--; int n = sc.nextInt(); int a[]=new int[n]; int i=0; int x=0; while(i<n){ a[i]=sc.nextInt(); if(i>=1){ if(a[i]%a[0]!=0 ){ x=1; } } i++; } if(x==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 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
32a72040690da7382f118459a035d8c5
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 in = new Scanner(System.in); //OutputStream outputStream = System.out; //PrintWriter w = new PrintWriter(outputStream); int i,j; int t=in.nextInt(); for(i=0;i<t;i++) { int n,m, x,c=0,k; n=in.nextInt(); int a[]=new int[n]; for(j=0;j<n;j++) { a[j]=in.nextInt(); } for(j=n-1;j>=0;j--) { if(a[j]%a[0]!=0) { System.out.println("NO"); break; } else { if(j!=1) continue; else { System.out.println("YES"); break; } } } } } }
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
973e3af625085f09b175baa08f1e73d8
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 final Scanner in= new Scanner(System.in); static Long globalInt; public static String Solve(int n){ int[] array = new int[n]; for(int i=0; i<n; i++) array[i] = in.nextInt(); int first = array[0]; for(int i=1 ;i<n; i++) if(array[i]%first != 0) return "NO"; return "YES"; } public static void main(String[] args){ int cases=in.nextInt(); for(int i=0; i<cases; i++){ int n=in.nextInt(); System.out.println(Solve(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 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
28c0d9133017f52522b866324f114212
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 scanner = new Scanner(System.in); int t = scanner.nextInt(); while(t-- > 0){ int n = scanner.nextInt(); String check = "YES"; int[] arr = new int[n]; for(int i = 0 ; i < n ; i++){ arr[i] = scanner.nextInt(); if (arr[i] % arr[0] != 0) { check = "NO"; } } System.out.println(check); } } }
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
bef67e7f501994b01de72799fd1b8b40
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 and Editor */ import java.util.*; public class HelloWorld { 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(); } boolean ans = true; if(arr[0] == 1) { System.out.println("YES"); continue; } for(int i = 1; i<n-1; i++) { if(arr[i]%arr[i-1] == 0)arr[i] = arr[i-1]; else arr[i] = arr[i]%arr[i-1]; } for(int i = n-1; i> 0; i--) { arr[i] = arr[i]%arr[i-1]; if(arr[i] != 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 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
2034e4c1acddda24c9766eaa0ea37f6d
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
/*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(); int arr[] = s.readIntArr(n); int gcd = arr[0]; for(int i = 1; i < n; i++){ gcd = (int)gcd(gcd, arr[i]); } for(int i = 0; i < n; i++){ arr[i] = arr[i]/gcd; } if(arr[0] == 1){ println("YES"); }else{ println("NO"); } } /*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/ public static void main(String[] args) throws IOException { int test = s.nextInt(); for(int t = 1; t <= test; t++) { solve(); } out.close(); } } // public static boolean allsame(int arr[]){ // for(int i = 0; i < arr.length-1; i++){ // if(arr[i] != arr[i+1]) return false; // } // return true; // } // // public static List<List<Integer>> permutation(int arr[], int i){ // if(i == 0){ // List<List<Integer>> ans = new ArrayList<>(); // List<Integer> li = new ArrayList<>(); // li.add(arr[i]); // ans.add(li); // return ans; // } // int temp = arr[i]; // List<List<Integer>> rec = permutation(arr, i-1); // List<List<Integer>> ans = new ArrayList<>(); // for(List<Integer> li : rec){ // for(int j = 0; j <= li.size(); j++){ // List<Integer> list = new ArrayList<>(); // for(int ele: li){ // list.add(ele); // } // list.add(j, temp); // ans.add(list); // } // } // 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 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
1f0b29183f2cdb52b787683141766eb8
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 StringBuilder sb; static dsu dsu; static long fact[]; static int mod = (int) (1e9 + 7); static void print(int[]arr){ for(int i=0;i<arr.length;i++){ System.out.print(arr[i]+" "); } System.out.println(); } static void print(int t){ System.out.println(t); } static long get1(long val,long p){ for(long i=0;i<p;i++){ val=(long)val/2; } return val; } static int get(long t, ArrayList<Long>temp, long[][]arr,long n){ if(t<n)return (int)t; for(int i=1;i<temp.size();i++){ long l=temp.get(i-1); long r=temp.get(i)-1; if(t>=l&&t<=r){ long dis=t-l; return get(arr[i-1][0]+dis,temp,arr,n); } } return -1; } static void solve() { int n=i(); int[]arr=new int[n]; for(int i=0;i<n;i++)arr[i]=i(); for(int i=1;i<n;i++){ int dif=arr[i]-arr[i-1]; if(arr[i]%arr[0]!=0){ sb.append("NO\n"); return; } } sb.append("YES\n"); } public static void main(String[] args) { sb = new StringBuilder(); int test =i(); fact=new long[(int)1e6+10]; fact[0]=fact[1]=1; for(int i=2;i<fact.length;i++) { fact[i]=((long)(i%mod)*(long)(fact[i-1]%mod))%mod; } while (test-- > 0) { solve(); } System.out.println(sb); } //**************NCR%P******{************ static long ncr(int n, int r) { if (r > n) return (long) 0; long res = fact[n] % mod; // System.out.println(res); res = ((long) (res % mod) * (long) (p(fact[r], mod - 2) % mod)) % mod; res = ((long) (res % mod) * (long) (p(fact[n - r], mod - 2) % mod)) % mod; // System.out.println(res); return res; } static long p(long x, long y)// POWER FXN // { if (y == 0) return 1; long res = 1; while (y > 0) { if (y % 2 == 1) { res = (res * x) % mod; y--; } x = (x * x) % mod; y = y / 2; } return res; } //**************END****************** // *************Disjoint set // union*********// static class dsu { int parent[]; dsu(int n) { parent = new int[n]; for (int i = 0; i < n; i++) parent[i] = i; } int find(int a) { if (parent[a] ==a) return a; else { int x = find(parent[a]); parent[a] = x; return x; } } void merge(int a, int b) { a = find(a); b = find(b); if (a == b) return; parent[b] = a; } } //**************PRIME FACTORIZE **********************************// static TreeMap<Integer, Integer> prime(long n) { TreeMap<Integer, Integer> h = new TreeMap<>(); long num = n; for (int i = 2; i <= Math.sqrt(num); i++) { if (n % i == 0) { int nt = 0; while (n % i == 0) { n = n / i; nt++; } h.put(i, nt); } } if (n != 1) h.put((int) n, 1); return h; } //****CLASS PAIR ************************************************ static class Pair implements Comparable<Pair> { int i; int l; Pair(int i, int l) { this.i = i; this.l = l; } public int compareTo(Pair o) { return (int) (this.l - o.l); } } //****CLASS PAIR ************************************************** static class InputReader { private InputStream stream; private byte[] buf = new byte[1024]; private int curChar; private int numChars; private SpaceCharFilter filter; public InputReader(InputStream stream) { this.stream = stream; } public int read() { if (numChars == -1) throw new InputMismatchException(); if (curChar >= numChars) { curChar = 0; try { numChars = stream.read(buf); } catch (IOException e) { throw new InputMismatchException(); } if (numChars <= 0) return -1; } return buf[curChar++]; } public int Int() { int c = read(); while (isSpaceChar(c)) c = read(); int sgn = 1; if (c == '-') { sgn = -1; c = read(); } int res = 0; do { if (c < '0' || c > '9') throw new InputMismatchException(); res *= 10; res += c - '0'; c = read(); } while (!isSpaceChar(c)); return res * sgn; } public String String() { int c = read(); while (isSpaceChar(c)) c = read(); StringBuilder res = new StringBuilder(); do { res.appendCodePoint(c); c = read(); } while (!isSpaceChar(c)); return res.toString(); } public boolean isSpaceChar(int c) { if (filter != null) return filter.isSpaceChar(c); return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } public String next() { return String(); } public interface SpaceCharFilter { public boolean isSpaceChar(int ch); } } static class OutputWriter { private final PrintWriter writer; public OutputWriter(OutputStream outputStream) { writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream))); } public OutputWriter(Writer writer) { this.writer = new PrintWriter(writer); } public void print(Object... objects) { for (int i = 0; i < objects.length; i++) { if (i != 0) writer.print(' '); writer.print(objects[i]); } } public void printLine(Object... objects) { print(objects); writer.println(); } public void close() { writer.close(); } public void flush() { writer.flush(); } } static InputReader in = new InputReader(System.in); static OutputWriter out = new OutputWriter(System.out); public static long[] sort(long[] a2) { int n = a2.length; ArrayList<Long> l = new ArrayList<>(); for (long i : a2) l.add(i); Collections.sort(l); for (int i = 0; i < l.size(); i++) a2[i] = l.get(i); return a2; } public static char[] sort(char[] a2) { int n = a2.length; ArrayList<Character> l = new ArrayList<>(); for (char i : a2) l.add(i); Collections.sort(l); for (int i = 0; i < l.size(); i++) a2[i] = l.get(i); return a2; } public static long pow(long x, long y) { long res = 1; while (y > 0) { if (y % 2 != 0) { res = (res * x);// % modulus; y--; } x = (x * x);// % modulus; y = y / 2; } return res; } //GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ public static long gcd(long x, long y) { if (x == 0) return y; else return gcd(y % x, x); } // ******LOWEST COMMON MULTIPLE // ********************************************* public static long lcm(long x, long y) { return (x * (y / gcd(x, y))); } //INPUT PATTERN******************************************************** public static int i() { return in.Int(); } public static long l() { String s = in.String(); return Long.parseLong(s); } public static String s() { return in.String(); } public static int[] readArrayi(int n) { int A[] = new int[n]; for (int i = 0; i < n; i++) { A[i] = i(); } return A; } public static long[] readArray(long n) { long A[] = new long[(int) n]; for (int i = 0; i < n; i++) { A[i] = l(); } return A; } }
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
c4a26e733cfd20569f035a0bbbea3772
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 ColumnSwapping { 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(); } int flg=0; for(int j=1;j<n;j++) { if(arr[j]%arr[0]!=0) { flg=1; break; } } if(flg==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 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
20efdbdc6070cab162f96770fa7ad7a7
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 ans{ 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 val = arr[0]; String ans = "Yes"; for(int i:arr){ if(i%val!=0){ ans="No"; break; } } System.out.println(ans); } } } // int n = sc.nextInt(); // size of n array; // int[] arr = new int[n]; // for(int i=0;i<n;i++){ // arr[i] = sc.nextInt(); // } // int pa6i string levva // int a = sc.nextInt(); // sc.nextLine(); // String str = sc.nextLine();
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
33d5f7ee119ff33f4e4652d2ce6c116e
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 input=new Scanner (System.in); int t=input.nextInt(); for(int i=0;i<t;i++) { boolean flag=true; int n=input.nextInt(); int arr[]=new int[n]; for(int m=0;m<n;m++) { arr[m]=input.nextInt(); } int first=arr[0]; for(int m=1;m<n;m++) { if(arr[m]%first==0) continue; 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 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
29825166d49379de840d66b2309b783f
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 CF808 { public static void main(String[] args) { // TODO Auto-generated method stub Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while(t-->0) { int n = sc.nextInt(); int[] a = new int[n]; for(int i =0 ; i<n;i++) { a[i] = sc.nextInt(); } if(n==2) { if(a[1]%a[0]==0) { System.out.println("YES"); } else { System.out.println("NO"); } } else { boolean c = true; if(a[1]%a[0]!=0) { c = false; } for(int i = 2;i<n;i++) { if((a[i])%a[0]!=0) { c = false; break; } } if(c) { 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
d0718f6b47d6373198efdb191d1d4c27
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 long mod = 1000000007; static int[]parent; static int[]rank; static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out)); public static void main(String[] args) throws IOException { FastReader sc = new FastReader(); int t = sc.nextInt(); while( t-- > 0) { int n=sc.nextInt(); int[]arr=new int[n]; for(int i=0;i<n;i++){ arr[i]=sc.nextInt(); } boolean f=true; for(int i=1;i<n;i++){ if(arr[i]%arr[0]!=0){ f=false; break; } } if(f){ out.println("yes"); }else{ out.println("No"); } } out.flush(); } public static int find(int x){ if(parent[x]==x){ return x; }else{ int px=parent[x]; int lead=find(px); parent[x]=lead; return lead; } } public static class Pair{ long first; long second; Pair(long first,long second){ this.first=first; this.second=second; } } public static void union(int l1,int l2){ if(rank[l1]>rank[l2]){ parent[l2]=l1; rank[l1]++; }else if(rank[l2]>rank[l1]){ parent[l1]=l2; rank[l2]++; }else{ parent[l1]=l2; rank[l2]++; } } public static void reverse(Long[]arr,int i,int j){ while(i<=j){ long temp=arr[i]; arr[i]=arr[j]; arr[j]=temp; i++; j--; } } public static int gcd(int n1,int n2){ while(n1%n2!=0){ int rem=n2%n1; n2=n1; n1=rem; } return n1; } // static class Pair{ // int val; // int bd; // Pair(int val,int bd){ // this.val=val; // this.bd=bd; // } // } public static int binarySearch(int[]arr,int data){ int left = 0; int right = arr.length - 1; int ceil = -1; int floor = 0; while(left <= right){ int mid = (left + right) / 2; if(data > arr[mid]){ left = mid + 1; floor = arr[mid]; } else if(data < arr[mid]){ right = mid - 1; ceil = mid+1; } else { floor = arr[mid]; ceil = mid+1; break; } } return ceil; } /* * WARNING -> DONT EVER USE ARRAYS.SORT() IN ANY WAY. * A B C are easy just dont give up , you can do it! * FIRST AND VERY IMP -> READ AND UNDERSTAND THE QUESTION VERY CAREFULLY. * WARNING -> DONT CODE BULLSHIT , ALWAYS CHECK THE LOGIC ON MULTIPLE TESTCASES AND EDGECASES BEFORE. * SECOND -> TRY TO FIND RELEVENT PATTERN SMARTLY. * WARNING -> IF YOU THINK YOUR SOLUION IS JUST DUMB DONT SUBMIT IT BEFORE RECHECKING ON YOUR END. */ public static boolean ifpowof2(long n ) { return ((n&(n-1)) == 0); } static boolean isprime(long x ) { if( x== 2) { return true; } if( x%2 == 0) { return false; } for( long i = 3 ;i*i <= x ;i+=2) { if( x%i == 0) { return false; } } return true; } static boolean[] sieveOfEratosthenes(long n) { boolean prime[] = new boolean[(int)n + 1]; for (int i = 0; i <= n; i++) { prime[i] = true; } for (long p = 2; p * p <= n; p++) { if (prime[(int)p] == true) { for (long i = p * p; i <= n; i += p) prime[(int)i] = false; } } return prime; } public static int[] nextLargerElement(int[] arr, int n) { Stack<Integer> stack = new Stack<>(); int rtrn[] = new int[n]; rtrn[n-1] = -1; stack.push( n-1); for( int i = n-2 ;i >= 0 ; i--){ int temp = arr[i]; int lol = -1; while( !stack.isEmpty() && arr[stack.peek()] <= temp){ if(arr[stack.peek()] == temp ) { lol = stack.peek(); } stack.pop(); } if( stack.isEmpty()){ if( lol != -1) { rtrn[i] = lol; } else { rtrn[i] = -1; } } else{ rtrn[i] = stack.peek(); } stack.push( i); } return rtrn; } static void mysort(int[] arr) { for(int i=0;i<arr.length;i++) { int rand = (int) (Math.random() * arr.length); int loc = arr[rand]; arr[rand] = arr[i]; arr[i] = loc; } Arrays.sort(arr); } static void mySort(Long[] arr) { for(int i=0;i<arr.length;i++) { int rand = (int) (Math.random() * arr.length); long loc = arr[rand]; arr[rand] = arr[i]; arr[i] = loc; } Arrays.sort(arr); } static long gcd(long a, long b) { if (a == 0) return b; return gcd(b % a, a); } static long lcm(long a, long b) { return (a / gcd(a, b)) * b; } static long rightmostsetbit(long n) { return n&-n; } static long leftmostsetbit(long n) { long k = (long)(Math.log(n) / Math.log(2)); return k; } static HashMap<Long,Long> primefactor( long n){ HashMap<Long ,Long> hm = new HashMap<>(); long temp = 0; while( n%2 == 0) { temp++; n/=2; } if( temp!= 0) { hm.put( 2L, temp); } long c = (long)Math.sqrt(n); for( long i = 3 ; i <= c ; i+=2) { temp = 0; while( n% i == 0) { temp++; n/=i; } if( temp!= 0) { hm.put( i, temp); } } if( n!= 1) { hm.put( n , 1L); } return hm; } static ArrayList<Integer> allfactors(int abs) { HashMap<Integer,Integer> hm = new HashMap<>(); ArrayList<Integer> rtrn = new ArrayList<>(); for( int i = 2 ;i*i <= abs; i++) { if( abs% i == 0) { hm.put( i , 0); hm.put(abs/i, 0); } } for( int x : hm.keySet()) { rtrn.add(x); } if( abs != 0) { rtrn.add(abs); } return rtrn; } public static int[][] prefixsum( int n , int m , int arr[][] ){ int prefixsum[][] = new int[n+1][m+1]; for( int i = 1 ;i <= n ;i++) { for( int j = 1 ; j<= m ; j++) { int toadd = 0; if( arr[i-1][j-1] == 1) { toadd = 1; } prefixsum[i][j] = toadd + prefixsum[i][j-1] + prefixsum[i-1][j] - prefixsum[i-1][j-1]; } } return prefixsum; } 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 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
8926233817488462cbbb765242018537
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 edu130 { static class FastReader { BufferedReader br; StringTokenizer st; public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDouble() { return Double.parseDouble(next()); } String nextLine() { String str = ""; try { str = br.readLine().trim(); } catch (Exception e) { e.printStackTrace(); } return str; } } static class FastWriter { private final BufferedWriter bw; public FastWriter() { this.bw = new BufferedWriter(new OutputStreamWriter(System.out)); } public void print(Object object) throws IOException { bw.append("" + object); } public void println(Object object) throws IOException { print(object); bw.append("\n"); } public void close() throws IOException { bw.close(); } } public static void main(String[] args) { try { FastReader sc = new FastReader(); FastWriter out = new FastWriter(); 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 cc=false; for(int i=1;i<n;i++){ if(arr[i]%arr[0]!=0){ cc=true; break; } } if(cc){ if(arr[0]==1){ System.out.println("YES"); }else{ System.out.println("NO"); } }else{ System.out.println("YES"); } } }catch(Exception e){ return; } } private static void swap(long[] arr, int i, int ii) { long temp=arr[i]; arr[i]=arr[ii]; arr[ii]=temp; } public static int lcm(int a,int b){ return (a/gcd(a,b))*b; } private static int gcd(int a, int b) { if(b==0)return a; return gcd(b,a%b); } static class Pair { int a; int b; Pair(int a, int b) { this.a = a; this.b = b; } } }
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
9192a93e7d3506518b605fc2eae3bb38
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; public class Solution { public static String solve(int n, int[] a) { int f = a[0]; for(int i = 1; i<n; i++) { if(a[i]%f!=0) { return "NO"; } } return "YES"; } public static void main(String[] args) throws IOException { Reader r = new Reader(); BufferedWriter o = new BufferedWriter(new OutputStreamWriter(System.out)); int t = r.readI(); outer: for(int z = 1; z<=t; z++){ int n = r.readI(); int[] a = r.readArrI(); o.write(solve(n, a)+"\n"); o.flush(); } } } class Reader { BufferedReader r; Reader(){ r = new BufferedReader(new InputStreamReader(System.in)); } public int[] readArrI() throws IOException { String[] s = r.readLine().split(" "); int n = s.length; int[] a = new int[n]; for(int j=0; j<n;j++) { a[j] = Integer.parseInt(s[j]); } return a; } public long[] readArrL() throws IOException { String[] s = r.readLine().split(" "); int n = s.length; long[] a = new long[n]; for(int j=0; j<n;j++) { a[j] = Long.parseLong(s[j]); } return a; } public int readI() throws IOException { String[] s = r.readLine().split(" "); int n = Integer.parseInt(s[0]); return n; } public long readL() throws IOException { String[] s = r.readLine().split(" "); long n = Long.parseLong(s[0]); return n; } public String readS() throws IOException { String s = r.readLine(); return s; } }
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
31ce48398c68318a0547bbbe5958c5e1
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_1 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int T = sc.nextInt(); for(int f=1; f<=T; f++) { int n = sc.nextInt(); int arr[] = new int[n]; for(int i=0; i<n; i++) { arr[i] =sc.nextInt(); } boolean ans = true; for(int i=1; i<n; i++) { if(arr[i]%arr[i-1]!=0) { arr[i] = arr[i]%arr[i-1]; } } for(int i=1; i<n; i++) { if(arr[i]%arr[i-1]==0) { arr[i] = arr[i-1]; } } for(int i=1; i<n; i++) { if(arr[i]!=arr[i-1]) { 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
8082da9dfa7a5f3d191acfb3d3474abd
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 Prob1 { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out)); int testCases = Integer.parseInt(br.readLine()); while (testCases--!=0) { int n = Integer.parseInt(br.readLine()); StringTokenizer st = new StringTokenizer(br.readLine()); int[] nums = new int[n]; for (int i = 0; i < n; i++) { nums[i] = Integer.parseInt(st.nextToken()); } boolean yes = true; for (int i = 1; i < n; i++) { if (nums[i] % nums[0] != 0) { pw.println("NO"); yes = false; break; } } if (yes) { 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