Reorder Codeforces Round #678 SOLUTION

A. Reorder Codeforces Round #678 SOLUTION

For a given array aa consisting of nn integers and a given integer mm find if it is possible to reorder elements of the array aa in such a way that ∑ni=1∑nj=iajj∑i=1n∑j=inajj equals mm? It is forbidden to delete elements as well as insert new elements. Please note that no rounding occurs during division, for example, 52=2.552=2.5.Input

The first line contains a single integer tt — the number of test cases (1≤t≤1001≤t≤100). The test cases follow, each in two lines.

The first line of a test case contains two integers nn and mm (1≤n≤1001≤n≤100, 0≤m≤1060≤m≤106). The second line contains integers a1,a2,…,ana1,a2,…,an (0≤ai≤1060≤ai≤106) — the elements of the array.Output

For each test case print “YES”, if it is possible to reorder the elements of the array in such a way that the given formula gives the given value, and “NO” otherwise.

Example

input

2
3 8
2 5 1
4 4
0 1 2 3

output

YES
NO

Note

In the first test case one of the reorders could be [1,2,5][1,2,5]. The sum is equal to (11+22+53)+(22+53)+(53)=8(11+22+53)+(22+53)+(53)=8. The brackets denote the inner sum ∑nj=iajj∑j=inajj, while the summation of brackets corresponds to the sum over ii.

Solution

Program C++:

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
 
 
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
 
  int t;
  cin >> t;
  while (t--) {
    int n, s = 0, w;
    cin >> n >> w;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      s += x;
    }
    if (s == w) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
 
}
#include <iostream>
 
using namespace std;
 
int main()
{
    int a[100], t, n, m, i, s, b;
    cin >> t;
    for (i = 0; i < t ; i++){
        cin>>n>>m;
        s = 0;
        for (int j = 0; j < n; j++) {
            cin >> b;
            s += b;
        }
        cout << (s == m ? "YES" : "NO") << endl;
    }
    return 0;
}

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