Algorithm
You are given an array a consisting of n integers.
In one move, you can choose two indices 1≤i,j≤n such that i≠j and set ai:=aj. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace ai with aj).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 2) sum of elements.
You have to answer t independent test cases.
The first line of the input contains one integer t (1≤t≤2000) — the number of test cases.
The next 2t lines describe test cases. The first line of the test case contains one integer n (1≤n≤2000) — the number of elements in a. The second line of the test case contains n integers a1,a2,…,an (1≤ai≤2000), where ai is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2000 (∑n≤2000).
For each test case, print the answer on it — "YES" (without quotes) if it is possible to obtain the array with an odd sum of elements, and "NO" otherwise.
5 2 2 3 4 2 2 8 8 3 3 3 3 4 5 5 5 5 4 1 1 1 1
YES NO YES NO NO
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int x;
int odd = 0;
for(int i = 0; i < n; i++){
cin>>x;
(x & 1) ? odd++ : odd += 0;
}
(((odd == n) || (odd == 0) )&& !(odd & 1)) ? cout<<"NO"<<endl : cout<<"YES"<<endl;
}
}
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Input
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
NO
YES
NO
NO
Demonstration
Codeforcess solution 1296-A A. Array with Odd Sum C++, Java, Js and Python, 1296-A,Codeforcess solution