You are given an array aa consisting of nn integers.
In one move, you can choose two indices 1≤i,j≤n1≤i,j≤n such that i≠ji≠j and set ai:=ajai:=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 ii and jj and replace aiai with ajaj).
Your task is to say if it is possible to obtain an array with an odd (not divisible by 22) sum of elements.
You have to answer tt independent test cases.
Input
The first line of the input contains one integer tt (1≤t≤20001≤t≤2000) — the number of test cases.
The next 2t2t lines describe test cases. The first line of the test case contains one integer nn (1≤n≤20001≤n≤2000) — the number of elements in aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤20001≤ai≤2000), where aiai is the ii-th element of aa.
It is guaranteed that the sum of nn over all test cases does not exceed 20002000 (∑n≤2000∑n≤2000).
Output
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.
Example
Input
5
2
2 3
4
2 2 8 8
3
3 3 3
4
5 5 5 5
4
1 1 1 1
Output
YES
NO
YES
NO
NO
思路：在偶数的时候只要不是全奇数或者全偶数就行了。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=2e3+100;
int a[maxx];
int n;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int sum=0;
int sumj=0;
int sumo=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
if(a[i]%2) sumj++;
else sumo++;
}
if(sum&1) cout<<"YES"<<endl;
else if(sumj&&sumo) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
努力加油a啊，(o)/~
