题目链接：A. Array with Odd Sum

题目

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.

输入

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).

输出

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.

样例输入

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

样例输出

YES
NO
YES
NO
NO

题意

你可以把数列里的一个数的值变成数列里另一个数的值，随便变，变几次都行。然后如果你能使这 n 个数的和为奇数，那你就成功了。如果你不能把数列变得求和为奇数，那就输出 NO 。

解题思路

其实计算一下原数列中奇数和偶数的个数即可。
要我们得出和为奇数，其实就是需要有 “奇数个奇数” ，所以我们只需要判断三种情况：

1. 有奇数个奇数 - 那就说明总和肯定是奇数；
2. 有奇数有偶数 - 轮到这种情况说明奇数肯定是偶数个，这种情况下我们只需要把一个偶数变成奇数（或者把一个奇数变成偶数）即可，所以也是满足要求的；【顺便其实只要有奇有偶，那就肯定可以利用偶数来获得奇数个奇数
3. 如果上面两种情况都不满足，那就说明这个数列 “全为奇数而且总共偶数个” 或者 “全为偶数” ，显然是怎么变都徒劳无功的，所以输出 NO 。

代码

#include <iostream>

using namespace std;

int main() {
        ios::sync_with_stdio(0); cin.tie(0);
        int t; cin >> t;
        while (t--) {
                int n; cin >> n;
                int o = 0, e = 0;
                while (n--) {
                        int a; cin >> a;
                        if (a % 2) o += 1;
                        else e += 1;
                }
                if (o % 2) cout << "YES" << endl;
                else if (o && e) cout << "YES" << endl;
                else cout << "NO" << endl;
        }
        return 0;
}