技術標籤：競賽

B. Fair Division
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Alice and Bob received n candies from their parents. Each candy weighs either 1 gram or 2 grams. Now they want to divide all candies among themselves fairly so that the total weight of Alice’s candies is equal to the total weight of Bob’s candies.

Check if they can do that.

Note that candies are not allowed to be cut in half.

Input
The first line contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.

The first line of each test case contains an integer n (1≤n≤100) — the number of candies that Alice and Bob received.

The next line contains n integers a1,a2,…,an — the weights of the candies. The weight of each candy is either 1 or 2.

It is guaranteed that the sum of n over all test cases does not exceed 105.

Output
For each test case, output on a separate line:

“YES”, if all candies can be divided into two sets with the same weight;

Example
inputCopy
5
2
1 1
2
1 2
4
1 2 1 2
3
2 2 2
3
2 1 2
outputCopy
YES
NO
YES
NO
NO
Note
In the first test case, Alice and Bob can each take one candy, then both will have a total weight of 1.

In the second test case, any division will be unfair.

In the third test case, both Alice and Bob can take two candies, one of weight 1 and one of weight 2.

In the fourth test case, it is impossible to divide three identical candies between two people.

In the fifth test case, any division will also be unfair.

本題題意很明顯，本題就是對於一系列給定的重為1或2的糖果是否能進行均分，本題的坑點就是兩個重為1的糖果剛好可以抵消重為2的糖果的影響

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int count1=0,count2=0;
        int n;
        int x;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>x;
            if(x&1)
               count1++;
            if(x==2)
               count2++;
        }
        if(count2%2==0)
        {
            if(count1%2==0)
              cout<<"YES"<<endl;
            else
            {
                cout<<"NO"<<endl;
            }
            
        }
        else
            {
                count1-=2;//容易忽略的點
                if(count1>=0&&count1%2==0)
                   cout<<"YES"<<endl;
                else
                   cout<<"NO"<<endl;
            }
        
    }
}