题解

暴力枚举真好用awa

#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n,a[10001],s1=0,s2=0;
        bool b=0;
        cin>>n;
        for(int i=1;i<=n;i++) cin>>a[i];
        for(int i=1;i<=n;i++){
            for(int j=1;j<=i;j++) s1+=a[j];
            for(int j=i+1;j<=n;j++) s2+=a[j];
            if(s1==s2){
                cout<<"Yes\n";
                b=1;
                break;
            }else s1=0,s2=0;
        }if(!b) cout<<"No\n";
    }
    return 0;
}

如果不想暴力枚举，请移步下面的前缀和法

#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n,a[10001]={0};
        long long s[10001]={0};
        bool b=0;
        cin>>n;
        for(int i=1;i<=n;i++) cin>>a[i];
        for(int i=1;i<=n;i++) s[i]=s[i-1]+a[i];
        for(int i=1;i<=n-1;i++){
            if(s[i]==s[n]-s[i]){
                cout<<"Yes\n";
                b=1;
                break;
            }
        }if(!b) cout<<"No\n";
    }
    return 0;
}

还想要更好的解法，请到这个地方：
https://www.bilibili.com/video/BV1hq4y1s7VH/?spm_id_from=333.337.search-card.all.click