题目链接

题意：



题解：










AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int a[110];
bool dp[25010];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int _;cin>>_;
    while(_--){
        int n;cin>>n;
        for(int i=1;i<=n;i++)cin>>a[i];
        sort(a+1,a+1+n);
        memset(dp,0,sizeof dp);
        int ans=0;
        dp[0]=1;
        for(int i=1;i<=n;i++){
            if(!dp[a[i]])ans++;
            else continue;
            for(int j=a[i];j<=25000;j++)dp[j]|=dp[j-a[i]];
        }
        cout<<ans<<endl;
    }
    return 0;
}