f[N][M]在前i个物品中选,重量恰好是j的次数,无非就是0,1 二维存的是左右的差值
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 110;
const int M = 2e5 + 10;
int n, m;
int v[N];
int f[N][M];//前i个物品中选,重量为j的次数
signed main()
{
cin >> n;
for (int i = 1; i <= n; i++)cin >> v[i], m += v[i];
f[0][0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 0; j <= m; j++)
f[i][j] = f[i - 1][j] | f[i - 1][abs(j - v[i])] | f[i - 1][j + v[i]];
int res = 0;
for (int i = 1; i <= m; i++)
if (f[n][i])res++;
cout << res << endl;
return 0;
}