01背包确定方案数量（板子）
ti

 #include<bits/stdc++.h>using namespace std;int n;
int a[25];
int dp[45][45];int main(){cin>>n;for(int i = 1; i <= n; i ++ ){cin>>a[i];}//注意前i个物品凑体积为0的方案数都为1 for(int i = 0; i <= 20; i ++ ) dp[i][0] = 1;for(int i = 1; i <= n; i ++ ){  for(int j = 0; j <= 40; j ++ ){  if(j < a[i]) dp[i][j] = dp[i - 1][j];  //当第i个物品加不进去时 else dp[i][j] = dp[i - 1][j] + dp[i - 1][j - a[i]];  //当第i个物品能加进去时 }}cout<<dp[n][40]<<endl;return 0;}