Subset Sums

链接

分析:dp[i][j]表示前i个数能够组成j的对数，可得dp[i][j]=dp[i-1][j]+dp[i-1][j-i]，所以最后dp[n][sum/2]既是所求

/*
    PROB:subset
    ID:wanghan
    LANG:C++
*/
#include "iostream"
#include "cstdio"
#include "cstring"
#include "string"
using namespace std;
const int maxn=50;
const int maxm=1000;
int a[maxn],n,ans,cnt;
int dp[maxn][maxm];
int main()
{
    freopen("subset.in","r",stdin);
    freopen("subset.out","w",stdout);
    cin>>n;
    int sum=0;
    for(int i=1;i<=n;i++){
        sum+=i;
    }
    if(sum%2){
        cout<<"0"<<endl;
    }else{
        ans=sum/2;
        for(int i=1;i<=n;i++)
            dp[i][0]=1;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=ans;j++)
                dp[i][j]=dp[i-1][j]+dp[i-1][j-i];
        }
        cout<<dp[n][ans]<<endl;
    }
}