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=;
 const int maxm=;
 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=;
     for(int i=;i<=n;i++){
         sum+=i;
     }
     if(sum%){
         cout<<""<<endl;
     }else{
         ans=sum/;
         for(int i=;i<=n;i++)
             dp[i][]=;
         for(int i=;i<=n;i++){
             for(int j=;j<=ans;j++)
                 dp[i][j]=dp[i-][j]+dp[i-][j-i];
         }
         cout<<dp[n][ans]<<endl;
     }
 }