USACO-Section2.2 Subset Sums

2017-9-1

题目描述

将1到n总共n个数分成两个数总和相同的集合，求出所有的种数

解答

一看到直接深搜，共2^n次方，剪枝只到36就超时了，最后用动归

代码

/*ID: 18795871PROG: subsetLANG: C++*/#include<iostream>#include<fstream> #include<cstring>using namespace std;const int N = 800;ifstream fin("subset.in");ofstream fout("subset.out");long dp[N+1][N+1];int main(){    int i,j,n;    int sum;    fin>>n;    sum=(n)*(n+1)/2;    if (n==1||n==2||sum%2!=0){        fout<<"0"<<endl;        return 0;    }    sum/=2;dp[0][0]=1;    for (i=1;i<=n;i++){        for (j=1;j<=sum;j++){            dp[i][j]=dp[i-1][j-i]+dp[i-1][j];        }    }    fout<<dp[n][sum]<<endl;    return 0;}

一开始写的超时代码

/*ID: 18795871PROG: subsetLANG: C++*/#include<iostream>#include<fstream> #include<cstring>using namespace std;ifstream fin("subset.in");ofstream fout("subset.out");int n;long cnt=0;int sum;void dfs(int m,long s1,long s2){    if (m<=0) return ;    if (s1+m*(m+1)/2<sum/2||s2+m*(m+1)/2<sum/2) return ;    if (s1==sum/2||s2==sum/2){        cnt++;        return ;    }    if (s1+m<=sum/2) dfs(m-1,s1+m,s2);    if (s2+m<=sum/2)dfs(m-1,s1,s2+m);    return ;}int main(){    fin>>n;    sum=(n)*(n+1)/2;    if (n==1||n==2||sum%2!=0){        fout<<"0"<<endl;        return 0;    }    dfs(n,0,0);    fout<<cnt/2<<endl;    return 0;}