usaco Subset Sums

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动态规划， dp[j] 表示1-N中任意挑选几个元素加起来和为 j 的种类数，所以dp[j]=∑dp[ j - i ]，初始化dp[1]=1;

code

/*ID: yueqiqLANG: C++TASK: subset*/#include <set>#include <map>#include <ctime>#include <queue>#include <cmath>#include <stack>#include <limits>#include <vector>#include <bitset>#include <string>#include <cstdio>#include <cstring>#include <fstream>#include <string.h>#include <iostream>#include <algorithm>#define ls rt<<1#define rs rt<<1|1#define Si set<int>#define LL long long#define pb push_back#define PS printf(" ")#define Vi vector<int>#define LN printf("\n")#define SD(a) scanf("%d",&a)#define PD(a) printf("%d",a)#define SET(a,b) memset(a,b,sizeof(a))#define FF(i,a) for(int i(0);i<(a);i++)#define FD(i,a) for(int i(a);i>=(1);i--)#define FOR(i,a,b) for(int i(a);i<=(b);i++)#define FOD(i,a,b) for(int i(a);i>=(b);i--)#define readf freopen("subset.in","r",stdin)#define writef freopen("subset.out","w",stdout)const double pi = acos(-1.0);const int maxn = 50;const int BigP = 99999999;const int INF = 99999999;const int dx[]={0,1,0,-1};const int dy[]={1,0,-1,0};using namespace std;int N;int dp[1600];int main (){    readf;    writef;    SD(N);    int lim=(1+N)*N/2;    if(lim&1){        puts("0");        return 0;    }else{        dp[1]=1;        FOR(i,2,N) FOD(j,lim/2,i) dp[j]+=dp[j-i];        PD(dp[lim/2]);LN;    }return 0;}