学步园

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: yueqiq
LANG: 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;
}