1 solutions

  • 0
    @ 2025-3-3 16:21:29

    C++ :

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #define ll long long
    #define N 10000000 + 10
    using namespace std;
    ll n,m,dp[N]; 
    int main()
    {
        scanf("%lld",&n);
        m=(n*(n+1))/2;
        if(m%2)
        {
            cout<<0;
            return 0;
        }
        m/=2;
        dp[0]=1;
        for(int i=1;i<=n;i++)
            for(int j=m;j>=i;j--)//数组压缩
                dp[j]+=dp[j-i];
        cout<<dp[m]/2;//一定注意最后除以2
        return 0;
    }
    
    • 1

    Information

    ID
    9986
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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