3 solutions

  • 1
    @ 2025-3-15 15:12:15
    #include<bits/stdc++.h>
    using namespace std;
    const int N=200010;
    vector<int> g[N];
    int main()
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n/i;j++)
            {
                g[i*j].push_back(i); //将i放到i的倍数后面 
            }
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d:",i);
            for(int j=0;j<g[i].size();j++)
            {
                printf("%d ",g[i][j]);
            }
            printf("\n");
        }
        return 0;
    }

    Information

    ID
    182
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    3
    Tags
    # Submissions
    89
    Accepted
    15
    Uploaded By