3 solutions

  • -1
    @ 2024-12-28 14:09:12

    可行的新方法

    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
          int a,b;
          cin>>a>>b;
          for(int i=a;i<=b;i++)
          {
                int t=0;
                for(int j=1;j<=i;j++)
                {
                      if(i%j==0)
                      {
                            t++;
                      }
                }
                if(t==2)//质数只有2个因数
                {
                      cout<<i<<endl;
                }
          }
          return 0;
    }
    

    Information

    ID
    55
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    3
    Tags
    (None)
    # Submissions
    131
    Accepted
    47
    Uploaded By