1 solutions

  • 1
    @ 2024-11-12 17:43:02
    #include<bits/stdc++.h>
    using namespace std;
    int main(){
        int n,k,a=0,b=0,n1=0,n2=0;cin>>n>>k;
        for(int i=1;i<=n;i++){
            if(i%k==0) {a+=i;n1++;}
            else {b+=i;n2++;}
        }
        printf("%.1lf %.1lf",a*1.0/n1,b*1.0/n2);
        return 0;
    }
    
    • 1

    Information

    ID
    423
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    5
    Tags
    (None)
    # Submissions
    90
    Accepted
    35
    Uploaded By