1 solutions
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0
C :
#include <stdio.h> void main(){ int n; scanf("%d",&n); printf("%.1f%%",(2*n-1+4) * 100.0 / (n * n)); }
C++ :
#include<bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; int zonggong =n*n; int fense = 2*n+3; double r = (double)fense / zonggong * 100; cout<<fixed<<setprecision(1)<<r<<"%"; return 0; }
Python :
n=int(input()) s=(2*n+3)/(n*n)*100 print('{:.1f}%'.format(s))
- 1
Information
- ID
- 10654
- Time
- 1000ms
- Memory
- 16MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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