1 solutions
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0
C :
#include<stdio.h> void main() { int n; double m; scanf("%d",&n); n=n*1.0; m=(n-2.0)*180/n; printf("%.1f",m); }
C++ :
#include<bits/stdc++.h> using namespace std; int main(){ double n; cin>>n; cout<<fixed<<setprecision(1)<<((n-2)*180)/n<<endl; return 0; }
Python :
#!/usr/bin/env python3 n = int(input()) s = (n-2)*180 f = s/n print("%.1f"% f)
- 1
Information
- ID
- 10626
- Time
- 1000ms
- Memory
- 16MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By