1 solutions

  • 0
    @ 2025-3-3 16:33:39

    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
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