1 solutions

  • 0
    @ 2025-3-3 16:28:34

    C :

    #include<stdio.h>
    int main()
    {
    	int a;
    	scanf("%d",&a);
    	printf("%.2f",(a+a/100+a%100)*1.0/3);
    	return 0;
    }
    

    C++ :

    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	int n;
    	cin>>n;
    	cout<<fixed<<setprecision(2)<<(n+n/100+n%100)/3.0;
    }
    
    

    Java :

    import java.util.Scanner;
    
    public class Main {
    	public static void main(String[] args) {
    		Scanner sc = new Scanner(System.in);
    		int x = sc.nextInt();
    		int s = x / 100 ;
    		int g = x % 100;
    		double sum = (double)(x + s + g) / 3;
    		System.out.println(String.format("%.2f", sum));
    	}
    }
    

    Python :

    n=int(input())
    a=n//1000
    b=n//100%10
    c=n//10%10
    d=n%10
    print("%.2f"%((n+(a*10+b)+(c*10+d))/3))
    
    • 1

    Information

    ID
    10189
    Time
    1000ms
    Memory
    64MiB
    Difficulty
    (None)
    Tags
    # Submissions
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