1 solutions
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0
C :
#include<stdio.h> void main(){ int x; double p; scanf("%d",&x); p=0.9*x; printf("%.1f",p); }
C++ :
#include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; cout<<fixed<<setprecision(1)<<(n+n*0.8+n*0.9) / 3; }
Java :
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); double x = sc.nextDouble(); double a = x * 0.8; double b = x * 0.9; double p =(double) (x + a + b) / 3; System.out.println(String.format("%.1f", p)); } }
Python :
a=int(input()) b=a*0.8 c=a*0.9 print("%.1lf"%((a+b+c)/3))
- 1
Information
- ID
- 10184
- Time
- 1000ms
- Memory
- 64MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By