1 条题解

  • 0
    @ 2023-6-11 12:22:23

    C :

    #include<stdio.h>
    int main(){
    	int n;
    	double m,s=0.0;
    	scanf("%d%lf",&n,&m);
    	if(n<=40) s=n*m;
    	if(n>40 && n<=50) s=40*m+(n-40)*1.5*m;
    	if(n>50) s=40*m+15*m+(n-50)*2*m;
    	printf("%.2f",s);
    	return 0;
    }
    

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    int main(){
       int i;
       cin>>i;
       double s,t,a;
       cin>>a;
       if(i <= 40){
       	 t = i * a;
       }else if(i > 40 && i <= 50){
       	 s = i - 40;
       	 t = a * 40 + a * 1.5 * s;
       }else if(i > 50){
       	 s = i - 50;
       	 t = a * 40 + a * 1.5 * 10 + a * 2 * s;
       }
       cout<<fixed<<setprecision(2)<<t<<endl;
       return 0;
    }
    

    Java :

    import java.util.Scanner;
    public class Main{
    public static void main (String[] args){
    	Scanner sc = new Scanner(System.in);
    		int n = sc.nextInt();			
    		double m = sc.nextDouble();
    		    double s = 0.0;
    		    if(n <= 40) {
    		        s = n * m;
    		    }else if(n <= 50 && n > 40) {
    		        s = m * 40 + (n - 40) * m * 1.5;
    		    }else if(n > 50) {
    		        s = m * 40 + 10 * m * 1.5 + (n - 50) * 2 * m;
    		    }
    		    System.out.println(String.format("%.2f",s));
    		
    	 }
    		
    				
         }
    
    • 1

    信息

    ID
    2730
    时间
    1000ms
    内存
    512MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者