1 条题解

  • 0
    @ 2023-6-11 12:20:41

    C :

    #include<stdio.h>
    int main()
    {
        double m,n,t,s;
        scanf("%lf%lf%lf",&n,&m,&t);
        s=m+n+t;
        if(s>=200)
            printf("%.1lf",s*0.8);
        else
            printf("%.1lf",s*0.9);
        return 0;
    }
    
    

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    int main(){
        double a,b,c,r;
        cin>>a>>b>>c;
        r = a + b + c;
        if(r >= 200){
        	r = r * 0.8;
    	}else{
    		r = r * 0.9;
    	}
        cout<<fixed<<setprecision(1)<<r;
    }
     
    
    

    Java :

    import java.util.Scanner;
    
    public class Main {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    		Scanner sc = new Scanner(System.in);
    		double a = sc.nextDouble();
    		double b = sc.nextDouble();
    		double c = sc.nextDouble();
    		double sum;
    		sum = a + b + c;
    		double s;
    		if(sum < 200){
    			s = sum * 0.9;
    			System.out.println(String.format("%.1f", s));
    		}else if(sum >= 200){
    			s = sum * 0.8;
    			System.out.println(String.format("%.1f", s));
    		}
    	}
    
    }
    

    Python :

    n = input().split()
    a = float(n[0])
    b = float(n[1])
    c = float(n[2])
    su = a + b + c
    if su >= 200:
        print("%.1f" % (su * 0.8))
    else:
        print('%.1f' % (su * 0.9))
    
    • 1

    信息

    ID
    2569
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者