1 条题解

  • 0
    @ 2025-10-10 15:45:27

    C :

    #include <stdio.h>
    #include <stdlib.h>
    
    /* run this program using the console pauser or add your own getch, system("pause") or input loop */
    
    void main(int argc, char *argv[]) {
    	int a[10];
    	int i,n,y;
    	double s;
    	s=0;
    	scanf("%d",&n);
    	for(i=0;i<n;i++){
    		scanf("%d",&a[i]);
    	}
    	for(i=0;i<n;i++){
    		s=s+a[i];
    		
    	}
    
    	for(i=0;i<n;i++){ 
    	  if(s>100&&a[i]==a[n-1])
    			printf("%.2f",(s-100)*0.9+100);
    	}
    	}
    
    

    C++ :

    #include <iostream>
    #include <iomanip>
    #include <cmath>
    
    using namespace std;
    
    int main(){
    	int a[100],n,s=0,m,x;
    	cin>>n;
    	for(int i=0;i<n;i++){
    		cin>>a[i];
    		s+=a[i];
    	}
    	cout<<fixed<<setprecision(2)<<100+((s-100)*0.9)<<endl;
    	
    }
    

    Pascal :

    var
    
    s:real;
    
    n,m,i:longint;
    
    begin
    
    readln(n);
    
    for i:=1 to n do
    
    begin
    
    read(m);
    
    s:=s+m;        //读入并累加
    
    end;
    
    if s>100 then s:=s-(s-100)+(s-100)*0.9;
    
    writeln(s:0:2);
    
    end.
    
    

    Java :

    import java.util.Scanner;
    
    public class Main {
    	public static void main(String args[]){
    		Scanner sc = new Scanner(System.in);
    		int n=sc.nextInt();
    		int[] a=new int[n];
    		for(int i=0;i<n;i++){
    			a[i]=sc.nextInt();
    		}
    		double s=0.00;
    		for(int i=0;i<n;i++){
    			s+=a[i];
    		}
    		if(s<=100){
    			System.out.println(String.format("%.2f", s));
    		}else{
    			System.out.println(String.format("%.2f", (s-100)*0.9+100));
    		}
    	}
    	
    }
    
    

    Python :

    n=int(input())
    l1=list(map(int,input().split()))
    def qiuHe(n):
        s=0
        for i in l1:
            s+=i
        return s
    s=qiuHe(l1)
    if s>100:
        s=(s-100)*0.9+100
        print('%.2f'% s)
    else:
        print(s)
    
    • 1

    信息

    ID
    155
    时间
    1000ms
    内存
    512MiB
    难度
    10
    标签
    递交数
    3
    已通过
    3
    上传者