方法不唯一哟~~~

$so$ $easy$

代码：

#include <bits/stdc++.h>
using namespace std;
int a[15];
double sum;
int main(){
	for(int i=1;i<=10;i++){
        cin>>a[i];
        switch(i){
            case 1:
                sum+=a[i]*28.9;
                break;
            case 2:
                sum+=a[i]*32.7;
                break;
            case 3:
                sum+=a[i]*45.6;
                break;
            case 4:
                sum+=a[i]*78;
                break;
            case 5:
                sum+=a[i]*35;
                break;
            case 6:
                sum+=a[i]*86.2;
                break;
            case 7:
                sum+=a[i]*27.8;
                break;
            case 8:
                sum+=a[i]*43;
                break;
            case 9:
                sum+=a[i]*56;
                break;
            case 10:
                sum+=a[i]*65;
                break;
        }
    }
    printf("%0.1lf",sum);
}

c++

#include<bits/stdc++.h>
using namespace std;
int main()
{
	double a[10]={28.9,32.7,45.6,78,35,86.2,27.8,43,56,65}; 
	double x,s=0;
	for(int i=0;i<10;i++) 
	{
		cin>>x;  
		s+=a[i]*x;
	}
	printf("%.1lf\n",s); 
	return 0;
}

c++

#include <iostream>	
#include <cstdio>	
using namespace std;
int main()
{
	double a[10]={28.9,32.7,45.6,78,35,86.2,27.8,43,56,65};//价格存放 
	double x,s=0;
	for(int i=0;i<10;i++) 
	{
		cin>>x; //每次输入本数 
		s+=a[i]*x;//对应本数乘对应价格 累加 
	}
	printf("%.1lf\n",s);//保留一位小数输出 
	return 0;
}