题目描述

输入

输出

样例输入

样例输出

源代码

#include <stdlib.h>
#include <stdio.h>
int main()
{
    
    int m;
    int n;
    int k;
    int i;
    int a[1001],b[1001];
    double sum=0;
    double p[1001];  //
    double max = 0;
    
    while(scanf("%d %d",&m,&n)&&m!=-1&&n!=-1)
    {
    
        
        sum=0;  //
        for(i=0;i<n;i++)
        {
    
            scanf("%d %d",&a[i],&b[i]);
            p[i]=(double)a[i]/b[i];
        }
        while(m!=0&&n!=0)
        {
    
            
            max=0;
            for(i=0;i<n;i++)
            {
    
                if(p[i]>max)
                {
    
                    max=p[i];
                    k=i;
                }
            }
            if(m-b[k]>=0)
            {
    
                m=m-b[k];
                sum+=a[k];
                p[k]=0;
            }
            else
            {
    
                sum+=(double)p[k]*m;
                m=0;
            }
        }
        printf("%.3lf\n",sum);
    }
    
    return 0;
}

关于这题

For optimal solution 就要知道 在一个房间中 Meet one of the cat can get the number of cheese This is a key to the problem solving
m 磅 猫食
n 个 房间
a[i]Each room number of cheese b[i] The demand of cat
p[i] 在一个房间中 Meet one of the cat can get the number of cheese
k Marked which one room
max Compare which one room Meet a cat can get the maximum number of cheese
注：在计算除法时 要转double 类型