题解|#FatMouse's Trade#（简单贪心）

FatMouse's Tradehttps://oj.youdao.com/problem/658?from=problems

题目描述

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.

The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

输入描述

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

输出描述

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

#include<iostream>
#include<algorithm>
using namespace std;

struct javabean{
    int weight;
    int cost;
};

javabean s[10010];

bool compare(javabean a,javabean b){
    return (1.0*a.weight/a.cost>1.0*b.weight/b.cost);
}

int main(){
    int M,N;
    while(scanf("%d %d",&M,&N)!=EOF){
        
        if(M<0&&N<0)break;
        for(int i=0;i<N;i++)scanf("%d %d",&s[i].weight,&s[i].cost);
        sort(s,s+N,compare);
        double sum=0;
        for(int i=0;i<N;i++){
            if(M>=s[i].cost){
                sum+=s[i].weight;
                M-=s[i].cost;
            }
            else {
                sum+=M*1.0/s[i].cost*s[i].weight;
                break;
            }
        }
        printf("%.3lf\n",sum);
        
    }
    return 0;
}