题目1433：FatMouse 九度OJ_综合

题目1433：FatMouse

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：4875

解决：2159

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：

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：

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.

样例输入：

样例输出：

13.333
31.500

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;struct goods{double pound;double price;double ratio;
}goods[1005];
bool cmp(struct goods g1,struct goods g2){return g1.ratio<g2.ratio;
}
int main()
{   int m,n;while(scanf("%d%d",&m,&n)!=EOF){if(m==-1&&n==-1)break;for(int i=0;i<n;i++){cin>>goods[i].pound>>goods[i].price;goods[i].ratio=goods[i].price/goods[i].pound;//cout<<"goods["<<i<<"].ratio="<<goods[i].ratio<<endl;}sort(goods,goods+n,cmp);//for(int i=0;i<n;i++){//	cout<<goods[i].pound<<"  "<<goods[i].price<<endl;//}int house=0;//从房间0开始访问double javabeans=0.0;while(m&&house!=n){if(goods[house].price<=m){//手里的钱足够购买当前房间里所有的食品javabeans+=goods[house].pound;m-=goods[house].price;}else if(goods[house].price>m){javabeans+=goods[house].pound*(m/goods[house].price);m=0;}house++;}printf("%.3lf\n",javabeans);}return 0;
}

贪心算法。