【九度】题目1433：FatMouse

题目1433：FatMouse

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：3692

解决：1635

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：

5 37 24 35 220 325 1824 1515 10-1 -1

样例输出：

13.33331.500

// ConsoleApplication6.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <algorithm>using namespace std;struct Info{double weight;double value;double vpw;bool operator <(const Info&A)const{return  A.vpw<vpw;}}buf[1000];int _tmain(int argc, _TCHAR* argv[]){  double m;int n;while (scanf("%lf%d",&m,&n)&&m!=-1&&n!=-1){for(int i =0;i<n;i++){scanf("%lf%lf",&buf[i].weight,&buf[i].value);buf[i].vpw=buf[i].weight/buf[i].value;}sort(buf,buf+n);int indx=0;double total=0;while (m>0&&indx<n){if(m>buf[indx].value){total+=buf[indx].weight;m-=buf[indx].value;}else{total+=m*buf[indx].vpw;break;}
//简单的贪心算法。
indx++;}printf("%.3lf\n",total);//double保留三位小数。}return 0;}