hdoj1009FatMouse' Trade(贪心）

Problem Description

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.

 

Input

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.

 

Output

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.

 

Sample Input

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

 

Sample Output

13.33331.500

 

Author

CHEN, Yue

 

题意：

康康准备了 M 斤的食物, 准备跟舍长交换哲学之宝 ♂

舍长有 N 个房间. 第 i 个房间有 J[i] 的 ♂ 需要 F[i] 斤的食物. 康康可以不换完整个房间的♂ ,

他可以用 F[i]* a% 的食物 换 J[i]* a% 的 ♂ 

现在给你一个实数 M 问你康康最多能获得多少的 ♂

代码如下：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{double a;double b;double c;}arr[1100];bool cmp(node x,node y){return x.c>y.c;}int main(){int m,i;double n;double sum;while(scanf("%lf%d",&n,&m)!=EOF){if(n==-1&&m==-1)break;sum=0;for(i=0;i<m;i++){scanf("%lf %lf",&arr[i].a,&arr[i].b);arr[i].c=arr[i].a/arr[i].b;}sort(arr,arr+m,cmp);for(i=0;i<m;i++){if(n>=arr[i].b){sum+=arr[i].a;n-=arr[i].b;}else{sum+=n*arr[i].c;n=0;//刚开始wa是因为这一行没加.}}printf("%.3lf\n",sum);}}