acm之贪心算法题目10

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 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output

13.333
31.500

代码：

#include<stdio.h>#include<stdlib.h>#include<algorithm>using namespace std;#define Max 1000struct room{    double j,f;    double rate;}Room[Max];bool cmp(room a,room b){    return a.rate>b.rate;}int main(){    int N;     double M,sumJ;    int i;    while(scanf("%lf %d",&M,&N))    {        if(M==-1&&N==-1)            break;        for(i=0;i<N;i++)        {            scanf("%lf%lf",&Room[i].j,&Room[i].f);            Room[i].rate=(double)Room[i].j/Room[i].f;        }        sort(Room,Room+N,cmp);         sumJ=0;        for(i=0;i<N;i++)        {            if(M>=Room[i].f)            {                sumJ+=Room[i].j;                M-=Room[i].f;            }            else            {                sumJ+=(Room[i].rate)*M;                break;            }        }        printf("%.3lf\n",sumJ);    }    return 0;}