【杭电 1009】 FatMouse' Trade

FatMouse’ Trade

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<algorithm>#include<string.h>using namespace std;struct node{    double a,b,c;}st[10000+11]={0};int cmp(node i,node j){    return i.c<j.c;}int main(){    int n,i;    double m;    while(scanf("%lf%d",&m,&n)!=EOF&&(n!=-1||m!=-1))    {        for(i=0;i<n;i++)        {        scanf("%lf%lf",&st[i].a,&st[i].b);        st[i].c=st[i].b/st[i].a;        }        sort(st,st+n,cmp);        double sum=0;        for(i=0;i<n;i++)        {            if(m>=st[i].b)            {                sum+=st[i].a;                m-=st[i].b;            }            else            {                sum+=m/st[i].c;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}