题目1433：FatMouse

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

代码：

#include <stdio.h>#include <algorithm>using namespace std;struct Fuck {    float j; //javabean    float f; //cat food} r[1001];bool cmp(Fuck a, Fuck b) {    return (a.j/a.f) > (b.j/b.f);}int main() {    int n,i;    float m;    while(scanf("%f %d",&m,&n)!=EOF) {        if(m==-1&&n==-1)            break;        for(i=0;i<n;i++)            scanf("%f %f",&r[i].j,&r[i].f);        sort(r,r+n,cmp);        i = 0;        float sum = 0.0;        while(m>r[i].f) {            sum += r[i].j;            m -= r[i].f;            i++;        }        sum += m * (r[i].j/r[i].f);        printf("%.3f\n",sum);    }    return 0;}

在自己的机器上测试都正常，不过提交时结果是: Runtime Error!

？？？？？