FatMouse' Trade

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.

示例输入

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

示例输出

13.33331.500

提示

#include
    
     struct bean{    int contain, demand;    double bi;}a[1000], b;void quick_sort(bean s[], int l, int r){    if(l < r){        int i=l, j=r;        double x=s[l].bi;        b = s[l];        while(i < j){            while(i < j && s[j].bi >= x)  j--;            if(i < j)  s[i++] = s[j];            while(i < j && s[i].bi < x)  i++;            if(i < j)  s[j--] = s[i];        }        s[i] = b;        quick_sort(s, l, i-1);        quick_sort(s, i+1, r);    }}int main(){    int m, n;    while(scanf("%d%d", &m, &n) && m!= -1 && n!= -1){        int i;        for(i=0; i
     
      0 && i>=0) {            if(m - a[i].demand >= 0) {                amount += a[i].contain;                m -= a[i].demand;                i--;            }            else {                amount += m * a[i].bi;                break;            }        }        printf("%.3lf\n", amount);    }    return 0;}