FatMouse' Trade 贪心

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

示例输出

13.333
31.500

提示

hdoj1009

#include<stdio.h>
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<n; i++){
            scanf("%d%d", &a[i].contain, &a[i].demand);
            a[i].bi = a[i].contain *1.0 / a[i].demand;
        }
        quick_sort(a, 0, n-1);
        i = n-1;
        double amount = 0;
        while(m>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
", amount);
    }
    return 0;
}