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最新推荐文章于 2021-09-02 16:17:36 发布

翻译最新推荐文章于 2021-09-02 16:17:36 发布 · 273 阅读

文章标签：

#javabeans #homework #struct #each #input #output

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

Sample Output

  
   13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

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#include<stdio.h>
#include<stdlib.h>

#define size 1001

int m;
int n;

double max;

struct room {
    double j;
    double f;
    double price;
} room[size];//房间结构，j为JavaBeans的数量，f为猫食，price为JavaBeans的价格

int cmp(const void*p1, const void*p2) {
    return ((struct room *)p2)->price < ((struct room *)p1)->price?1:-1;
}//快排函数，price小的放在前面

int main(void) {
    while (scanf("%d%d", &m, &n), m + 1) {
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &room[i].j, &room[i].f);
            room[i].price = room[i].f / room[i].j;
        }
        qsort(room, n, sizeof (room[0]),cmp);//对room数组进行排序
        max = 0;
        for (int i = 0; i < n; i++) {
            if (m > room[i].f) {
                max += room[i].j;
                m -= room[i].f;
            } else {
                max += m / room[i].price;
                break;
            }
        }
        printf("%.3lf\n", max);
    }
    return 0;
}