ACM — FatMouse' Trade

Link

http://acm.hdu.edu.cn/showpro...

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

ACed Code

#include <stdio.h>

int main() {
    int a, b, k;
    double n[1000][3], temp;
    // a 總共錢數 b 房間數
    while (scanf("%d%d", &a, &b) != EOF && (a != -1) || (b != -1)) {
        int i = 0, j = 0;
        double sum = 0;
        for (i = 0; i < b; i++) {
            // 0 豆子總數 1 費用總數 2 單位價格
            scanf("%lf%lf", &n[i][0], &n[i][1]);
            n[i][2] = n[i][0] / n[i][1];
        }

        // 選擇排序法 以單位價格爲索引 最小的放最前
        for (i = 0; i < b - 1; i++) {
            k = i;
            for (j = i + 1; j < b; j++) {
                if (n[k][2] < n[j][2])
                    k = j;
            }
            for (int l = 0; l < 3; l++) {
                temp = n[k][l];
                n[k][l] = n[i][l];
                n[i][l] = temp;
            }

        }

        // 開始買豆子
        for (i = 0; i < b; i++) {
            if (a >= n[i][1]) {
                sum += n[i][0];
                a = a - n[i][1];
            } else {
                sum += a * n[i][2];
                break;
            }
        }

        printf("%.3f\n", sum);
    }
}

Reference

Jason Lee