HDU

HDU - 1009FatMouse' Trade

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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

Source ZJCPC2004

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #define MAX 1010
 4 
 5 struct food{
 6     double j, f, a;
 7 };
 8 
 9 food fd[MAX];
10 
11 int comp(const void *a, const void *b) {
12     food *_a = (food *)a;
13     food *_b = (food *)b;
14     if(_a->a > _b->a)
15         return -1;
16     else if(_a->a < _b->a)
17         return 1;
18     else
19         return 0;
20 }
21 
22 int main()
23 {
24     double m = 0, ans = 0;
25     int n = 0;
26     while(scanf("%lf%d", &m, &n) != EOF) {
27         if(m == -1 && n == -1) break;
28         for(int i = 0; i < n; i++) {
29             scanf("%lf%lf", &fd[i].j, &fd[i].f);
30             fd[i].a = fd[i].j/fd[i].f;
31         }
32 
33         qsort(fd, n, sizeof(food), comp);
34 
35         for(int i = 0; i < n && m > 0; i++) {
36             if(m >= fd[i].f) {
37                 ans += fd[i].j;
38                 m -= fd[i].f;
39             }
40             else {
41                 ans += m*fd[i].a;
42                 m = 0;
43                 break;
44             }
45         }
46         printf("%.3lf
", ans);
47         m = 0; n = 0; ans = 0;
48     }
49 
50     return 0;
51 }