Fatmouse 'trade (hangdian 1009)

Fatmouse 'tradetime limit: 2000/1000 ms (Java/other) memory limit: 65536/32768 K (Java/other) total submission (s): 1 accepted submission (s): 1 Font: times New Roman | verdana | georgiafont size: Regular → problem descriptionfatmouse 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.
Inputthe 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.
Outputfor 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 37 24 35 220 325 1824 1515 10-1 -1

Sample output

13.33331.500

// This question is a simple greedy question. It means that a mouse has a cat grain in the M-listed list. Data [I] is sent to the cat food. B can be exchanged for data [I]. a. How much food can I redeem at most. // Simply sort the values of data [I]. A/data [I]. B in ascending order to solve the problem. # Include <stdio. h> struct st {double A; Double B; double C;} data [1000]; int main () {int I, j, M, N; struct st data [1000], T; while (scanf ("% d", & M, & N) & (M! =-1 | n! =-1) {double sum = 0.000; for (I = 0; I <n; I ++) {scanf ("% lf ", & Data [I]. a, & Data [I]. b) ;}for (I = 0; I <n; I ++) {for (j = I + 1; j <n; j ++) {If (data [I]. a/data [I]. b) <data [J]. a/data [J]. b) {T = data [I]; data [I] = data [J]; data [J] = T ;}}for (I = 0; I <N; I ++) {If (m-data [I]. B >=0.001) {sum + = data [I]. a; m-= data [I]. b;} else {sum = sum + M * Data [I]. a/data [I]. b; break ;}} printf ("%. 3lf \ n ", sum);} return 0 ;}

Fatmouse 'trade (hangdian 1009)