HDU-1009-FatMouse 'trade

 

FatMouse 'trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 33514 Accepted Submission (s): 10902

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
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

 

 

Code:

# Include "cstdio" # include "cstdlib" # include "algorithm" # include "iostream" using namespace std; typedef struct changeinfo {int j; int f;} changeinfo; // The structure of the room information, bool cmp (changeinfo a, changeinfo B) {return 1.0 *. j/. f> 1.0 * B. j/B. f; // multiply by 1.0 to make it double} int main () {int m, n, I, k; double sum; changeinfo a [1001]; while (scanf ("% d", & m, & n), (m + 1) & (n + 1) {sum = 0; for (I = 0; I <n; I ++) scanf ("% d", & a [I]. j, & a [I]. f); sort (A, a + n, cmp); // sort the values from large to small Based on the division result, the parameters are the first and last addresses involved in sorting and the cmp comparison function // for (I = 0; I <n; I ++) // printf ("% d \ n", a [I]. j, a [I]. f); if (m = 0) // =! The question must be considered in both cases before AC {for (I = 0; I <n; I ++) if (a [I]. f = 0) sum = sum + a [I]. j;} if (n = 0) sum = 0; if (m! = 0 & n! = 0) {for (I = 0, k = m; I <n, k> = 0; I ++) {if (a [I]. f> k) // sum = (1.0 * a [I]. j/a [I]. f) * k + sum; else sum = a [I]. j + sum; k = k-a [I]. f ;}} printf ("%. 3lf \ n ", sum);} return 0 ;}