HDUOJ1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15808Accepted Submission(s): 4746

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

Author

CHEN, Yue

#include<iostream> #include<algorithm> using namespace std; struct fat { //定义结构体 int x, y; double k; } p[1001]; bool cmp(fat a, fat b) { return a.k > b.k; } int main() { int i, j, m, n; while (cin >> m >> n, m + n >= 0) { for (i = 0; i < n; i++) { cin >> p[i].x >> p[i].y; p[i].k = p[i].x * 1.0 / p[i].y; } sort(p, p + n, cmp); // 排序 double sum = 0, tot = 0; for (i = 0; i < n; i++) { sum += p[i].x; tot += p[i].y; if (tot == m) break; if (tot > m) { tot -= p[i].y; sum -= p[i].x; sum += (((m - tot) * 1.0 / p[i].y) * p[i].x); break; } } printf("%.3lf\n", sum); } return 0; }