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
贪心法水题，
个人觉得这句话难理解：he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food
这样表达可以购买几分之几的。
#include
#include
#include
using namespace std; struct twoInts { int j, f; bool operator<(const twoInts two) const { double a = (double)j / (double)f; double b = (double)two.j / (double)two.f; return a > b; } }; int main() { int M, N; while (scanf("%d %d", &M, &N) && -1 != M) { vector
vt(N); for (int i = 0; i < N; i++) { scanf("%d", &vt[i].j); scanf("%d", &vt[i].f); } sort(vt.begin(), vt.end()); double maxBean = 0.0; for (int i = 0; i < N; i++) { if (M >= vt[i].f) { maxBean += vt[i].j; M -= vt[i].f; } else { maxBean += (double)vt[i].j * M / (double)vt[i].f; break; } } printf("%.3lf\n", maxBean); } return 0; }