首页 > 代码库 > HDU 1009 FatMouse' Trade题解
HDU 1009 FatMouse' Trade题解
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.
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 <stdio.h> #include <vector> #include <algorithm> 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<twoInts> 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; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。