题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1009 FatMouse’ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 96673 Accepted Submission(s): 33689
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
题目大意：一共有n个房子，每个房子里有老鼠喜欢吃的javabeans,但是每个房间里的javabeans的价格不一样。老鼠用m元，问m元最多可以买多少javabeans，每个房间里的javabeans可以被分割。直接贪。 AC代码：
#include<bits/stdc++.h> typedef struct trade_s { //注意这里只能是整数 int javaBean, catFood; double normalize; } trade_t; static trade_t w[1100]; //用qsort函数必须注意这个函数返回的是int类型，所以这里会返回1和-1 int cmp(const void * x, const void * y) { //注意这里直接用减法并且返回1和-1 return ((trade_t*)x)->normalize - ((trade_t*)y)->normalize > 0 ? -1 : 1; } int main() { //n是行数，m是总重量 int n, m, i; double sum; while (scanf("%d %d", &m, &n)) { //输入-1结束 if (m == -1 && n == -1) break; //重置sum sum = 0; //输入n行数据 for (i = 0; i <= n - 1; i++) { //得到javaBean需要付出catFood scanf("%d %d", &w[i].javaBean, &w[i].catFood); //计算单价购买量 w[i].normalize = ((double)w[i].javaBean) / w[i].catFood; } //排序 qsort(w, n, sizeof(trade_t), cmp); for (i = 0; i < n; i++) { //如果支付得起，则购买 if (m >= w[i].catFood) { //printf("food %d %d\n", w[i].catFood, w[i].javaBean); //增加获得的重量 sum = sum + w[i].javaBean; //减去付出 m -= w[i].catFood; } else { //支付不起，以单价购买量乘以重量 sum = sum + w[i].normalize * m; break; } } printf("%.3lf\n", sum); cout<<fixed<<setprication(3)<<sum<<endl; } return 0; }