贪心-HDU 1009
描述：
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.
说明：
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元最多可以卖多少javabeans，其中每个房间里的javabeans可以被分割。
输入：
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.
输出：
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.
源码：
#include<stdio.h>
#include <stdlib.h>
//http://acm.hdu.edu.cn/showproblem.php?pid=1009
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);
}
return 0;
}