九度OJ 1433 FatMouse -- 贪心算法

题目地址：http://ac.jobdu.com/problem.php?pid=1433

题目描述：

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.

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.

样例输入：

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

样例输出：

13.333
31.500

#include 
#include 
 
typedef struct fj{
    int food;
    int javabean;
    double rate;
}FtoJ;
 
int compare(const void * p, const void * q){
    FtoJ * p1 = (FtoJ *)p;
    FtoJ * q1 = (FtoJ *)q;
    if (p1->rate - q1->rate > 0) return -1;
    if (p1->rate - q1->rate == 0) return 0;
    if (p1->rate - q1->rate < 0) return 1;
}
 
int main(void){
    int m, n;
    FtoJ trade[1000];
    int i;
    double sum;
 
    while ((scanf ("%d %d", &m, &n) != EOF) && (m != -1) && (n != -1)){
        for (i=0; i 0 && i= trade[i].food){
                sum += (double)trade[i].javabean;
                m -= trade[i].food;
            }
            else{
                sum += (double)trade[i].javabean * ((double)m / (double)trade[i].food);
                m = 0;
            }
            ++i;
        }
        printf ("%.3lf\n", sum);
    }
 
    return 0;
}