九度OJ 1433 FatMouse -- 贪心算法
创始人
2024-02-10 04:46:14
0

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;
}

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