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:
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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:
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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:
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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:
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5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1
- 样例输出:
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13.333 31.500
#include <stdio.h> #include <stdlib.h> 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<n; ++i){ scanf ("%d %d", &trade[i].javabean, &trade[i].food); trade[i].rate = (double)trade[i].javabean / (double)trade[i].food; } qsort (trade, n, sizeof(FtoJ), compare); i = 0; sum = 0; while (m > 0 && i<n){ if (m >= 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; }