HDU 1.3.1 FatMouse' Trade_e:fatmouse的交易-优快云博客

本文链接：https://blog.youkuaiyun.com/hwj_wayne/article/details/50848908

本篇介绍了FatMouse如何通过最优策略用猫粮交换仓库中的JavaBeans。面对多个房间不同比例的交换条件，需要通过算法确定最大收益。文章提供了一个具体的编程实现方案。

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FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4921 Accepted Submission(s): 1476

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

Sample Output

13.333
31.500

#include <stdio.h>  
#include <stdlib.h>  
int main()  
{  
int M, N;  
while (scanf("%d%d", &M, &N) != EOF)  
{  
int i, j, J[1000], F[1000], temp1;  
double ratio[1000], JavaBeans = 0, temp;  
if ((M == -1) && (N == -1))  
{  
return 0;  
}  
for (i = 0; i<N; i++)  
{  
scanf("%d%d", &J[i], &F[i]);  
ratio[i] = (J[i] * 1.0) / F[i];  
}  
for (i = 0; i<N; i++)  
{  
for (j = 0; j<N - i - 1; j++)  
{  
if (ratio[j]<ratio[j + 1])  
{  
temp = ratio[j];  
ratio[j] = ratio[j + 1];  
ratio[j + 1] = temp;  
temp1 = F[j];  
F[j] = F[j + 1];  
F[j + 1] = temp1;  
temp1 = J[j];  
J[j] = J[j + 1];  
J[j + 1] = temp1;  
}  
}  
}  
for (i = 0; i<N; i++)  
{  
if (M >= F[i])  
{  
JavaBeans = J[i] + JavaBeans;  
M = M - F[i];  
}  
else  
{  
JavaBeans = JavaBeans + ratio[i] * M;  
break;  
}  
}  
printf("%.3lf\n", JavaBeans);  
}  
return 0;  
}