码迷,mamicode.com
首页 > 其他好文 > 详细

hdu 1009 FatMouse' Trade

时间:2016-08-17 01:24:10      阅读:168      评论:0      收藏:0      [点我收藏+]

标签:

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67291    Accepted Submission(s): 22911


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

 

Sample Output
13.333
31.500
 
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 using namespace std;
 5 typedef struct {
 6     int j, f;
 7     double aver;
 8 }food;
 9 
10 bool cmp(food a, food b){
11     return a.aver >= b.aver;
12 }
13 
14 food javabean[1005];
15 int main(){
16     int m, n;
17     while(scanf("%d%d", &m, &n) != EOF){
18         if((m == -1) && (n == -1))
19             break;
20         
21         for(int i = 0; i < n; i++){
22             scanf("%d %d", &javabean[i].j, &javabean[i].f);
23             javabean[i].aver = (double)javabean[i].j / javabean[i].f;
24         }
25         sort(javabean, javabean+n, cmp);
26         double sum = 0;
27         for(int i = 0; i < n; i++){
28             if(m >= javabean[i].f){
29                 sum += javabean[i].j;
30                 m -= javabean[i].f;
31             }
32             else {
33                 sum += (double)javabean[i].j / javabean[i].f * m;
34                 break;
35             }
36         }
37         printf("%.3lf\n", sum);
38     }
39     return 0;
40 }

 

hdu 1009 FatMouse' Trade

标签:

原文地址:http://www.cnblogs.com/qinduanyinghua/p/5778321.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!