FatMouse‘ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103515 Accepted Submission(s): 36159
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
Author
CHEN, Yue
Source
题意：
这一道题意思就是老鼠用猫食物换取自己最喜爱的食物javaBean的过程，当然换取的最终结果是保证最后的JavaBean是最多的，
而且是当自己手中的猫食物小于每个仓库所需交换的猫食物时候，可以手中有多少就交换多少。
所以在解这道题时候要想到按照每个仓库javaBean最大的比率排序才能保证最后的交换的javaBean是最大的。
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 using namespace std;
5 const int maxn = 1e3+10;
6 int m,n;
7 struct node{
8 int j,f;
9 double r;
10 } food[maxn];
11 class cmp{
12 public:
13 bool operator()(node a,node b)const{
14 return a.r>b.r;
15 }
16 };
17 int main(){
18 while(~scanf("%d%d",&m,&n)){
19 if(m==-1) break;
20 for(int i=0; i<n; i++ ){
21 cin>>food[i].j>>food[i].f;
22 food[i].r=1.0*food[i].j/food[i].f;
23 }
24 sort(food,food+n,cmp());
25 double ans=0;
26 for( int i=0; i<n; i++ ){
27 if(m>=food[i].f){
28 ans+=food[i].j;
29 m-=food[i].f;
30 }
31 else{
32 if(m==0) break;
33 else{
34 ans+=m*food[i].r;
35 break;
36 }
37 }
38 }
39 printf("%.3f\n",ans);
40 }
41 return 0;
42 }
FatMouse' Trade（杭电ACM---1009）
原文：https://www.cnblogs.com/Bravewtz/p/10520251.html
