FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 97478    Accepted Submission(s): 33916

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  

題意：給出貓糧的數量，用貓糧作為貨幣買java(?)豆，在豆子充足的情況下，有多少錢就能買多少豆子

題解：根據單價從小到大排序，先買便宜的。寫得有點亂。。。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 struct node {
 4     int weight;//豆子數量 
 5     int mao;//貓糧數量 
 6     double weight1;//單價 
 7 } a[1005];
 8 bool cmp(node x,node y) {
 9     return x.weight1>=y.weight1;
10 }
11 void init() {
12     for(int i=0; i<1005; i++) {
13         a[i].weight=0;
14         a[i].weight1=0;
15         a[i].mao=0;
16     }
17 }
18 int main() {
19     int n;
20     double m;
21     while(~scanf("%lf %d",&m,&n)) {
22         if(n==-1&&m==-1)break;
23         init();
24         for(int i=0; i<n; i++) {
25             scanf("%d %d",&a[i].weight,&a[i].mao);
26             a[i].weight1=a[i].weight*1.0/a[i].mao;
27         }
28         sort(a,a+n,cmp);
29         double ans=0;
30         for(int i=0; i<n; i++) {
31             if(a[i].mao<=m) {
32                 ans+=a[i].weight;
33                 m-=a[i].mao;
34             } else {
35                 ans+=m*a[i].weight1;
36                 break;
37             }
38         }
39         printf("%.3lf\n",ans);
40     }
41     return 0;
42 }