基础贪心题 HDOJ 1007

Posted 2020-10-03

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 80535    Accepted Submission(s): 27842

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

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1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 int main()
 6 { 
 7 int M,N;
 8 double sum;
 9 int J[1005],F[1005];
10 double JF[1005];
11 while(cin>>M>>N)
12 {
13 if(M==-1&&N==-1)
14 break;
15 sum=0;
16 for(int i=1;i<=N;i++)
17 {
18 cin>>J[i]>>F[i];
19 JF[i]=1.0*J[i]/F[i]; //给个1.0很重要，光定义整形是不行的 
20 } 
21 for(int i=1;i<N;i++)
22 {
23 for(int j=i+1;j<=N;j++)
24 {
25 if(JF[i]<JF[j])
26 {    
27 double t; //当初一个傻逼在这一天都找不到错误。    
28 t=J[i];J[i]=J[j];J[j]=t;
29 t=F[i];F[i]=F[j];F[j]=t;
30 t=JF[i];JF[i]=JF[j];JF[j]=t;//是根据Jf排序，这个当然要交换拉 ,
31 } //否则影响后面的排序。