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