hdu 1009 FatMouse' Trade(贪心)

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500
 
#include <iostream>#include <stdlib.h>#include <iomanip>using namespace std;struct catfood{ int a; int b; double c;};int main(){ int M,N; while(cin>>M>>N) {  if(M==-1&&N==-1)   break;  struct catfood *p,temp;  p=(struct catfood*)malloc(sizeof(struct catfood)*N);  int i,j,k;  for(i=0;i<N;i++)  {   cin>>p[i].a>>p[i].b;   p[i].c=1.0*p[i].a/p[i].b;  }  for(i=0;i<N-1;i++)  {   k=i;   for(j=i+1;j<N;j++)    if(p[j].c>p[k].c)     k=j;   temp=p[k];   p[k]=p[i];   p[i]=temp;  }  double sum=0;  for(i=0;i<N;i++)  {   if(M>=p[i].b)   {    sum=sum+p[i].a;    M=M-p[i].b;   }   else   {    sum=sum+M*1.0*p[i].a/p[i].b;    M=0;   }   if(M==0)    break;  }  cout<<setiosflags(ios::fixed)<<setprecision(3);  cout<<sum<<endl; } return 0;}