杭电 hdu 1009

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32406    Accepted Submission(s): 10471 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   Author CHEN, Yue   Source ZJCPC2004   Recommend JGShining       贪心算法，直接套用模板     #include<iostream> #include<algorithm> #include<stdlib.h> #include<iomanip> using namespace std; struct trade {  int j,f;  double avg; }q[1008]; bool comp(trade x,trade y) {  return x.avg>y.avg; } int main() {  int n,m,i;  double sum;  while(cin>>m>>n&&n!=-1&&m!=-1)  {   for(i=0;i<n;i++){    cin>>q[i].j>>q[i].f;    q[i].avg=(double)q[i].j/q[i].f;    }   sort(q,q+n,comp);   sum=0;   i=0;   while(m!=0&&i<n)   {    if(m>=q[i].f)    {    m-=q[i].f;    sum+=q[i].j;     }    else{     sum+=m*q[i].avg;     m=0;    }    i++;   }   cout<<setiosflags(ios::fixed)<<setprecision(3)<<sum<<endl;  }  return 0; }