HDU 1009 FatMouse' Trade（贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 76068    Accepted Submission(s): 26058

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

题意：我们要用M磅猫粮去换豆子，分别能用F [ i ] * a % 磅猫粮换到 J  [ i ]  * a %磅豆子，问我们最多可以换多少豆子

#include<bits/stdc++.h>using namespace std;const int N = 1000 + 10;struct xx{    int val,vol;    double p; ///p为性价比} a[N];int cmp(xx a, xx b){    return a.p>b.p;}int main(){    int m,n;    while(scanf("%d%d",&m,&n)==2)    {        if(m==-1&&n==-1)            break;        for(int i=0; i<n; i++)        {            scanf("%d%d",&a[i].val,&a[i].vol);            a[i].p=(double)a[i].val/a[i].vol;        }        sort(a,a+n,cmp);        double ans=0,sum=0;        for(int i=0; i<n; i++)        {            if(sum+a[i].vol<=m)            {                sum+=a[i].vol;                ans+=a[i].val;            }            else            {                ans+=(m-sum)*a[i].p;                break;            }        }        printf("%.3f\n",ans);    }}