贪心专题 HDU

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 77018    Accepted Submission(s): 26456

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，这里有一个奇葩的老鼠和一群奇葩的猫，老鼠喜欢吃javabean，但是他手里只有猫咪爱 吃的食物。每一个猫咪看守一部分javabean，允许老鼠拿他爱吃的食物来换，多少食物换多javabean，不同的猫咪要求的食物及提供的javabean不一样，让你算算老鼠最大可以换多少javabean。

贪心题，老鼠肯定是先换取最实惠的，他那么聪明

ac代码（提交使用G++的时候，要记得double输出使用%f）

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;struct node{    int jav,foo;    double ave;}nn[1005];bool cmp(node x,node y){    return x.ave > y.ave;}int m,n;int main(){    while(~scanf("%d%d",&m,&n),(m!=-1 || n!=-1)){        for(int i=1;i<=n;i++){            scanf("%d%d",&nn[i].jav,&nn[i].foo);            nn[i].ave=nn[i].jav*1.0/nn[i].foo;        }        sort(nn+1,nn+1+n,cmp);        double ans=0;        for(int i=1;i<=n;i++){            if(m > nn[i].foo){                ans+=nn[i].jav;                m=m-nn[i].foo;            }            else if(m > 0){                ans+=nn[i].ave*m;                break;            }            else                break;        }        printf("%.3f\n",ans);    }    return 0;}