HDU 1009 FatMouse' Trade（经典贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34320    Accepted Submission(s): 11228

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

 

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JGShining

先讲讲题目意思。讲的是老鼠为了到仓库里里偷吃仓粮（因为有猫守着）为猫准备了M镑猫食，仓库里有N个房间，每个房间里有一只猫，但是如果老鼠想吃每i-th个房间里的f份量仓粮的话，要给猫j份量的猫食，但是老鼠可以不吃完，按照仓粮乘以A%比例吃。。。问最后老鼠拿着M份猫粮最多能吃到多少仓粮. 

看到题目当时就懵了，A%是啥意思。。以为是输入值（但是也没有输入）一直未懂   

解题思路:    A%即是第I-TH房间老鼠能吃得到的仓粮和要给猫的猫粮比值即是。。=F/J   对所有房间的比值进行从大到排序即可  然后从大依次加上每个房间的能吃到的仓粮直至到N个房间并上猫食没有了即可，在这里处里如果剩于的M镑猫食少于当前房间需要的猫食，所以这里就只能取到M*（F/J）镑的猫粮了。。。。。。。

下面贴上自己的代码（为了熟悉结构体排序，采用多种排序方式，都AC过）

#include<iostream>#include<algorithm>#include<iomanip>using namespace std;struct mouse{int f;int j;double v;}cat[1000];//用于qsort比较函数int  cmp(const void *a,const void *b){struct mouse *aa=(mouse*)a;struct mouse *bb=(mouse*)b;   if(aa->v<bb->v)//为降序的话要改为小于   return 1;   else return -1;}//用于STL中的比较函数 >表示从大到小,<表示从小到大bool cmp1(struct mouse a,struct mouse b){   return a.v>b.v;}int main(){int m,n;while(cin>>m>>n){        if(m==-1&&n==-1)break;//输入仓粮和猫粮for(int i=0;i<n;i++){cin>>cat[i].f>>cat[i].j;cat[i].v=(double)cat[i].f/cat[i].j;}//整个冒泡排序试试  /*    for(i=1;i<=n;i++){for(int j=0;j<n-i;j++){if(cat[j].v<cat[j+1].v){double t=cat[j].v;  cat[j].v=cat[j+1].v;  cat[j+1].v=t;int k=cat[j].f;  cat[j].f=cat[j+1].f;  cat[j+1].f=k;  k=cat[j].j;  cat[j].j=cat[j+1].j;  cat[j+1].j=k;}}}*/        //整个qsort试试/* qsort(cat,n,sizeof(cat[0]),cmp);*///整个STL里面的QOSORT试试sort(cat,cat+n,cmp1);//MOUSE最多能吃的仓粮for(double sum=i=0;m&&i<n;i++){if(cat[i].j<=m){                sum+=cat[i].f;   m-=cat[i].j;}else{                sum+=m*cat[i].v;m=0;}}//输出MOUSE 最多能吃的仓粮cout<<fixed<<setprecision(3)<<sum<<endl;}return 0;}