HDU 1009 FatMouse' Trade【贪心】

/*
中文题目 胖老鼠的贿赂
中文翻译-大意 话说胖老鼠准备了M磅猫食，准备去贿赂猫警官，去换取它喜欢的咖啡豆。在仓库里面有N个房间，每个房间里面有J磅咖啡豆和F磅猫食。胖老鼠如果不能得到仓库里面的所有咖啡豆，它就选择能兑换多少是多少，让你求出老鼠最多能兑换多少咖啡豆
解题思路：将用猫食兑换咖啡豆的性价比排序。如果先排猫食，如果猫食相同按降序排咖啡豆是错误的，因为要使所获得的咖啡豆最多，要从性价比上面分析
难点详解：将性价比c排序，然后计算就可以了
关键点：排序，贪心
解题人：lingnichong
解题时间：2014-09-01 16:53:45
解题体会：排序，还要考虑怎样排序才是对的
*/

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 43633    Accepted Submission(s): 14588

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

 

#include<stdio.h>#include<algorithm>using namespace std;struct node{    int a;    int b;    double c;}arr[1100]; int cmp(node x,node y) {     return x.c > y.c; }     int main(){    int m,n;    int i,j;    double sum;    while(~scanf("%d%d",&m,&n))    {        if(m==-1&&n==-1)   break;        for(i=0;i<n;i++)        {            scanf("%d%d",&arr[i].a,&arr[i].b);            arr[i].c=(double)arr[i].a/arr[i].b;        }            sort(arr,arr+n,cmp);              //  puts("");//        for(i=0;i<n;i++)//        printf("%d %d\n",arr[i].a,arr[i].b);//        puts("");                sum=0;        for(i=0;i<n;i++)        {            if(m >= arr[i].b)            sum+=arr[i].a;        else        sum+=m*(double)arr[i].a/arr[i].b;    m-=arr[i].b;    if(m<=0)    break;        }            printf("%.3lf\n",sum);    }        return 0;}     /*5 37 24 35 27 25 24 313.33320 325 1824 1515 1024 1515 1025 1831.500*/