HDU 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 78423    Accepted Submission(s): 27010

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
5 2
4 3
25 18
20 3
15 10
24 15
-1 -1

 

Sample Output

13.333
31.500

解题思路：这题就是一个典型的贪心。。。

代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#define LL long longusing namespace std;const int maxn = 1000000007;struct P{    int x,y;    double z;}a[1010];bool cmp(P a,P b){    return a.z>b.z;}int main(){    int n,m;    while(~scanf("%d %d",&n,&m))    {        if(n==-1 && m==-1) break;        for(int i=0;i<m;i++)        {            scanf("%d %d",&a[i].x,&a[i].y);            a[i].z=1.0*a[i].x/(a[i].y*1.0);        }        sort(a,a+m,cmp);        double sum=0;        for(int i=0;i<m;i++)        {            if(n>=a[i].y) {sum+=a[i].x; n-=a[i].y;}            else {sum+=a[i].z*n; n=0;}            if(n==0) break;        }        printf("%.3lf\n",sum);    }    return 0;}