HDU 1009 FatMouse' Trade

FatMouse’ Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53708 Accepted Submission(s): 17954

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

Sample Output
13.333
31.500

hud 1009

题目大意 老鼠可以用猫粮换吃的，他总共有M个猫粮，对于每一个房间的东西可以使用ji单位的猫粮换Fi的食物问最多可以换多少（可以不全部拿走）

基础贪心 计算每一个的性价比然后先换性价比高的

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;struct mon{    int m,j;    double x;}num[2000];bool cmp(mon x,mon y){    return x.x>y.x? 1:0;}int main(){    int n,m,i;    while(cin>>m>>n,n!=-1)    {        int a,b;        memset(num,0,sizeof(num));        for(i=0;i<n;i++)        {            cin>>a>>b;            num[i].m=a;            num[i].j=b;            num[i].x=(double)a/b;        }        sort(num,num+n,cmp);        i=0;        double sum=0;        while(m)        {            if(m>=num[i].j)            {                m-=num[i].j;                sum+=num[i].m;            }            else            {                sum+=num[i].x*m;                m=0;            }            //cout<<num[i].m<<"  asd  "<<num[i].j<<endl;            i++;        }        printf("%.3f\n",sum);    }    return 0;}