hdu 1009 FatMouse' Trade



                                             FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53745    Accepted Submission(s): 17970

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

 

/题意：就是给你m猫粮，让你你换鼠粮，怎样换最多，注意，仓库里鼠粮只有j[i],也就是只能换一次，换东西当然靠考虑性价比，相同猫粮换鼠粮多的才划算，也就是就性价比高的开始算，性价比算法就是对应仓库鼠粮比上猫粮，再排序*/

代码如下

#include<iostream>#include<algorithm>#include<cstdio>#define maxn 1000+10using namespace std;struct node{int j;  //鼠粮int f;  // 猫粮粮double value;//性价比 }; bool rule(node x,node y){return x.value>y.value;}int main(){node stu[maxn];int n,m;while(scanf("%d %d",&m,&n)&&m!=-1&&n!=-1){for(int i=0;i<n;i++) {scanf("%d %d",&stu[i].j,&stu[i].f);stu[i].value=(double)stu[i].j/stu[i].f;}sort(stu,stu+n,rule);double sum=0;for(int i=0;i<n;i++) {if(m>=stu[i].f) sum+=stu[i].j,m-=stu[i].f;else{   sum+=m*stu[i].value;    break;}}printf("%.3lf\n",sum);}return 0;}