【HDU】 1009 FatMouse' Trade

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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题意：m pounds食物换取JavaBean，给出兑换关系，求换取的最多，类似背包问题。

AC代码：

#include<cstdio>#include<algorithm>using namespace std;struct Vote{int p,w;double s;     }q[10000+5];bool cmp(Vote x,Vote y){return x.s>y.s;}int main(){int m,n;while(~scanf("%d %d",&m,&n)&&(m!=-1||n!=-1)){double sum=0;              //用double... for(int i=1;i<=n;i++){scanf("%d %d",&q[i].p,&q[i].w);q[i].s=(double)(q[i].p)/q[i].w;}sort(q+1,q+n+1,cmp);for(int i=1;i<=n;i++){if(m>=q[i].w){sum+=q[i].p;m-=q[i].w;}else{sum+=m*q[i].s;break;}}printf("%.3lf\n",sum);}return 0; }