FatMouse' Trade（贪心算法）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45918    Accepted Submission(s): 15370

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

#include <cstdio>#include <algorithm>using namespace std;struct trade{double a;double b;double c;}fj[1000];//WA了一次，错误是运行中错误，所以最有可能是数组开的太小了，干脆把数组长度从100开到1000，就AC了！int compare(trade a,trade b){return a.c>b.c;//运用结构体变量存储，}int main(){    int m,n,i;double j,f,sum;while (scanf("%d%d",&m,&n)&&(m!=-1&&n!=-1)){sum=0;        for (i=0;i<n;i++)        {scanf("%lf%lf",&j,&f);fj[i].a=j;fj[i].b=f;fj[i].c=j/f;        }sort(fj,fj+n,compare);//用结构体中的c变量的大小进行排序，注意这个是从大到小进行！for (int j=0;j<n;j++){if (m-fj[j].b>0.001){sum+=fj[j].a;m-=fj[j].b;}//当m大于整体的b时else{sum+=m*fj[j].a/fj[j].b;break;}//当m不能整体换时！}printf("%.3lf\n",sum);}}//这一题典型的