zju 2109 FatMouse' Trade

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1109

 

FatMouse' Trade

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

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

---

Author: CHEN, Yue
Source: Zhejiang Provincial Programming Contest 2004

关键是读懂题目的意思，然后就很容易求了。

先购买单位F中J最多的那个room的J。如果M<J  则按比例购买。

#include <stdio.h>struct  change{  int f,j;  double s;}cat[1002];int n,m;void print(){for(int i=1;i<=n;i++)printf("%d %d %lf\n",cat[i].j,cat[i].f,cat[i].s);}int main(){int i,j;double max;while(scanf("%d%d",&m,&n)==2&&m>=0){max=0;for(i=1;i<=n;i++){scanf("%d%d",&cat[i].j,&cat[i].f);cat[i].s=cat[i].j*(1.0)/cat[i].f;    j=i;while(j>1&&cat[j].s>cat[j-1].s)    //根据比例排序{struct change tmp;tmp=cat[j];cat[j]=cat[j-1];cat[j-1]=tmp;j--;}}//print();for(i=1;i<=n;i++){if(cat[i].f>m){max+=((double)cat[i].j*m*(1.0)/cat[i].f);break;}else {m-=cat[i].f;max+=cat[i].j;}}printf("%.3lf\n",max);}  return 0;}

 

typedef struct   {      double v;      int j,f;  }Bean;  Bean bean[1005];  int cmp( Bean x,Bean y )  {     if( x.v >= y.v )         return 1;     else         return 0;  }    sort(bean,bean+n,cmp);  结构体用sort函数排序