FatMouse' Trade

来源:互联网 发布:mac本怎么退出全屏显示 编辑:程序博客网 时间:2024/06/11 04:33
A - FatMouse' Trade
Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u
SubmitStatus

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
#include<stdio.h>#include<string.h>int main(){    int m,n,j[1005],f[1005],z,t=0;    double e[1005];    scanf("%d%d",&m,&n);      while(m!=-1&&n!=-1)    {           t++;        z=0;        for(int x=0;x<n;x++)          scanf("%d%d",&j[x],&f[x]);        double a[1005],s=0,c; int k;    for (int i=0;i<n;i++)    a[i]=(double)j[i]/f[i];    for(int i=0;i<n-1;i++)      for(int w=i+1;w<n;w++)      if(a[i]<a[w])        {   c=a[i];a[i]=a[w];a[w]=c;            k=j[i];j[i]=j[w];j[w]=k;            k=f[i];f[i]=f[w];f[w]=k;        }       for(int i=0;i<n;i++)        if(f[i]<=m) { s=s+j[i];m=m-f[i];if(m==0) break;}        else {s=s+(double)m/f[i]*j[i];break;}     e[t-1]=s;    memset(j,0,sizeof(j));    memset(f,0,sizeof(f));          scanf("%d%d",&m,&n)    ;                }    for(int d=0;d<t;d++)    printf("%.3f\n",e[d]);        return 0;}

0 0
原创粉丝点击