hdu_1009 FatMouse' Trade----贪心

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 20004 Accepted Submission(s): 6223



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

[align=left]Sample Output[/align]

13.333
31.500

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[align=left] 还是一道贪心算法的题。刚接触贪心，所以趁热打铁继续做几道题。有点晚了 有点累 。[/align]
[align=left] 这道题 的思路是，总是先选得到与付出比最大的。[/align]
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#include<stdio.h>
int main()
{
//freopen("1.in","r",stdin);
void swap(double *,double *);  //声明了一个交换函数用于排序。顺便回顾了一下用指针改变数组的值。
int m,n,i,j;
double ave[1000],get[1000],pay[1000],sum;
while(scanf("%d%d",&m,&n)!=EOF&&(m!=-1||n!=-1))
{
sum=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&get[i],&pay[i]);
ave[i]=get[i]/pay[i];         //计算得到付出之比。
}

//下面是按得到付出之比 从大到小排好序。
for(i=1;i<n;i++)
for(j=0;j<n-i;j++)
{
if(ave[j]<ave[j+1])
{
swap(ave+j,ave+j+1);
swap(get+j,get+j+1);
swap(pay+j,pay+j+1);
}
}

//计算能得到的最大值。
for(i=0;i<n;i++)
{
if(m>=pay[i])
sum+=get[i];
else
{
sum+=ave[i]*m;
break;
}
m-=(int)pay[i];
}
printf("%.3lf\n",sum);
}
return 0;
}
void swap(double *a,double *b)
{
double t;
t=* a;
* a=* b;
* b=t;
}

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