FatMouse' Trade
时间： 1ms 内存：64M
描述：
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.
输入：
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.
输出：
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.
示例输入：
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1示例输出：
13.333
31.500提示：
参考答案（内存最优[776]）：
#include"stdio.h"
double a[1001][3]={0};
int found(int n)
{
double max=0;
int j=0,i;
for(i=0;i<n;i++)
{
if(a[i][2]>max)
{
max=a[i][2];
j=i;
}
}
if(max==0)
return -1;
else
return j;
}
int main()
{
double m,sum=0;
int n,t=0,i;
int f, j;
scanf("%lf %d",&m,&n);
while(m!=-1&&n!=-1)
{
for(i=0;i<n;i++)
{
scanf("%lf %lf",&a[i][0],&a[i][1]);
a[i][2]=a[i][0]/a[i][1];
}
while(m!=0)
{
t=found(n);
if(t==-1)
{
break;
}
if(m>=a[t][1])
{
m-=a[t][1];
sum+=a[t][0];
a[t][2]=0;
}
else
{
sum+=((double)m)/((double)a[t][1])*a[t][0];
break;
}
}
printf("%.3lf\n",sum);
scanf("%lf %d",&m,&n);
sum=0;
}
return 0;
}
参考答案（时间最优[0]）：
#include"stdio.h"
double a[1001][3]={0};
int found(int n)
{
double max=0;
int j=0,i;
for(i=0;i<n;i++)
{
if(a[i][2]>max)
{
max=a[i][2];
j=i;
}
}
if(max==0)
return -1;
else
return j;
}
int main()
{
double m,sum=0;
int n,t=0,i;
int f, j;
scanf("%lf %d",&m,&n);
while(m!=-1&&n!=-1)
{
for(i=0;i<n;i++)
{
scanf("%lf %lf",&a[i][0],&a[i][1]);
a[i][2]=a[i][0]/a[i][1];
}
while(m!=0)
{
t=found(n);
if(t==-1)
{
break;
}
if(m>=a[t][1])
{
m-=a[t][1];
sum+=a[t][0];
a[t][2]=0;
}
else
{
sum+=((double)m)/((double)a[t][1])*a[t][0];
break;
}
}
printf("%.3lf\n",sum);
scanf("%lf %d",&m,&n);
sum=0;
}
return 0;
}
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