cy
BAN USER
the ith number in level x 's index is :
(x-1)*x/2 + i, and it's child is (x+1)*x/2 +i, (x+1)*x/2 + i +1;
use the same method just like lucas.eustaquio .
int cup_pyramid(int c, int l, int kth)
{
double a[kth*2+3];
for(int i=0; i<sizeof(a)/sizeof(double); i++)
a[i] = 0.0;
a[1] = l;
for(int level=1; 1; level++)
{
int pos = level*(level-1)/2;
int nextpos = level*(level+1)/2;
int haveover = 0;
for(int index=1;index <=level;index++)
{
if(a[pos+index] - c > 0.00001) {
double over = (a[pos+index] -c )/2;
a[pos+index] = c;
a[nextpos+index] += over;
a[nextpos+index+1] += over;
haveover = 1;
}
}
if(pos > kth) break;
if(!haveover) break;
}
return a[kth];
}
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For part 1 of the problem, use min-heap of size k.
- cy August 14, 2013For part 2 of the problem,
1. get the first k number from the stream, and the smallest number of these k number, say mink
2. store the first k number into temp file
3. continue scan the stream, put the number larger than mink into memory buffer
4. if memory buffer is full (size larger than m), merge the buffer into temp file
1. only keep the top k larger number
1. get the new mink (k-th large number)
5. empty the memory buffer, go back to step 3 until the stream end