Hello, my program does this:

i find the number of numbers with sum S and number of digits n/2, and store it in position SUM[S]
The total number of solutions is summation of SUM[S]*SUM[S] over all possible S

But I get the answer 669 in this case....where am I wrong?

Here's my code:
main()
{
 int N , sums[100] , end , t , t2 , cursum , total = 0;
 scanf("%d" , &N);
 for(t = 0 ; t <= 36 ; t++)
  sums[t] = 0;
 end = pow(10 , N / 2);
 for(t = 0 ; t < end ; t++)
 {
  cursum = 0;
  t2 = t;
  while(t2 > 0)
  {
   cursum += t2 % 10;
   t2 /= 10;
  }
  sums[cursum]++;
 }
 for(t = 0 ; t <= 36 ; t++)
  total += sums[t] * sums[t];
 printf("%d" , total);
}

for n=4 answer is 670.
one more ticket you didn't count is 0000

i have same problem,
but i fixed it
problem is in function pow:
pow(10,2) returns 99
::))) (stupid problem with pow)
do it with your hands:))

use powl(long, long) from <math.h>

Edited by author 15.11.2009 01:12