nCr using Dynamic programming

Here is a technique to calculate nCr using DP.

From pascal's triangular relation, we get,
nCr =( n-1)C(r) + (n-1)C(r-1)

The basic idea lies behind DP-
nCn = 1
nC1 = n
nC0 = 1
and
nCr = n-1Cr + n-1Cr-1

So the code forms like following-

#define i64 unsigned long long

i64 dp[66][33];

i64 nCr(int n, int r)
{
 if(n==r) return dp[n][r] = 1;
 if(r==0) return dp[n][r] = 1;
 if(r==1) return dp[n][r] = (i64)n;
 if(dp[n][r]) return dp[n][r];
 return dp[n][r] = nCr(n-1,r) + nCr(n-1,r-1);
}