Integer Partitioning [Range]

long long intPart(int n, long long  *P = P)
{
    if(n == 0) return (P[0] = 1);
    if(n == 1) return (P[1] = 1);

    if(!P[n])
        for(int k = 1; (k * (3 * k - 1)) / 2 <= n; k++)
            for(int i = 1; i < 4 && k * (3 * k + i - 2) / 2 <= n; i += 2)
                P[n] += ((k & 1 ? 1: -1) * intPart(n - (k * (3 * k + i - 2) / 2), P));

    return P[n];
}