Day7 - G - Divisors POJ - 2992

Your task in this problem is to determine the number of divisors of C_n^k. Just for fun -- or do you need any special reason for such a useful computation?

Input

The input consists of several instances. Each instance consists of a single line containing two integers n and k (0 ≤ k ≤ n ≤ 431), separated by a single space.

Output

For each instance, output a line containing exactly one integer -- the number of distinct divisors of C_n^k. For the input instances, this number does not exceed 2 ⁶³ - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

思路：求因数个数，想到唯一分解定理
p = a1^s1+a2^s2+...., 因子总数=(s1+1)(s2+1)..... 每次都计算组合数再计算因子数显然会超时，范围只有431，可以预处理
先预处理出质数，由C[n][m] = n!/m!(n-m)!, 将每个阶乘中的质因子次数求出来，例如对于n!,求质数i的次数 = n/i+n/i^2+n/i^3+....
递推优化， a = n/i+n/i^2+...., b = a / i = n/i^2+n/i^3+....

typedef long long LL;
typedef pair<LL, LL> PLL;

const int maxm = ;

bool prime[maxm];
int num[maxm][maxm];
int jud[maxm], siz = ;
LL C[maxm][maxm];

void getprime() {
    for(int i = ; i * i <= maxm; ++i) {
        if(!prime[i]) {
            for(int j = i*i; j <= maxm; j += i)
                prime[j] = true;
        }
    }
    for(int i = ; i <= maxm; ++i)
        if(!prime[i]) {
            jud[siz++] = i;
        }
    for(int i = ; i < siz; ++i) {
        for(int j = ; j <= maxm; ++j)
            num[j][i] = j/jud[i] + num[j/jud[i]][i];
    }
    for(int i = ; i <= maxm; ++i) { // C[i][j]
        for(int j = ; j < i; ++j) {
            C[i][j] = ;
            for(int k = ; k < siz; ++k) {
                int d = num[i][k] - num[i-j][k] - num[j][k];
                if(d) C[i][j] *= (d+);
            }
        }
    }
}

int main() {
    getprime();
    int n, k;
    while(scanf("%d%d", &n, &k) != EOF) { // C(n, k) n!/k!(n-k)!
        if(n == k || k == )
            printf("1\n");
        else
            printf("%lld\n", C[n][k]);
    }
    return ;
}