poj 2992 Divisors

Divisors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7846		Accepted: 2163

Description

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 2⁶³ - 1.

Sample Input

5 1
6 3
10 4

Sample Output

2
6
16

#include <iostream>
#include <cstring>
#include <cstdio>
 
using namespace std;
const int MAXN = 431;
int pCnt, prime[MAXN / 4];
 
void getPrime(int n) {
    int i, j, notPrime[MAXN + 1];
    memset(notPrime, 0, sizeof(notPrime));
    pCnt = 0;
    for(i = 2; i <= n; i ++) {
        if(! notPrime[i]) prime[pCnt ++] = i;
        for(j = 0; j < pCnt && prime[j] * i <= n; j ++) {
            notPrime[prime[j] * i] = prime[j];
            if(i % prime[j] == 0)
                break;
        }
    }
}
int cal(int n, int p) {
    if(n < p) return 0;
    else return n / p + cal(n / p, p);
}
int main() {
    getPrime(MAXN);
    int i, n, k;
    while(scanf("%d%d", &n, &k) != EOF) {
        __int64 cnt = 1;
        //if(k > n / 2) k = n - k;
        for(i = 0; prime[i] <= n && i < pCnt; i ++) {
            int t = cal(n, prime[i]) - cal(n - k, prime[i]) - cal(k, prime[i]);
            cnt *= (t + 1);
        }
        printf("%I64d\n", cnt);
    }
    return(0);
}