Divisors POJ - 2992

题目链接

---

Your task in this problem is to determine the number of divisors of Cnk. 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 Cnk. 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

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题意:

求组合数的因子个数.

思路:

大概就是将组合数写成阶乘的形式, 然后统计每个质数因子的个数即可.

代价:

#include
#include
#include
#include
using namespace std;

typedef long long LL;
const int maxn = 500;
int pri[maxn];
int a[maxn];
int b[maxn];
bool vis[maxn];
int num;

void prime(){
    num = 0;
    memset(vis, false, sizeof(vis));
    for(int i = 2; i < maxn; ++i){
        if (!vis[i]) pri[++num] = i;
        for(int j = 1; j <= num && i * pri[j] < maxn; ++j){
            vis[ i* pri[j]] = true;
            if (i % pri[j] == 0) break;
        }
    }
}

int solve(int n, int t){
   int ans = 0;
   int x = t;
   while(t <= n){
         ans += n / t;
         t *= x;
   }return ans;
}

int main(){
    //ios::sync_with_stdio(false);
    int n, m;
    prime();
    while(scanf("%d %d", &m, &n) != EOF){
     //   if(n > m / 2) n = m - n;
        LL ans = 1;
       for(int i = 1; i <= num &&  pri[i] <= m; ++i)
            ans *=(solve(m, pri[i]) - solve(n, pri[i]) - solve(m - n, pri[i]) + 1);
      printf("%lld\n",ans);
    }return 0;
}