Divisors POJ

题目链接

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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 16 310 4

Sample Output

2616

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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<cstdio>#include<cstring>#include<iostream>#include<algorithm>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;}