思路:
C(2n,n)=( (2n)! ) / (n!*n!)
求上式可以被素数p整除多少次,也就是求上式中素数p的幂
所以统计2n!的幂和(n!*n!)的幂,相减就可以了,但要注意统计幂的时候要用long long
定理:n! 的素因子分解中的素数p的指数(幂)为【n/p】+【n/p^2】+【n/p^3】+.......
代码实现:
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
using namespace std;
const double eps = 1e-8;
typedef long long ll;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN=5010;
const int MAXM=100010;
const int M=50010;
ll cal(ll n,ll p)
{
ll num=0;
while (n)
{
num+=n/p;
n=n/ p;
}
return num;
}
int main()
{
ll n,p;
int t;
cin>>t;
while(t--)
{
scanf("%lld%lld",&n,&p);
printf("%lld\n",cal(2*n,p)-2*cal(n,p));
}
return 0;
}