nefu119组合素数

/*
题目描述：给出正整数n和素数p（1<n, p<1e9），问C(2*n , n)能整除p多少次

方法：C(2*n , n) = (2*n)!/(n! * n!)，因此答案就是2*n分解后p的指数减去2倍n分解后p的指数
根据定理，n！分解后p的指数为 k = [n / p] + [n / p^2] + [n / p^3] +... + 0
*/
#pragma warning(disable:4786)
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#include<string>
#include<sstream>
#define LL long long
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define lson l,m,x<<1
#define rson m+1,r,x<<1|1
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps=1e-8;
LL factor_times(LL n , LL p)
{
LL temp = p , sum = 0 ;
while( n / temp >0){
sum += n / temp;
temp *= p;
}
return sum;
}
int main()
{
int T;
LL n,p;
scanf("%d",&T);
while(T--){
scanf("%lld %lld",&n , &p);
LL ans1 = factor_times(2 * n , p);
LL ans2 = factor_times(n , p);
printf("%lld\n",ans1 - 2 * ans2);
}
return 0 ;
}