SJTU 机试 数学

题目描述

给定n，a求最大的k，使n！可以被a^k整除但不能被a^(k+1)整除。

输入描述:

两个整数n(2<=n<=1000)，a(2<=a<=1000)

输出描述:

一个整数.

示例1

输入

复制

6 10

输出

复制

1
分解质因数取min即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-11
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int pme[maxn];
int tot;
int vis[maxn];
int cnt[maxn];
void init() {
	for (int i = 2; i < maxn; i++) {
		if (!vis[i]) {
			pme[++tot] = i;
			for (int j = i; j < maxn; j += i) {
				vis[j] = 1;
			}
		}
	}
}
int num[maxn];

int main() {
	int  n, a;
	scanf("%d%d", &n, &a);
	init();
	for (int i = 1; i <= tot; i++) {
		int tmp = n;
		while (tmp) {
			cnt[pme[i]] += tmp / pme[i]; tmp /= pme[i];
		}
	}
	int minn = 999999;
/*	    for(int i=1;i<=tot;i++){
			if(cnt[pme[i]]){
				cout<<pme[i]<<' '<<cnt[pme[i]]<<endl;
			}
		}
 */
	for (int i = 1; i <= tot; i++) {
		while (a%pme[i] == 0) {
			num[i]++; a /= pme[i];
		}
	}
	for (int i = 1; i <= tot; i++) {
		if(num[i])
		minn = min(minn, cnt[pme[i]] / num[i]);
	}
	cout << minn << endl;
}