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

6 10

输出

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;

}