洛谷 P1069 细胞分裂

题目传送门

将所给的数质因数分解,当细胞数增长到所有质因数的指数都比试管数中的大,即可平均分.

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

long long n,m1,m2,tot,ans = 20000000000000,c,oo;
long long x,num[45001],sum[45001],sum1[45001],num1[45001];
bool vis[45001];

inline long long mx() {
	long long s = 0,w = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		if(ch == '-') w = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9') {
		s = s * 10 + (ch - '0');
		ch = getchar();
	}
	return s * w;
}

inline void oula() {
	vis[1] = 1;
	for(int i = 2;i <= 45000; i++) {
		if(!vis[i]) num[++tot] = i;
		for(int j = 1;j <= tot; j++) {
			if(num[j] * i > 45000) break;
			vis[num[j]*i] = 1;
			if(i % num[j] == 0) break;
		}
	}
}

inline void fenjie(long long a) {
	for(int i = tot;i >= 1; i--) {
		bool _ok = 0;
		if(a % num[i] == 0) num1[++oo] = num[i],_ok = 1;
		while(a % num[i] == 0) {
			a = a / num[i];
			sum[oo]++;
		}
		if(_ok) sum[oo] = sum[oo] * m2;
	}
}

inline void fenjie1(long long a) {
	for(int i = tot;i >= 1; i--) {
		while(a % num[i] == 0) {
			a = a / num[i];
			sum1[num[i]]++;
		}
	}
}

int main() {
	oula();
	n = mx();
	m1 = mx();
	m2 = mx();
	if(m1 == 1) {
		printf("0");
		return 0;
	}
	fenjie(m1);
	for(int i = 1;i <= n; i++) {
		x = mx();
		memset(sum1,0,sizeof(sum1));
		c = -1;
		fenjie1(x);
		for(int j = 1;j <= oo; j++) {
			if(sum1[num1[j]] == 0) {
				c = -1;
				break;
			}
			if(sum1[num1[j]] >= sum[j]) {
				c = max(c,(long long)1);
				continue;
			}
			if(sum[j] % sum1[num1[j]] == 0)
				c = max(c,sum[j] / sum1[num1[j]]);
			else c = max(c,sum[j] / sum1[num1[j]] + 1);
		}
		if(c != -1)
			ans = min(c,ans);
	}
	if(ans == 20000000000000)
		printf("-1");
	else
		printf("%lld",ans);
	return 0;
}