CF 1445C

题目传送门

题目大意是,给定(p_i)和(q_i),找到一个最大的(x_i)使得(p_i mod x_i=0)且(x_i mod q_i eq 0)

分情况讨论:

1.(p_i < q_i)时,答案为(p_i).

2.(p_i mod q_i eq 0)时,答案为(p_i).

3.利用唯一分解定理分解(p_i)和(q_i):

(p_i = k_1^{t_1} * k_2^{t_2} * k_3^{t_3} * ... * k_n^{t_n})

(p_i = k_1^{r_1} * k_2^{r_2} * k_3^{r_3} * ... * k_n^{r_n})

其中,(forall r_i,t_i)都满足(t_i geq r_i)

而要找的(x),满足(x = k_1^{d_1} * k_2^{d_2} * k_3^{d_3} * ... * k_n^{d_n}) 且 (exists i) 使得(d_i < t_i)且(d_i geq 0).

所以最终要做的就是枚举(q_i)的所有质因子,使其次数减一,其它所有质因子次数与(p_i)里的一样(包括(q_i)不含的质因子).

注意:对于第三种情况,(q_i)可能是一个大质数,导致我们无法枚举到那个质因子,所以答案为(p_i)除去所有(p_i)后的余数.

下面给两份代码(一份Wa,一份AC),WA的原因是运用错误处理方法,使得对于一些因数为大质数的数分解错误.

//AC
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

long long t,tot,zs;
long long num[1000001],sum[1000001];
long long x,y,ans,mx;
bool vis[1000001];

inline void oula() {
	vis[1] = 1;
	for(long long i = 2;i <= 1000000; i++) {
		if(!vis[i]) num[++tot] = i;
		for(long long j = 1;j <= tot; j++) {
			if(num[j] * i > 1000000) break;
			vis[num[j]*i] = 1;
			if(i % num[j] == 0) break;
		}
	}
}

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

int main() {
	scanf("%lld",&t);
	oula();
	while(t--) {
		memset(sum,0,sizeof(sum));
		ans = 1;
		scanf("%lld%lld",&x,&y);
		if(x < y) {
			printf("%lld
",x);
			continue;
		}
		if(x % y != 0) {
			printf("%lld
",x);
			continue;
		}
		fenjie(y);
		for(long long i = tot;i >= 1; i--) {
			if(sum[i] == 0) continue;
			long long uu = x,vv = y,m1 = 0,m2 = 0;
			while(uu % num[i] == 0) uu /= num[i],m1++;
			while(vv % num[i] == 0) vv /= num[i],m2++;
			for(int j = 1;j <= m2 - 1; j++) uu *= num[i];
			ans = max(ans,uu);
		}
		if(ans == 1) {
			while(x % y == 0) x /= y;
			ans = x;
		}
		printf("%lld
",ans);
	}
	return 0;
} 

//WA
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

long long t,tot,zs;
long long num[1000001],sum[1000001];
long long x,y,ans,mx;
bool vis[1000001];

inline void oula() {
	vis[1] = 1;
	for(long long i = 2;i <= 1000000; i++) {
		if(!vis[i]) num[++tot] = i;
		for(long long j = 1;j <= tot; j++) {
			if(num[j] * i > 1000000) break;
			vis[num[j]*i] = 1;
			if(i % num[j] == 0) break;
		}
	}
}

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

int main() {
	scanf("%lld",&t);
	oula();
	while(t--) {
		zs = 0;
		memset(sum,0,sizeof(sum));
		ans = 1;
		scanf("%lld%lld",&x,&y);
		if(x < y) {
			printf("%lld
",x);
			continue;
		}
		if(x % y != 0) {
			printf("%lld
",x);
			continue;
		}
		fenjie(y);
		for(long long i = tot;i >= 1; i--) {
			if(sum[i] == 0) continue;
			long long uu = y / num[i];
			long long vv = x / y;
			while(vv % num[i] == 0) vv = vv / num[i];
			ans = max(ans,vv * uu);
		}
		if(zs <= 1) {
			while(x % y == 0) x /= y;
			ans = x;
		}
		printf("%lld
",ans);
	}
	return 0;
}