2019 HDU 多校赛第四场 HDU 6623 Minimal Power of Prime 1e18 合数分解+分类讨论

#include <bits/stdc++.h>
using namespace std;
const int MAXN=10000+10;
int prime[MAXN+1];
int is_prime[MAXN+1];
void getPrime() {
	memset(is_prime,0,sizeof(is_prime));
	for(int i=2; i<=MAXN; i++) {
		if(!is_prime[i])prime[++prime[0]]=i;//第prime[0]个素数是i
		for(int j=1; j<=prime[0]&&prime[j]<=MAXN/i; j++) {
			is_prime[prime[j]*i]=1;
			if(i%prime[j]==0) break;
		}
	}
}
long long sqrt3(long long n) {
	long long l=1,r=(long long)pow(n*1.0, 1.0 / 3) + 1,mid;
	while(l<=r) {
		mid=(l+r)>>1;
		if(mid*mid*mid==n) return mid;
		else if(mid*mid*mid>n) r=mid-1;
		else l=mid+1;
	}
	return -1;
}
long long factor[1000][2];
int fatCnt;
int getFactors(long long x) {
	fatCnt=0;
	int ans = 100;
	long long tmp=x;
	for(int i=1; i<=prime[0]; i++) {
		if(1ll*prime[i]*prime[i]>tmp)
			break;
		factor[fatCnt][1]=0;
		if(tmp%prime[i]==0) {
			factor[fatCnt][0]=prime[i];
			while(tmp%prime[i]==0) {
				factor[fatCnt][1]++;
				tmp/=prime[i];
			}
			if(factor[fatCnt][1]<ans) ans = factor[fatCnt][1];
			fatCnt++;
		}
		if(ans==1)return 1;
	}
	long long k1,k2;
	if(tmp>1) {
		k1 = sqrt(tmp);
		if(k1*k1==tmp) {
			k2 = sqrt(k1);
			if(k2*k2==k1&&ans>4)ans=4;
			else if(k2*k2!=k1&&ans>2)ans=2;
		}else{
			k1 = sqrt3(tmp);
			if(k1*k1*k1==tmp&&ans>3)ans=3;
			else ans = 1;
		}
	}
	return ans;
}
int main() {
	getPrime();
	int t;
	scanf("%d",&t);
	long long n;
	while(t--) {
		scanf("%lld",&n);
		printf("%d\n",getFactors(n));
	}
}