杭电2019多校第四场 HDU——6623 Minimal Power of Prime(思维)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 150    Accepted Submission(s): 28

 

Problem Description

You are given a positive integer n > 1. Consider all the different prime divisors of n. Each of them is included in the expansion n into prime factors in some degree. Required to find among the indicators of these powers is minimal.

Input

The first line of the input file is given a positive integer T ≤ 50000, number of positive integers n in the file. In the next T line sets these numbers themselves. It is guaranteed that each of them does not exceed 10^18.

Output

For each positive integer n from an input file output in a separate line a minimum degree of occurrence of a prime in the decomposition of n into simple factors.

Sample Input

5

2

12

108

36

65536

Sample Output

1

1

2

2

16

 

题意：把一个数质因数分解，然后求所有质因数的最小指数

题解：打印1-10000之间的素数表然后质因数分解，分解完剩下的那个数，最多是某个质数的4次方（因为这个质数必然大于1e4），所以讨论一下就好了，当然要注意：先看是4次方再看2次方（因为如果满足4次方一定满足2次方，但是满足4次方也满足2次方，2次方的话就不是质因数了），3次方无所谓，因为开3次方会损失精度，所以就二分一下~~最后如果2,3,4次方都不满足，那么它就是一个质数~~

别问我为什么是打印1-10000不是别的区间，1-20000就超时，不信你试试~~

上代码：

#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int MAX = 1e4+10;
ll p[MAX];
bool vis[MAX];
int cnt;
void init(){
	vis[0]=vis[1]=true;
	for (int i = 2; i < MAX;i++){
		if(!vis[i]) p[cnt++]=i;
		for (int j = 0; j < cnt&&i*p[j] < MAX;j++){
			vis[i*p[j]]=true;
			if(i%p[j]==0) break;
		}
	}
}
ll check(ll mid){
	return mid*mid*mid;
}
bool solve(ll n){//判断3次方是否满足
	ll l=10010,r=1000000;//因为10009是最侯一个素数，这里左区间就从10010开始，因为是3次方所以右区间是1e6
	while(l<=r){
		ll mid=(l+r)>>1;
		ll cao=check(mid);
		if(cao>n) r=mid-1;
		else if(cao10009){//10009是我打的表的最后一个素数
			ll sq1=(ll)sqrt(n);
			ll sq2=(ll)(sqrt(sqrt(n)));
			if(sq2*sq2*sq2*sq2==n) ans=min(ans,4);//先判断4次方
			else if(sq1*sq1==n) ans=min(ans,2);//再判断2次方
			else if(solve(n)) ans=min(ans,3);//判断3次方
			else ans=min(ans,1);//如果都不满足，说明是个质数
		}
		printf("%d\n",ans);
	}
	return 0;
}