题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=6623

题目：

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

 

解题思路：

质因子分解，直接做绝对会T。

即然不行，那呢，1e18开五次根=3980，不到3981，大概估算一下，如果统计完数x的1-4000内的质因子，一共550个质因子，若x有大于4000的质因子，那么x的因子可以分解成，或， 或，p为大于4000的质数，最高p的四次，否则就超过1e18， 所以在对x进行4000以内的质因子分解之后，剩下的数x'直接分情况讨论即可。

能分解成，或， 或，说明x'开4，或3， 或2次之后为整数，重点是要判断开次方后的结果是不是整数，这就涉及到精度的问题了，尤其是pow（x, 1.0/3）时结果更加不准确。且如果用pow（a,b）时如a为位数超过16的long long 类型的数，那么long long 转化成double 的结果其实是错误的，有误差的，可以用powl或者自己写个pow函数（判断函数）。

和这两种情况有点重复，要先判断开完4次是否为整数，再判断开2次的。

时间复杂度O(5e4*550)=O(3e7)

 

ac代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 4000;
typedef long long ll;
bool vis[maxn+10] = {false};
int prime[maxn+10];
int cnt, ans;
ll x;
//线性筛法
void get_prime()
{
    for(int i = 2; i <= maxn; i++)
    {
        if(!vis[i])
            prime[++cnt] = i;
        for(int j = 1; j <= cnt && (ll)i * prime[j] <= maxn; j++)
        {
            vis[(ll)i * prime[j]] = 1;
            if(i % prime[j] == 0)
                break;
        }
    }
}
void div(ll &x)
{
    for(int i = 1; i <= cnt && prime[i] <= x; i++)
    {
        if(x % prime[i] == 0)
        {
            int tmp = 0;
            while(x % prime[i] ==0)
            {
                tmp++;
                x /= prime[i];
            }
            ans = min(ans, tmp);
        }
    }
}
ll mul(ll a, ll b)
{
    ll res = 1;
    while(b--) res *= a;
    return res;
}
bool judge(int k)
{
    //或者将所有的(ll)powl换成mul,超过16位的long long 转 double 会出错，可转成long double
    ll p = (ll)powl(x, 1.0/k);
    while((ll)powl(p+1, k) <= x)
        p++;
    return (ll)powl(p, k) == x;
}
int main()
{
    //freopen("/Users/zhangkanqi/Desktop/11.txt","r",stdin);
    get_prime();
    int t;
    scanf("%d", &t);
    while(t--)
    {
        ans = 100;
        scanf("%lld", &x);
        if (x == 1) ans = 0;
        div(x);
        if (x != 1) // 没有质因子分解完全
        {
            bool flag = 0;
            for(int k = 4; k >= 2; k--)  //从k=4先开始判断，p^4 优先 (p^2)^2
                if(judge(k))
                {
                    ans = min(ans, k);
                    flag = 1;
                    break;
                }
            if(!flag) ans = 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}