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We say that x is a perfect square if, for some integer b, x = b 2 . Similarly, x is a perfect cube if, for some integer b, x = b 3 . More generally, x is a perfect pth power if, for some integer b, x = b p . Given an integer x you are to determine the largest p such that x is a perfect pth power. Input Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing ‘0’ follows the last test case.

Output For each test case, output a line giving the largest integer p such that x is a perfect pth power.

Sample Input

17

1073741824

25

0

Sample Output

1

30

2

对于一些整数b，n = b^p，（b为正整数）若p最大时，n为完美平方数，给出一个数n，求使n为完美平方数时，最大的p值

完全平方数的因子个数为奇数。

思路：p从31到1遍历，求n的p次开方，转为int型的t，再求t的p次方，转为int型的x

若x和n相等，则求得的p为最大值。

注意：求n的p次开方用pow()求，因为pow()函数得到的为double型，double型数据注意精度问题。

忘记了n为负数的时候，p不可能为奇数，所以循环时减2，导致wa了一次。

#include 
   
    using namespace std;int main(){    ios::sync_with_stdio(false);    int n;    while(cin>>n&&n)    {        if(n>0)        {            for(int i=31;i>=1;i--)            {                int t=pow(n*1.0,1.0/i)+0.1;                int x=pow(t*1.0,1.0*i)+0.1;                if(n==x)                {                    cout<
    <
     
      =1;i-=2)            {                int t=pow(n*1.0,1.0/i)+0.1;                int x=pow(t*1.0,1.0*i)+0.1;                if(n==x)                {                    cout<
      <
     
   

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