Poj1730_perfect PTH powers]

Perfect PTH powerstime limit: 1000 ms memory limit: 10000 ktotal submissions: 16699 accepted: 3786 description


We say that X is a perfect square if, for some integer B, x = B2. Similarly, X is a perfect cube if, for some integer B, x = B3. more generally, X is a perfect PTH power if, for some integer B, x = bp. 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 magn=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
Source

Waterloo local 2004.01.31

For integer B, n = B ^ P (B is a positive integer), if p is the maximum, n is the perfect number of integers.

Returns the maximum P value of the number n when the number n is the perfect limit.

Idea: P traverses from 31 to 1, calculates the p-start of N, converts it to t of int type, and then returns the P power of T to the X of int type.

If X and n are equal, the obtained p is the maximum value, and the break output Loop

Note: Evaluate the P times of N using POW (), because the POW () function obtains the double type, while the double type data

Accuracy problems. For example, 4 can be expressed as 3.99999 ...... Or 4.0000001, so when it is converted to int type + 0.1

Reference blog: http://blog.csdn.net/wangjian8006/article/details/7831478

#include<stdio.h>#include<math.h>int main(){    int n;    while(~scanf("%d",&n) && n)    {        if(n > 0)        {            for(int i = 31; i >= 1; i--)            {                int t = (int)(pow(n*1.0,1.0/i) + 0.1);                int x = (int)(pow(t*1.0,1.0*i) + 0.1);                if(n == x)                {                    printf("%d\n",i);                    break;                }            }        }        else        {            n = -n;            for(int i = 31; i >= 1; i-=2)            {                int t = (int)(pow(n*1.0,1.0/i) + 0.1);                int x = (int)(pow(t*1.0,1.0*i) + 0.1);                if(n == x)                {                    printf("%d\n",i);                    break;                }            }        }    }    return 0;}

Poj1730_perfect PTH powers]