J - Perfect Pth Powers解题报告（张宇）

J - Perfect Pth Powers

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1730

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 = 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

171073741824250

Sample Output

1302

题意：输入一个数x，若为0则结束。否则，我们知道这个数x必定存在相应的a，b，使得x等于a的b次方；这样的a，b有至少一对。你所要做的就是求出最大的b。

#include<iostream>#include<cmath>using namespace std;int main(){long long n;     //int不能满足要求了...while(scanf("%lld",&n)!=EOF){if(n==0) break;long long  m;int i;int ans=1;if(n<0)       //n可能为为负数{m=-n;i=3;}    //负数当然得要是某个负数的奇次幂咯else{m=n;i=2;}double eps=1e-12;   //精度很重要的,不能太小double p;for(;i<=32;){p=pow(m*1.0,1.0/i);    //pow(int,int)提交报错....改为floatif(fabs(int(p+eps)-p)<eps)    //x等于a的b次方,b是整数,a也要是整数哦.ans=i;if(n<0)i+=2;     //负数当然得要是某个负数的奇次幂咯elsei++;  }printf("%d\n",ans);}return 0;}