J - Perfect Pth Powers解题报告

J - Perfect Pth Powers

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

Submit Status

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

我发现写解题报告还是很有好处的，所以从今天起好好的写解题报告。
这个题目的大意就是给出一个数a，找到一个数的p次幂等于这个数，要求是p是最大的。
我首先将输入的数a开根号，因为除了1，p的最小的数就是2了（负数就另当别论了），然后取t=2~sqrt（a），求t的p次幂，因为t是从小到大，如果有p满足就一定是最大的，所以找到了p就马上退出，输出。
负数的话，还是先开根号，不过要注意的是负数的p一定是奇数哈，完了，有错误请纠正。
代码如下：

#include<stdio.h>#include<math.h>#include<iostream>using namespace std;int main(){long double a,i,j,p,sum,num;while(cin>>a&&a!=0){ sum=1,num=1; if(a>0) { for(i=2;i<=sqrt(a)+1;i++)//去根号a，缩小范围{for(j=2;;j++){for(p=0;p<j;p++)//求i的j次幂{sum=sum*i;}if(sum==a)//判断是否满足{num=j;goto end;//满足的话就退出}if(sum>a){goto top;}sum=1;}top:sum=1; } } else {  sum=1,num=1; a=-a;  for(i=2;i<=sqrt(a)+1;i++){for(j=3;;j+=2)//负数和正数不一样，只能为奇数{for(p=0;p<j;p++){sum=sum*i;}if(sum==a){num=j;goto end;}if(sum>a){goto top1;//不满足的话继续进行}sum=1;}top1:sum=1;} }end:;cout<<num<<endl;}return 0;}