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 = 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 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

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;
}