poj1730 素数 Perfect Pth Powers

Description

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 p th 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 p th power.

Sample Input

17

1073741824

25

0

Sample Output

1

30

2

解法一：

素数打表，计算所有素因子个数的最大公约数

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn=50009;//只需计算sqrt(三十二位范围内）
int prime[maxn+1];
void getprime()
{
for(int i=2; i<maxn; i++)
{
if(!prime[i]) prime[++prime[0]]=i;
for(int j=1; j<=prime[0]&&(long long)prime[j]*i<maxn; j++)
{
prime[prime[j]*i]=1;
if(i%prime[j]==0) break;
}
}
}
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
int solve(int n)
{
int i;
int ans=-1;
for(i=1; i<=prime[0]; i++)//这里不可以像求质因子那样将（prime[i]*prime[i]<=n作为判断条件，因为我们这里要计算个数，而不是简单地求出因子！
{
if(n%prime[i]==0)
{
int cnt=0;
while(n%prime[i]==0)
{
cnt++;
n/=prime[i];
}
if(ans==-1) ans=cnt;
else ans=gcd(ans,cnt);//直接计算结果，刚开始还傻傻的想着将结果记录下来。。。
}
}
if(ans==-1)
ans=1;
if(n<0)
{
while(ans%2==0)//注意负数的平方不可能是偶数，所以一直减半至结果为一奇数
ans/=2;
}
return ans;
}
int main()
{
int n;
getprime();
while(scanf("%d",&n)&&n)
{
cout<<solve(n)<<endl;
}
return 0;
}

解法二：

然而还可以直接计算！！！！

就是要考虑一下精度的问题；

正向结果与反向结果要相同才行；

#include <stdio.h>
#include <cmath>
int main(){
int p;
int n,i;
int t;
while(scanf("%d",&n) && n){
if(n>0){
for(i=31;i>=1;i--){
t=(int)(pow(1.0*n,1.0/i)+0.1);//加0.1是为了保证精度误差
p=(int)(pow(1.0*t,1.0*i)+0.1);
if(n==p){
printf("%d\n",i);
break;
}
}
}
else{
n=-n;
for(i=31;i>=1;i-=2){//负数就直接计算奇数的幂即可
t=(int)(pow(1.0*n,1.0/i)+0.1);//开指定数的方；如果是1/3则是开立方
p=(int)(pow(1.0*t,1.0*i)+0.1);
if(n==p){
printf("%d\n",i);
break;
}
}
}
}
return 0;
}