[暑假集训--数论]poj1730 Perfect Pth Powers

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

给个n，把它表示成a^b的形式，问b最大是多少，n有负数

正数简单，负数要打个标记，最后答案b要除到没有2的因子为止

比如-1073741824=-(2^30)=(-4)^15，但是不能是(-2)^30

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 #define LL long long
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL n;
 bool mk[];
 int p[],len;
 inline LL LLabs(LL a){return a<?-a:a;}
 inline void getp()
 {
     for (int i=;i<=;i++)
     {
         if (!mk[i])
         {
             p[++len]=i;
             for (int j=*i;j<=;j+=i)mk[j]=;
         }
     }
 }
 inline void work()
 {
     int flag=(n<),ans=;n=LLabs(n);
     for (int i=;i<=len;i++)
     {
         if ((LL)p[i]*p[i]>n)break;
         if (n%p[i]!=)continue;
         int now=;while (n%p[i]==)n/=p[i],now++;
         if (!ans)ans=now;else ans=__gcd(ans,now);
     }
     if (n!=)ans=;
     if (flag)while (ans%==)ans/=;
     printf("%d\n",ans);
 }
 int main()
 {
     getp();
     while (~scanf("%lld",&n)&&n)work();
 }

poj 1730