#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    long long mm,i,n;
    double m,p;
    while(1)
    {
        scanf("%I64d",&n);
        if(!n)break;
        
        if(n>0)
        {
            i=(long long)(log((double)n)/log(2.0000000))+1;
            for(;i>1;--i)
            {
                m=pow((double)n,1.0000000/(double)i);
                mm=(long long)m;
                if(m-(double)mm<1e-12||(double)mm+1-m<1e-12)
                {
                    printf("%I64d\n",i);
                    break;
                }
            }
            if(i==1)
                printf("1\n");
        }
        else
        {
            n=-n;
            i=(long long)(log((double)n)/log(2.0000000))+1;
            if(i%2==0)--i;
            for(;i>1;i-=2)
            {
                m=pow((double)n,1.0000000/(double)i);
                mm=(long long)m;
                if(m-(double)mm<1e-12||(double)mm+1-m<1e-12)
                {
                    printf("%I64d\n",i);
                    break;
                }
            }
            if(i==1)
                printf("1\n");
        }
    }
    return 0;
}
直接枚舉。
貌似pow()函數(shù)比log()快。之前用log()函數(shù)枚舉對(duì)數(shù)的底，超時(shí)。
對(duì)輸入要分正負(fù)。枚舉的范圍是[1，log((double)n)/log(2.0000000)+1]。
當(dāng)x為正數(shù)時(shí)，p取以上區(qū)間所有整數(shù)，在符合的p中取最大。
當(dāng)x為負(fù)數(shù)時(shí)，將x轉(zhuǎn)化為正數(shù)處理，p取以上區(qū)間所有奇數(shù)(因?yàn)橹挥衅鏀?shù)次冪才可能等于負(fù)數(shù))，
在符合的p中取最大。