ACM 1730 Perfect Pth Powers

#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;
}

直接枚举。
貌似pow()函数比log()快。之前用log()函数枚举对数的底，超时。
对输入要分正负。枚举的范围是[1，log((double)n)/log(2.0000000)+1]。
当x为正数时，p取以上区间所有整数，在符合的p中取最大。
当x为负数时，将x转化为正数处理，p取以上区间所有奇数(因为只有奇数次幂才可能等于负数)，
在符合的p中取最大。