hdu4282A very hard mathematic problem 暴力枚举

//给出k//找x,y,z使得x^z+y^z+x*y*z = k//x,y,z都为正整数x<y,z>1问有多少种方法//当z = 2时，可以看到左边是一个完全平方//而当z>=3时，可以暴力枚举x,y//由于k<2^31所以x<2^(31/3)枚举复杂度可以过#include<cstdio>#include<cstring>#include<iostream>#include<cmath>using namespace std ;const int maxn = 100 ;const int inf = 0x7fffffff ;typedef __int64 ll;ll pow(ll a , ll b){    ll c = 1;    while(b)    {        if(b&1)c*=a;        if(c > inf)return -1 ;        b >>= 1;        a *= a ;    }    return c ;}int main(){    int k;    while(scanf("%d" ,&k) && k)    {        int ans = 0 ;        int t = (int)sqrt((double)k) ;        if(t*t == (double)k)        ans += (int)(t-1)/2 ;        for(ll i = 1;i*i*i < k ;i++)           for(ll j = i + 1;j*j*j< k;j++)             for(ll s = 3;;s++)             {                 ll t1 = pow(i,s) ;ll t2 = pow(j , s) ;                 if(t1 == -1 || t2 == -1)                 break;                 if(t1 + t2 + i*j*s ==k)                 ans++ ;                 if(t1 + t2+ i*j*s > k)                 break;             }        printf("%d\n" ,ans) ;    }    return  0 ;}