hdu4282 x^z+y^z+x*y*z=k 解的个数

题意:

     x^z + y^z + x*y*z = k; (x < y ,z > 1),给你一个k问有多少组解.

思路:

     
 暴力枚举z,y,然后二分查找x.注意一点最好用快速幂,别用pow,不然有可能会超时,如果先把z=2的处理了会快一点.应该会0ms.....

#include<stdio.h>

__int64 quickp(__int64 a,__int64 n)
{
   __int64 aa=1;
   while(n)
   {
       if(n&1)
       aa*=a;
       a*=a;
       n>>=1;
   }
   return aa;
}
int main ()
{
   __int64 x ,y ,z ,i ,j ,k;
   __int64 low ,up ,mid;
   while(~scanf("%I64d" ,&k) && k)
   {
      __int64 sum = 0;
      for(z = 2 ;z <= 31 ;z ++)
      {
         for(y = 2 ;y <= 46341 ;y ++)
         {
            if(quickp(y ,z) > k) break;
            low = 1;
            up = y-1;
            __int64 mk = 0;
            while(low <= up)
            {
               mid = (low + up) / 2;
               if(quickp(mid ,z) + quickp(y ,z) + mid*y*z >= k)
               {
                  up = mid - 1;
                  mk = mid;
               }
               else
               low = mid + 1;
            }
            if(quickp(mk ,z) + quickp(y ,z) + mk*y*z == k)
            sum ++;
         }
      }
      printf("%I64d\n" ,sum);
   }
   return 0;
}