hdu 4282 A very hard mathematic problem(二分+枚举)

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2581    Accepted Submission(s): 747

Problem Description

　　Haoren is very good at solving mathematic problems. Today he is working a problem like this: 

　　Find three positive integers X, Y and Z (X 1) that holds

　　 X^Z + Y^Z + XYZ = K

　　where K is another given integer.

　　Here the operator “^” means power, e.g., 2^3 = 2 * 2 * 2.

　　Finding a solution is quite easy to Haoren. Now he wants to challenge more: What’s the total number of different solutions?

　　Surprisingly, he is unable to solve this one. It seems that it’s really a very hard mathematic problem.

　　Now, it’s your turn.

Input

　　There are multiple test cases. 

　　For each case, there is only one integer K (0 　　K = 0 implies the end of input.

　　

Output

　　Output the total number of solutions in a line for each test case.

Sample Input

9
53
6
0

Sample Output

Hint

Source

2012 ACM/ICPC Asia Regional Tianjin Online

Recommend

liuyiding

思路：枚举y和次数z然后二分查找是否有符合条件的x出现y*y

#include
#include
using namespace std;
const int mm=1000;
long long yz;
bool ok(long long z,long long y,long long k)
{
  long long l=1,r=y-1,mid,xx,sum;
  while(l>k&&k)
  {
    ans=0;x=1;y=2;z=2;
    ///枚举y和次数z
    for(y=2;y*yk)
        break;
      yz=y*z;
      ///y,z确定，二分查找是否有符合条件的x出现
      if(ok(z,y,k-yy))ans++;
    }
    cout

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