A very hard mathematic problem HDU

A very hard mathematic problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7411 Accepted Submission(s): 2232

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 < Y, Z > 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 < 2^31) in a line.
　　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

1
1
0
　　
Hint

9 = 1^2 + 2^2 + 1 * 2 * 2
53 = 2^3 + 3^3 + 2 * 3 * 3

Source
2012 ACM/ICPC Asia Regional Tianjin Online

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1.枚举+快速幂

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define maxx 1e9
using namespace std;
typedef long long ll;
ll powmod(ll a, ll b)//快速幂
{ll ans = 1;while(b){if(b%2==1)ans = (ans * a);b=b/2;a = (a*a);}return ans;
}
int main()
{int i,j;int k;int z;while(cin>>k, k){int ans = 0;ll tx,ty;ll tk = (int)sqrt(k*1.0);if(tk*tk == k)ans += (tk-1)/2;for(z=3; z<31; z++){for(ll x=1; ; x++){ll tx = powmod(x,z);if(tx >= k/2) break;for(ll y=x+1; ; y++){ll ty = powmod(y, z);if( tx + ty + x*y*z > k) break;else if(tx + ty + x*y*z == k){ans++;break;}}}}printf("%d\n", ans);}return 0;
}

2.暴搜+二分

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define maxx 1e9
using namespace std;
typedef long long ll;
ll powmod(ll a, ll b)
{ll ans = 1;while(b){if(b%2==1)ans = (ans * a);b=b/2;a = (a*a);}return ans;
}
int main()
{int i,j;int k;int z;while(cin>>k, k){int ans = 0;ll tx,ty;ll lt,rt;ll tk = (int)sqrt(k*1.0);if(tk*tk == k)ans += (tk-1)/2;for(z=3; z<31; z++){for(ll y=2; ;y++){ty = powmod(y, z);if(ty > k) break;lt =1; rt=y-1;while(lt<=rt){ll x= (lt+rt)/2;ll ta=pow(x, z) + pow(y, z) + x*y*z;if(ta < k){lt++;}else if(ta > k){rt--;}else{ans++;break;}}}}printf("%d\n", ans);}return 0;
}

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