hdu4282 A very hard mathematic problem

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

Sample Output

1
1
0

Hint

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

分析：

如果用三重循环依次枚举的话必定超时。所以我们考虑只枚举x和z，然后再二分找符合的y值。

先打表，可以缩短时间。

注意double类型二分函数的写法与int类型二分函数的写法又和不同。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<math.h>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<queue>
#include<iomanip>
using namespace std;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
int num[50000][35] = {0};
int n;
bool jud(int x,int z,int miu)
{int l = x+1, r= 45000;while(l<=r){int mid = (l+r)>> 1;if(num[mid][z] <= 0){r = mid-1;continue;}if(pow(mid,z)+x*mid*z > miu){r = mid-1;                  //here wa  以 2和4 为例 ！！  切记！}else if(pow(mid,z)+x*mid*z < miu) {l = mid+1;                  // wa}else return true;}return false;
}
int main()
{for(int x=1;x<45000;x++){num[x][1] = x;for(int z=2;z<31;z++){num[x][z] = num[x][z-1]* x;if(num[x][z] >= 2e9){break;}}}while(scanf("%d",&n) != EOF){if( n== 0 ) break;int cnt = 0;for(int x=1;x<45000&&x<n;x++){for(int z=2;z<31;z++){if(num[x][z] <= 0 ) break;int miu = n - num[x][z];if(miu <= 0) break;if(jud(x,z,miu)){cnt++;}}}printf("%d\n",cnt);}
}

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