Day8 二分-----A very hard mathematic problem

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
.

.
.

思路： 使用二分查找，但对x，y，z三个数中的哪一个进行查找需要判断，为了使时间复杂度最小，对三个数中范围最大的进行二分；
首先可得z的大致范围为2~31；
x为1~pow(k,1.0/z);
y范围最大，所以对y进行二分
.
.

代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int Max;
int k;bool ok(int x,int z){int l = x+1,r = Max;while(l<=r){int mid = (l+r)/2;if(pow(x*1.0,z*1.0)+pow(mid*1.0,z*1.0)+x*mid*z<k)l = mid+1;else if(pow(x*1.0,z*1.0)+pow(mid*1.0,z*1.0)+x*mid*z>k)r = mid-1;elsereturn true;}return false;
}int main()
{while(cin >> k && k!=0){int ans = 0;for(int z=2;z<31;z++){Max = pow(k,1.0/z);for(int x=1;x<=Max;x++){if(ok(x,z))ans++;}}cout << ans << endl;}return 0;
}

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Day8 二分-----A very hard mathematic problem

A very hard mathematic problem

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