A very hard mathematic problem

E - A very hard mathematic problemTime Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d& %I64uSubmit Status Practice HDU4282Description　　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

95360 Sample Output

110　　Hint

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

思路：

这题思路是听队友说的，我感觉自己好像患了一个毛病，就是看到暴力过不了就放弃这种方法，但是不会去想可不可以优化

一下。这个毛病必须改，不然很多暴力思路错过了，就走错解题方向了。其实如果肯自信一点，凡事别一下子否定了，肯定

收获巨大。改，改，改。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;#define T 100005typedef long long ll;#define CRL(a) memset(a,0,sizeof(a))ll PoW(int a,int b){ll k=1;while(b){k*=a;b--;}return k;}bool find(int Y,int Z,int K){int X,ln=1,rn=Y;ll sum;while(ln<rn){X = (ln+rn)/2;sum = PoW(X,Z) + PoW(Y,Z) + X*Y*Z;if(sum == K)return true;if(sum > K)rn = X;elseln = X+1;}return false;}int main(){/*freopen("input.txt","r",stdin);*/int n,i,j,k;while(scanf("%d",&n),n){k=0;for(i=1;i<50000;++i)for(j=2;j<=32&&PoW(i,j)<n;++j)if(find(i,j,n))k++;printf("%d\n",k);}return 0;}