hdu4282 A very hard mathematic problem

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3017 Accepted Submission(s): 875

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

95360

Sample Output

110　　
Hint
9 = 1^2 + 2^2 + 1 * 2 * 253 = 2^3 + 3^3 + 2 * 3 * 3
 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

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liuyiding

暴力就可以过！

#include <iostream>#include <stdio.h>#include <math.h>using namespace std;__int64 pow(__int64 a,__int64 b){    __int64 i,ans=1;    for(i=0;i<b;i++)    {        ans*=a;    }    return ans;}int main(){   __int64 k,ans,a1,a2,sum;   __int64 x,y,z;   while(scanf("%I64d",&k)!=EOF&&k)   {       ans=0;        __int64 temp =(__int64) sqrt(k);       if(temp*temp==k)//加这里，就是250ms,不加，就超时了！            ans += (temp-1)>>1;       for(z=3;z<31;z++)       {           for(x=1;;x++)           {               a1=pow(x,z);               if(a1>k/2)               break;               for(y=x+1;;y++)               {                   a2=pow(y,z);                   if(a2>k)                   break;                   sum=0;                   sum=a1+a2+x*y*z;                   if(sum>k)                   break;                   if(sum==k)                   {                       ans++;                       break;                   }               }           }       }       printf("%I64d\n",ans);   }    return 0;}