HDU 4282 A very hard mathematic problem 第37届ACM/ICPC长春赛区网络赛1005题 (暴力)

A very hard mathematic problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 964    Accepted Submission(s): 279


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
 

 

Recommend
liuyiding
 
 
 
 
暴力枚举z就可以了。z肯定比较小的。
当z等于2时就是平方和公式。
#include<stdio.h>

#include<math.h>

#include<string.h>

#include<algorithm>

#include<iostream>

using namespace std;



long long pow(long long a,int n)

{

    long long ret=1;

    long long temp=a;

    while(n)

    {

        if(n&1)ret*=temp;

        temp*=temp;

        n>>=1;

    }

    return ret;

}

int main()

{

    long long K;

    long long ans;

    while(scanf("%I64d",&K),K)

    {

        ans=0;

        long long temp=(long long)sqrt(K);

        if(temp*temp==K)ans+=(temp-1)/2;

        for(int z=3;z<31;z++)

        {

            for(long long x=1;;x++)

            {

                long long u=pow(x,z);

                if(u*2>=K)break;

                for(long long y=x+1;;y++)

                {

                    long long v=pow(y,z);

                    if(u+v+x*y*z>K)break;

                    if(u+v+x*y*z==K)ans++;

                }

            }

        }

        printf("%I64d\n",ans);

    }

    return 0;

}

 

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