hdu 4282 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): 2489    Accepted Submission(s): 728

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

2012 ACM/ICPC Asia Regional Tianjin Online

 

 

Recommend

liuyiding

 

 

解法 枚举 z，x，二分查找 y
用pow（） 984ms （差点超时，吓死了）
自己用二进制快速次方算 531ms 
看来这个浮点运算真是挺慢滴
 
984MS

#include <iostream>

#include <map>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

using namespace std;

int k;

int del(__int64 x,__int64 y,int z)

{

    int m=(int)pow(k*1.0,1.0/z);

    if(y>m) return 1;

    __int64 sum=(__int64)pow(x*1.0,z*1.0)+(__int64)pow(y*1.0,z*1.0);

    sum+=x*y*z;

    if(sum>k) return 1;

    if(sum<k) return -1;

   // printf("%I64d %I64d %d %I64d ",x,y,z,sum);

    return 0;

}

bool ok(int x,int z)

{

    int l=x+1,r=50000,m,flag;

    while(l<=r)//二分查找

    {

        m=(l+r)>>1;

        flag=del(x,m,z);

        if(flag>0) r=m-1;

        else if(flag<0) l=m+1;

        else return true;

    }

    return false;

}

int main()

{

     __int64 x,z;



     while(scanf("%d",&k),k)

     {

         __int64 cnt=0;

         __int64 s;

         for(z=2;;z++)

         {

              s=(__int64)pow(2.0,z*1.0);

             if(s>=k) break;

             for(x=1;;x++)

             {

                 s=(__int64)pow(x*1.0,z*1.0);

                 s=s*2+(__int64)x*x*z;

                 if(s>=k) break;

                 if(ok(x,z)) cnt++;

             }

         }

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

     }

    return 0;

}


  531MS

#include <iostream>

#include <map>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

using namespace std; int k; __int64 Pw(__int64 a,int b) { __int64 t=1; for(;b>0;b=(b>>1),a=(a*a)) if(b&1) t=t*a; return t; } int del(__int64 x,__int64 y,int z) { int m=(int)pow(k*1.0,1.0/z); if(y>m) return 1; __int64 sum=Pw(x,z)+Pw(y,z);

    sum+=x*y*z; if(sum>k) return 1; if(sum<k) return -1; return 0; } bool ok(int x,int z) { int l=x+1,r=50000,m,flag; while(l<=r) {

        m=(l+r)>>1;

        flag=del(x,m,z); if(flag>0) r=m-1; else if(flag<0) l=m+1; else return true; } return false; } int main() { __int64 x,z; while(scanf("%d",&k),k) { __int64 cnt=0; __int64 s; for(z=2;;z++) {

              s=Pw(2,z); if(s>=k) break; for(x=1;;x++) {

                 s=Pw(x,z);

                 s=s*2+x*x*z; if(s>=k) break; if(ok(x,z)) cnt++; } }

         printf("%I64d\n",cnt); } return 0; }