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 
#include 

#include 
#include 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;
}

 

转载于:https://www.cnblogs.com/372465774y/archive/2012/10/22/2734097.html