HDU 1097 a hard puzzle

A hard puzzle

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24457    Accepted Submission(s): 8676

Problem Description

lcy gives a hardpuzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know thea^b.everybody objects to this BT problem,so lcy makes the problem easier thanbegin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.Buteverybody is too lazy to slove this problem,so they remit to you who is wise.

 

 

Input

There are mutipletest cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

 

 

Output

For each testcase, you should output the a^b's last digit number.

 

 

Sample Input

7 66

8 800

 

 

Sample Output

9

6

 

很容易发现 a^b 的规律，个位数每四个一循环，有的虽然周期是2，但是4个一定是一个周期。

2: 2 4 8 16 32 64

3: 3 9 27 81 243 729

4: 4 16 64 256 1024

另外，我开始没有发现题目说a ,b 都 大于零。

因此，我的代码支持 0 ^ 0 =1(貌似在哪见过), 0 ^ 1=0 , 1 ^ 0=1;  

 

#include "string"
#include "iostream"

using namespace std;

int main(  )
{int m,n,s,i,tem;
  
  while(cin>>m>>n)
  {
  if( m==1||n==0 )
    s=1;
  else if( m%10==0 )
    s=0;
  else if( n==1 )
    s=m%10;
  else{ tem=1;
        m=m%10;//只保留个位数
	if( n%4==0 )
  {
	tem=m*m*m*m;
	}
else
    for( i=0;i<n%4;i++ )   
      tem*=m;
    s=tem%10;
  }
 
    cout<<s<<endl; 
  }

  return 0;
}