HDOJ 1097 A hard puzzle

Problem Description

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

Input

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

Output

For each test case, you should output the a^b‘s last
digit number.

Sample Input

7 66 8 800

Sample Output

9 6

这道题也是一道求最后一个数字的，只是方式稍微改变了一下...

 1 #include <iostream>

 2 using namespace std;

 3 int main()

 4 {

 5     int a,b;

 6     int round=0;

 7     int r[100]={0};

 8     int i=0;

 9     while(cin>>a>>b)

10     {

11         r[1]=a%10;

12         r[2]=(a%10)*(a%10)%10;//这里是关键，开始是a*a%10，提交提示WA。

13         for(i=3;;i++)

14         {

15             r[i]=r[i-1]*(a%10)%10;

16             if(r[i]==r[1])

17             {round=i-1;break;}

18         }

19         if(b%round==0)

20             cout<<r[round]<<endl;

21         else cout<<r[b%round]<<endl;

22     }

23     return 0;

24 }

HDOJ 1097 A hard puzzle