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

#include<cstdio>  
int mod(int a,int b,int k){  
    int ans=1;  
    a=a%k;  
    while(b>0){  
        if(b%2==1) ans=(ans*a)%k;  
        b/=2;  
        a=(a*a)%k;  
    }  
    return ans;  
}  
int
 
 main(){  
    int a,b;  
    while(scanf("%d%d",&a,&b)==2)  
        printf("%d\n",mod(a,b,10));  
    return 0;  
}  

這是直接的求法。看別人的題解，竟然還有一個規律：

尾數為0,1,5,6的不管是多少次方尾數依然不變，而尾數為4和9的每2次迴圈，

2,3,7,8為每4次迴圈。迴圈結果如下：

0，1，5，6：位數永遠是0，1，5，6

2：6，2，4，8迴圈

3：1，3，9，7迴圈

4：6，4迴圈

7：1，7，9，3迴圈

8：6，8，4，2迴圈

9：1，9迴圈

好的，可以水過了：

#include<cstdio>  
int div[10]={1,1,4,4,2,1,1,4,4,2};  
int f[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};  
int main(){  
    int a,b;  
    while(scanf("%d%d",&a,&b)==2)  
        printf("%d\n",f[a%10][b%div[a%10]]);  
    return 0;  
}