HDU 1097.A hard puzzle【快速幂或规律】【8月12】

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.

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

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

7 66
8 800

9
6

#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;
}

#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;
}