hdu 1.2.5

求N^N的最后一位数，实在想不出来，看了晚上的解法。可以找规律，也可以硬求用二分幂的方法。

//找规律的方法
#include<stdio.h>
int main()
{
  long n,k;
  int i,m,remainder=0;
     scanf("%d",&m);
     for(i=0; i<m; i++)
    {
        scanf("%ld",&n);
        k = n;
        n = n%10;
        if(n == 1 || n == 5 || n == 6 || n == 0|| n == 9 )
        {
            remainder = n;
        }
        else if( n == 4)
        {
            remainder = 6;
        }
        else if(n == 2 || n == 8)
        {
            if(k%4 == 2)
            remainder = (n*n)%10;

            else if(k%4 == 0)
            remainder = (n*n*n*n)%10;
        }
        else if(n == 3 || n == 7)
        {
            if(k%4 == 1)
                remainder = n;
            else if(k%4 == 3)
                remainder = (n*n*n)%10;
        }
        printf("%d/n",remainder);
    }
    return 0;
}
//二分幂的方法
#include <stdio.h>
int binary_exp(int base, int exp) {
    if (exp == 0)
        return 1;
    if (exp & 1) {
        return (base * binary_exp(base, exp - 1)) % 10;
    } else {
        exp >>= 1;
        int tmp = binary_exp(base, exp) % 10;
        return (tmp * tmp) % 10;
    }
}

int main() {
    int test, n;
    scanf("%d", &test);
    while (test--) {
        scanf("%d", &n);
        int base = n % 10, exp = n;
        printf("%d/n", binary_exp(base, exp));
    }
    return 0;
}