HDOJ 1062 Rightmost Digit

【题意】求N^N，输出最右边的那一位。

【代码1：获得周期】

#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

int main()
{
    int N = 0;
    cin >> N;
    while (N--)
    {
        int mul = 1, n = 0, r = 0, i = 0;
        cin >> n;
        r = n%10;
        for (i = 0; i < n; i++)
        {
            mul *= r;
            mul = mul%10;
        }
        cout << mul << endl;
    }
    return 0;
}

<pre name="code" class="cpp">#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

int main()
{
    int N = 0;
    int ans[10][4] = {{0,0,0,0},{1,1,1,1},
                    {2,4,8,6},{3,9,7,1},
                    {4,6,4,6},{5,5,5,5},
                    {6,6,6,6},{7,9,3,1},
                    {8,4,2,6},{9,1,9,1}};
    cin >> N;
    while (N--)
    {
        int n = 0, r = 0, l = 0;
        cin >> n;
        r = n%10;
        l = (n-1)%4;
        cout << ans[r][l] << endl;
    }
    return 0;
}

来自：http://blog.csdn.net/lovelyloulou/article/details/5241471

这种方法用标志数组去标志周期，但是周期不是1的情况下，如果周期序列中连续出现了两个值相同这种方法就会有bug。但是某一个数连续相乘可能也不会出现这种问题。

#include <iostream>
#include <cstring>
#include <stdio.h>
using namespace std;

bool l[10];
int r[10];
int main()
{
    int t;
    while(cin>>t)
    {
        while(t--)
        {
            memset(l,0,sizeof(l));
            memset(r,0,sizeof(r));
            int n = 0;
            cin>>n;
            int a=n%10;
            int b=a;
            int i=1;
            l[b]=true;
            r[0]=b;
            b=(b*a)%10;
            while(!l[b])
            {
                l[b]=true;
                r[i++]=b;
                b=(b*a)%10;
            }
            cout<<r[(n-1)%i]<<endl;
        }
    }
    return 0;
}