杭电 1061 Rightmost Digit计算N^N次方的最后一位

Problem Description
Given a positive integer N, you should output the most right digit of N^N.

Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output
For each test case, you should output the rightmost digit of N^N.

Sample Input
2
3
4
Sample Output
7
6
Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

　　

我写的错误？？？错误的原因是超时，

//rightmost digit
#include<iostream>
using namespace std;
int main()
{
    int n,tmp,m;
    cin>>n;
        while(n--){
        cin>>m;
        tmp=1;
        m%=10;
//        cout<<"m-==="<<m<<endl;
        for(int i=0;i<m;i++)
        tmp=tmp*m%10;
        cout<<tmp<<endl;
        }
    }

正确代码：

#include<iostream>
using namespace std;
int main()
{
    int a,ans,n=0;
    double dval = 0;
    int count=0;
    cin>>n;
    while(n--)
    {
        cin>>a;
        ans=1;
        count=a;
        a = a%10;
        while (count)
        {
            if (count&1==1)
                ans=(ans*a)%10;
            a=(a*a)%10;
            count>>=1;
        }
        cout<<ans<<endl;
    }
    return 0;
}

 解法二：

读完题首先想到的是大数，但是这大数貌似也太大了，就算能放下，这么多大数乘法铁定超时，考虑优化，写个小程序打表观察能发现这样一个规律，n的次方是有周期性的，且周期为4，这样就好办了，对于给定的N,求N*N的个位数，只需算N的个位数的N%4次方，然后对10取余就是所要求的结果了

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
int t;
cin >> t;
while (t--)
{
int n, m;
cin >> n;
m = n % 4;
if (m == 0)
m = 4;
n = n % 10;
cout <<int( pow(n, m)) % 10 << endl;
}
return 0;
}

PS:

//利用二进制输出 输入的数字 
#include<iostream>
using namespace std;
int main()
{
    int test,t,n,ans;
    while(cin>>n){
        
    test=n;
    t=1;
    ans=0;
    while(test){
        if(test&1==1) ans+=t;
        t*=2;
        test>>=1;
    }
    cout<<ans<<endl;
}
}