HDU-1061-Rightmost Digit

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.

这道题主要是要求N^N的结果的最后一位，我在这里用的是快速幂算法，不会这算法的可以点这个链接http://blog.csdn.net/qq_38712932/article/details/76218952

这里需要注意的就是在快速幂中需要对这个形参a对10取余，后面的每部也需要注意取余

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

我看别人写的题解后，发现有位大佬写的算法贼六

#include <cstdio>
int main()
{
int n,sum=1,f,m;
scanf("%d",&m);
while(m --)
{
sum=1;
scanf("%d",&n);
f=n%10;
if(n%4==0)
n=4;
else
n=n%4;
for(int a = 0; a < n; a ++)
sum=sum*f%10;
printf("%d\n",sum);
}
return 0;
}

这个代码的意思是N^N次方从1到N和N相乘你测试时会发现最后一位数有一个周期为4，但是4，5等一系列的数的周期是2，5的周期是1，那么可以将其看作是周期是4的数。