【杭电 1061】 Rightmost Digit

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.

主要掌握模幂运算，而且求个位，一直求10的模

这个是大数问题吧，不过数据太大一定有技巧的，n^n(n<=1000000000), 题目要取的是个位数字，所以我们只有把（n）^n就可以了，因为进位对个位不影响，对于n的次方还是很大，我们就先求n/2，可是还是要考虑到n为奇数的时候 t=t*t*a，（a=n）；比如当n=5的时候n/2=2，t=(a^2)(a^2)(a)=a^5,但a^4即n=4；n/2=2；此时t=(a^2)*(a^2)

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int node(__int64 m)
{
int t=1,b;
b=m%10;
if(b==0)
return 0;
while(m)
{
if(m%2==1)
{
t*=b;
t%=10;
}
b*=b;
b%=10;
m/=2;
}
return t;
}
int main()
{
int t;
__int64 n;
scanf("%d",&t);
while(t--)
{
scanf("%I64d",&n);
printf("%d\n",node(n));
}
return 0;
}