Rightmost Digit（HDU 1061）
Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33823 Accepted Submission(s): 12912
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.
每个数的n^n的个位数最多是以4为周期的乘积，所以只要对每个数做最多四次乘积，由于数比较大，所以首先要求先求尾数在进行乘积，不然会爆。
代码：
#include<stdio.h>
int main()
{
int n,a,b,ans;
scanf("%d", &n);
while(n--)
{
scanf("%d", &a);
b = (a-1)%4+1; //最大以4为周期，所以对4取余
a = a%10; //必须求尾数之后再进行相乘，不然会爆
if(b==1)
ans = a;
if(b==2)
ans = a*a;
if(b==3)
ans = a*a*a;
if(b==4)
ans = a*a*a*a;
ans = ans%10; //最后再求尾数
printf("%d\n", ans);
}
return 0;
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