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Rightmost Digit（杭电1061）（求N^N的最低位）
Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33558 Accepted Submission(s): 12831
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).
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 6HintIn 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.
Author
Ignatius.L
/*求N^N的最低位。只能观察尾数的规律。
很容易发现0,1,5,6的任何次幂都为本身。
2,3,7,8是每四次幂循环一次。4,9均是每两次幂一循环
*/
#include<stdio.h>
int a[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};
int main()
{
int test,n;
scanf("%d",&test);
while(test--)
{
scanf("%d",&n);
if(n%10==0||n%10==1||n%10==5||n%10==6)
printf("%d\n",a[n%10][0]);
if(n%10==4||n%10==9)
printf("%d\n",a[n%10][n%2]);
if(n%10==2||n%10==3||n%10==7||n%10==8)
printf("%d\n",a[n%10][n%4]);
}
return 0;
}Rightmost Digit（杭电1061）（求N^N的最低位）
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