HDU1061 Rightmost Digit

1061
Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57033 Accepted Submission(s): 21556

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.

Author
Ignatius.L

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只求个位上的数，总结出各个数字的幂的个位变化规律就行，即都是<=4循环

#include<cstdio>#include<cmath>int main(){    int n,N;    scanf("%d", &N);        while (N--&&scanf("%d", &n)!=EOF)        {            int mi, temp = 1;            if (n % 4 == 0)mi = 4;            else mi = n % 4;            n = n % 10;                while (mi--)                    temp *= n;                printf("%d\n", temp % 10);        }    return 0;}