HDU 1061.Rightmost Digit "number theory and Method" "August 30"

Rightmost Digit

Problem Descriptiongiven A positive integer N, you should output the most right digit of n^n.

Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow.
Each test case is contains a single positive integer N (1<=n<=1,000,000,000).

Outputfor Each test case, you should output the rightmost digit of n^n.

Sample Input

234

Sample Output

HintIn the first case, 3 * 3 * 3 = rightmost, so the digit is 7.In the second case, 4 * 4 * 4 * 4 = The rightmost digit is 6.

give you a number, output the single digit of the n^n. This problem is much simpler than it is. The method of this problem uses the second method of HDU 1097, can refer to my blog .

the mantissa is 0,1,5,6 regardless of how many times the mantissa remains unchanged, while the mantissa is 4 and 9 for every 2 cycles,
2,3,7,8 for every 4 cycles. The loop results are as follows:
0,1,5,6: The number of digits is always 0,1,5,6
2:6,2,4,8 Cycle
3:1,3,9,7 Cycle
4:6,4 Cycle
7:1,7,9,3 Cycle
8:6,8,4,2 Cycle
9:1,9 Cycle

Find this law, this problem is not difficult. Otherwise, no idea.

The code is as follows:

#include <cstdio>int main () {    int t,n,f[10][10]={{1,0},{1,1},{4,2,4,8,6},{4,3,9,7,1},{2,4,6},{1,5},{1,6} , {4,7,9,3,1},{4,8,4,2,6},{2,9,1}};    scanf ("%d", &t);    while (t--) {        scanf ("%d", &n);        int m=n%f[n%10][0];        if (!m) m=f[n%10][0];        printf ("%d\n", F[n%10][m]);    }    return 0;}

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HDU 1061.Rightmost Digit "number theory and Method" "August 30"