Hdu1061_rightmost Digit "Fast power take-over surplus"

rightmost Digit

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) total submission (s): 33161 Accepted Submission (s): 12696
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 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).

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 = rightmost, so the-the-digit is 7.
In the second case, 4 * 4 * 4 * 4 = at the very rightmost digit is 6.

Author

Ignatius.l

To give you a n, calculate the number of digits on the n^n digit is how much

Idea: The common method time-out, the use of fast power to take the remainder calculation n^n%10, here paste a binary

Code for fast power-over redundancy

#include <stdio.h> #include <string.h>__int64 quickpow (__int64 A,__int64 p) {    __int64 r = 1,base = A;    __int64 m = ten;    while (p!=0)    {        if (P & 1)            R = r * Base% m;        Base = base * Base% m;        P >>= 1;    }    return r;} int main () {    __int64 N;    int T;    scanf ("%d", &t);    while (t--)    {        scanf ("%i64d", &n);        __int64 ans = quickpow (n,n);        printf ("%i64d\n", ans);    }    return 0;}

Hdu1061_rightmost Digit "Fast power take-over surplus"