acm--Fast Power--hdoj 1061--rightmost Digit

Hdoj Address: http://acm.hdu.edu.cn/showproblem.php?pid=1061

Fast Power algorithm Explanation: http://blog.csdn.net/qq_26891045/article/details/51334101

Rightmost DigitTime Limit: -/1 theMS (java/others) Memory Limit:65536 /32768K (Java/others)Total Submission (s):45746 Accepted Submission (s): 17221

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

 the Hint in the first case, 3 * 3 * 3 = digit, so the rightmost is 7.In the second case, 4 * 4 * 4 * 4 =, so the rightm OST digit is 6.

================================ born proud of the split line =============================

The meaning of the topic has been very clear, that is, to seek the single digit of n^n, it is equivalent to a single digit of the MoD 10 operation, here directly with the fast power algorithm, you can get the result, the topic given n may be relatively large, so a long long is more insurance.

#include <iostream> #include <stdio.h>using namespace std;/**  fast power algorithm */long long calculation (long long A, Long long B,long long c) {   int ans=1;   a=a%c;   while (b>0) {      if (b%2==1)        ans= (ans*a)%c;      B=B/2;      A= (a*a)%c;   }   return ans;} int main () {   int n,n;   cin>>n;   while (n--) {     cin>>n;     printf ("%d\n", Calculation (n,n,10));   }   return 0;}

Reference Blog: http://blog.csdn.net/whjkm/article/details/42803805

acm--Fast Power--hdoj 1061--rightmost Digit