ACM--快速幂--HDOJ 1061--Rightmost Digit

HDOJ地址：http://acm.hdu.edu.cn/showproblem.php?pid=1061

快速幂算法讲解：http://blog.csdn.net/qq_26891045/article/details/51334101

Rightmost Digit

Time Limit:2000/1000
MS (Java/Others)    Memory Limit:65536 /32768 K
(Java/Others)

Total Submission(s):45746  
  Accepted Submission(s):
17221

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.

================================天生骄傲的分割线=============================

题目的意思已经非常明确了，就是求n^n的个位数，求它的个位数就相当于进行mod 10运算，这里就直接用快速幂算法，就可以得出结果，题目给定的N可能会比较大，所以用long long比较保险。

#include<iostream>
#include<stdio.h>
using namespace std;
/**
  快速幂算法
*/
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;
}

参考博客：http://blog.csdn.net/whjkm/article/details/42803805