Rightmost Digit

浏览数：35 / 时间：2015年06月09日

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

Total Submission(s): 6667 Accepted Submission(s): 1721

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

思路： long long int 因为已知容易超出限制，所以此处一定要注意 ,另外，利用/2的方式进行这样可以大大缩短程序运行时间。

 1 #include<stdio.h> 
 2 //大数问题 
 3 long long  Solve(long long  n)
 4 {
 5      long long int count=n;
 6      long long int result=1;
 7      while(count)
 8      {
 9          if(count%2!=0)
10          {
11              result*=n;
12            if(result>=10)
13              result=result%10;
14         }
15         n*=n;
16         n=n%10;       
17          count/=2;
18      }
19      return result;
20 }
21 
22 
23 int main(int argc, char *argv[])
24 {
25     long long  T,N;
26     scanf("%lld",&T);
27     while(T--)
28     {
29         scanf("%lld",&N);
30         printf("%lld\n",Solve(N));
31         
32     } 
33     return 0;
34 }