HDOJ 题目1061Rightmost Digit（规律）

Jogging_Clown

于 2014-09-14 01:23:51 发布

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本文介绍了一种高效算法来解决给定正整数N时，计算N^N的最右位数的问题。通过分析数字的幂次方规律，利用取模运算减少计算复杂度，提供了一个简洁的C语言实现示例。

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Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32374 Accepted Submission(s): 12441

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

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.

ac代码

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int main()
{
    int n;int s,a,b,num;
    scanf("%d",&n);
    while(n--)
    {
        
        scanf("%d",&num);
        a=num%10;
        b=num%4;
        if(b==0)
            b=4;
        printf("%d\n",(int)pow((double)a,b)%10);
    }
}