hdu1061 rightmost digit（C语言）

最新推荐文章于 2018-08-24 19:21:02 发布

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最新推荐文章于 2018-08-24 19:21:02 发布

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本文详细介绍了快速幂取模算法，旨在解决求大数幂的模运算问题，避免溢出。通过逐步优化，从最初的朴素算法到最终的快速幂算法，时间复杂度降低到O(log b)。算法依赖于幂运算的性质，并采用二分法进行迭代，是程序设计中常用的高效算法。

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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.

Author

Ignatius.L

C语言AC代码

法一

#include <stdio.h>
#include <string.h>
#include<math.h>
int main()
{
    int a,d2=0,d3=0,d5=0,d7=0,sum,i,j,m,n,str[20]={0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};
    while(scanf("%d",&m)!=EOF)
    {
        for(i=0; i<m; i++)
        {
            scanf("%d",&n);
            n=n%20;
            printf("%d\n",str[n]);
        }
    }
}

思路：常规算法时间复杂度过高，所以可以用常规算法先找出规律，个位数20一循环，然后n%20即可，简单暴力。