HDU 1061 Rightmost Digit

 

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. 
    
   
   
   
   

 

 

题意:

求n^n模10

代码:

#include<stdio.h>

int exp_mod(int a,unsigned int n,int b)
{
	int t;
	if(n==0||n==1)
	return n==0?1:a;
	t=exp_mod(a,n/2,b);
	t=t*t;
	if(n%2==1)t=t*a;
	return t%b;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		long long n;
		scanf("%lld",&n);
		printf("%d\n",exp_mod(n%10,n,10));
	}
	return 0;
}

 

分析:

代码是在网上找的模版,第一次做是想找规律的可是实在是太繁琐勒(其实是我懒= =! ),就上网看看有没有更简单的方法,才知道这样的题有模版,就是快速幂取模.果断背下来....

下面是我经过在网上找的解析:

快速幂取模就是在O(logn)内求出a^n mod b的值。算法的原理是ab mod c=(a mod c)(b mod c)mod c