HDU 1061 Rightmost Digit(快速幂取模)

                                           Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45296    Accepted Submission(s): 17031

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

题解：快速幂取模！！！

#include<iostream>
#include<cstdio>
int qmod(long long a,long long b,int c)
{
	int ans=1;
	a=a%c;
	while(b>0){
		if(b&1)
		ans=(ans*a)%c;
		b=b>>1;
		a=(a*a)%c;
	}
	return ans;
}
int main()
{
	int n;  long long t;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%lld",&t);
		printf("%d\n",qmod(t,t,10));
	}
	
	return 0;
}

听说是ans的周期是20......(无语。。。)

#include<stdio.h>
int main()
{
	int T,n;
	int a[25]={0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
	
	scanf("%d",&T);
	
	while(T--)
	{
		scanf("%d",&n);
		printf("%d\n",a[n%20]);
		
	}
	return 0;
}