HDU 1061 Rightmost Digit（简单快速幂）

http://acm.hdu.edu.cn/showproblem.php?pid=1061

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39858    Accepted Submission(s): 15045

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：简单的快速幂题目。求a的b次方所得结果的最后一位数。

解题思路：这是一题较简单的快速幂。以下是我对快速幂的理解，具体看代码。

 快速幂思路：
• 求a^b，是不是要把a乘b次呢？这样也行， 只是慢了点 • 例如求a^8，先把a乘a一次，得到a^2，然后把 a^2和a^2乘一次，得到a^4，然后把a^4和a^4乘一 下就得到a^8了 • 假如要求a^10呢？10=8+2，a^10=a^8*a^2 • 把10化为二进制， 10=1010(2),1010(2)=1000(2)+10(2)

类似的题型有：http://blog.csdn.net/hellohelloc/article/details/47725353

<span style="font-size:18px;"><span style="font-size:24px;">#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int q_pow(int a,int b)
{
    int ans=1;//ans初始化为1
    while(b>0)
    {
        if(b&1) ans =(ans*a)%10;//如果b为奇数，把ans乘于a
        a =(a*a)%10;//把a变成a^a,并且%10,防止溢出
        b>>=1;//b右移一位，相当于除于2
    }
    return ans;
}
int main()
{
   int n,cas;
   cin>>n;
   while(n--)
   {
       cin>>cas;
       printf("%d\n",q_pow(cas%10,cas));//a%10在这是为了提速，原因自己想想
   }
   return 0;
}
</span>
</span>