HDU 1097 A hard puzzle （规律&&快速幂）

Problem Description

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

 

Input

There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

 

Output

For each test case, you should output the a^b's last digit number.

 

Sample Input

   
   
   
   

    
    
    
    7 66 8 800
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    9 6
   
   
   
   

 

题目大意：求x^n的末尾；

题解：证明得 ，只与x的个位有关， 相乘后也只和个位有关；

所以一定有规律or快速幂；

一：（打表法）

#include <bits/stdc++.h>
using namespace std ;
int dir[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};
int main()
{
    int a , b;
    while(cin>>a>>b)
    {
        int d = a % 10  ;
        if(d==0||d == 1 || d==5||d==6)
            printf("%d\n",d);
        else if(d==4||d==9)
            printf("%d\n",dir[d][b%2]);
        else if(d==2||d==3||d==7||d==8)
            printf("%d\n",dir[d][b%4]);
    }
    return 0 ;
}

二、（快速幂）

个位可以看做为：x^n%10;

#include <bits/stdc++.h>
using namespace std ;
int kkk(int a  , int b , int n )
{
    a%=n;//*取个位
    int ans ;
    ans = 1 ;//*定位
    while(b>0)
    {
        if(b%2==1) ans = ans*a%n; //*如果是奇数幂 ，则乘后变为偶次幂 ；
        b/=2; //*x^n 一定可以由  (x^2/n)^2得到；
        a = a*a%n;//*更新个位
    }
    return ans ;
}
int main()
{
    int a , b ;
    while(cin>>a>>b)
    {
        printf("%d\n",kkk(a,b,10));
    }
    return 0;
}