uva 10515 规律打表

Problem G Power et al. Input: Standard Input

Output: Standard Output

Finding the exponent of any number can be very troublesome as it grows exponentially J. But in this problem you will have to do a very simple task. Given two non-negative numbers m and n, you will have to find the last digit of mⁿ in decimal number system.

Input

The input file contains less than 100000 lines. Each line contains two integers m and n (Less than 10^101). Input is terminated by a line containing two zeroes. This line should not be processed.

Output

For each set of input you must produce one line of output which contains a single digit. This digit is the last digit of mⁿ.

 

Sample Input 

2 2

2 5

0 0                            

Output for Sample Input

4

2

题目大意：求m^n的最后一位数字

打表出0、1、2、3、4、5、6、7、8、9（50次方以内的数）

发现规律，最长的周期为4

（0） 0 0 0 0

（1）1 1 1 1

（2）2 4 8 6

（3）3 9 7 1

（4）4 6 4 6

（5）5 5 5 5

（6）6 6 6 6

（7）7 9 3 1

（8）8 4 2 6

（9）1 9 1 9

AC代码：

#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
string a,b;
int f[10][4]=
{
    0,0,0,0,
    1,1,1,1,
    6,2,4,8,
    1,3,9,7,
    6,4,6,4,
    5,5,5,5,
    6,6,6,6,
    1,7,9,3,
    6,8,4,2,
    1,9,1,9
};

int deal()
{
    int p=a[a.size()-1]-'0';
    int i,ret=0;
    for(i=0;i<b.size();i++)
        ret=(ret*10+b[i]-'0')%4;
    return f[p][ret];
}
int main()
{
    while(cin>>a>>b,!(a=="0" && a==b))
    {
        if(b=="0")
            cout<<1<<endl;
        else 
            cout<<deal()<<endl;
    }
    return 0;
}