HDU - 1097 A hard puzzle（快速幂----或找规律）

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本文介绍了一种高效算法来求解大数幂运算结果的最后一位数字。通过两种方法实现：一是利用周期性规律进行查找；二是采用快速幂取模算法。这两种方法均可有效解决大数幂次方尾数问题。

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

方法一：找规律，打表

代码如下：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int table[10][4] =
    {
        0,0,0,0,
        1,1,1,1,
        2,4,8,6,
        3,9,7,1,
        4,6,4,6,
        5,5,5,5,
        6,6,6,6,
        7,9,3,1,
        8,4,2,6,
        9,1,9,1,
    };
    int a,b;
    while(~scanf("%d%d",&a,&b))
    {
        a = a % 10;
        b = (b - 1) % 4;
        printf("%d\n",table[a][b]);
    }
    return 0;
}

方法二：用快速幂取模

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int fun(int a,int b)
{
    a %= 10;
    int t = 1;
    while(b > 0)
    {
        if(b % 2 == 1)
        t = t * a % 10;
        a = a * a % 10;
        b /= 2;
    }
    return t;
}
int main()
{
    int a,b;
    while(~scanf("%d%d",&a,&b))
    {
        printf("%d\n",fun(a,b));
    }
    return 0;
}