HDU 1097 A hard puzzle(快速幂模板||规律)

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14975    Accepted Submission(s): 5314

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

Author

eddy

Recommend

JGShinin

方法一：找规律

解题报告：此题是一个找规律的题目，结果至于尾数有关，而尾数从2到9的几次方有个规律，尾数是1,5，6的循环规律是1。尾数是4,9的循环规律是2，尾数是2,3，7,8，的循环规律是4.所以最大的公倍数是4；

方法一： 找规律：

代码如下：

#include<iostream>
#include <cstdio>
int main()
{
    int a[4],m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        a[1]=m%10;
        a[2]=(a[1]*a[1])%10;
        a[3]=(a[1]*a[2])%10;
        a[0]=(a[1]*a[3])%10;
        n=n%4;
        printf("%d\n",a[n]);
    }
    return 0;
}

方法二：快速幂模

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
//m^n  % k 的快速幂：
__int64 Expmod(__int64 m,__int64 n,int k)
{
    __int64 b=1;
    while(n>0)
    {
        if(n%2==1)
        {
            b=(b*m)%k;
        }
        n=n/2;
        m=(m*m)%k;
    }
    return b;
}
int main()
{
    __int64 x,y;
    while(scanf("%I64d%I64d",&x,&y)!=EOF)
    {
        printf("%I64d\n",Expmod(x,y,10));
    }
    return 0;
}