HDOJ 1097 A hard puzzle(循环节||快速幂)lcy gives a hard puzzle to

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

Sample Output

9
6

法一：快速幂（指数的形式而且数值较大，直接快速幂）

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL __int64
#define inf 0x3f3f3f3f
using namespace std;

LL n,m,e1,e2;
LL sf(LL a,LL b)
{
    LL ba=a,r=1;
    while(b!=0)//直到b除到0为止
    {
        if(b%2)//当b为奇数时才垒乘ba
        {
            r=(r*ba)%10;
        }
        ba=(ba*ba)%10;
        b>>=1;
    }
    return r;
}
int main()
{
    LL k,i,j,l;
    while(~scanf("%lld%lld",&n,&m))
    {
        l=sf(n,m);
        printf("%lld\n",l);
    }
    return 0;
}

法二：（找循环节）

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL __int64
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
    int n,m,j,k;
    while(~scanf("%d%d",&n,&m))
    {
        m%=4;
        n%=10;//就是底数没有取余。。而WA
        if(m==0)//当为0时再置为4
            m=4;
        k=1;
        while(m--)
        {
            k*=n;
        }
        printf("%d\n",k%10);
    }
}