hdu 1097 A hard puzzle 求a^b的个位数二分b

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本文介绍了一个算法问题，即如何快速计算a的b次方结果的最后一位数字。该问题通过一种高效的递归算法解决，适用于多个测试用例，并提供了C++代码实现。

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10738 Accepted Submission(s): 3691

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
#include<iostream>
#include<cstdio>
using namespace std;
int _pow(__int64 a,__int64 b)
{
    if(b==1) return a%10;
    if(b==0) return 1;
    int cnt=_pow(a,b>>1);
    cnt=cnt*cnt%10;
    if(b&1) cnt=cnt*a%10;
    return cnt;
}
int main()
{
    __int64 a,b;
    while(scanf("%I64d%I64d",&a,&b)==2)
    {
        int cnt=_pow(a,b);
        printf("%d/n",cnt);
    }
    return 0;
}