杭电1097-A hard puzzle（快速幂）

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本文介绍了一个经典的算法问题：如何快速计算a的b次方的最后一位数字。通过使用快速幂算法，该问题可以在限定的时间内得到解决。文章提供了一个C++实现示例，演示了如何通过迭代方式高效地计算出结果。

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 35177 Accepted Submission(s): 12627

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

这个题要用快速幂，否则会超时！

#include<cstdio>
int fun(int x,int y)
{
    x%=10;
    int t=1;
    while(y>0)
    {
        if(y&1)
        t=t*x%10;
        
        x=x*x%10;
        y/=2;
    }
    return t;
}
int main()
{
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        printf("%d\n",fun(a,b));
    }
    return 0;
}