nyoj473 A^B Problem(快速幂取模)

题目：

A^B Problem

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 2

描述

Give you two numbers a and b,how to know the a^b's the last digit number.It looks so easy,but everybody is too lazy to slove this problem,so they remit to you who is wise.

输入

There are mutiple test cases. Each test cases consists of two numbers a and b(0<=a,b<2^30)

输出

For each test case, you should output the a^b's last digit number.

样例输入

7 66
8 800

样例输出

9
6

提示

There is no such case in which a = 0 && b = 0。

代码：

#include 
#include
int pow_mod(int a,int b,int c)
{
    int s=1;
    a=a%c;
    while(b!=0)
    {
        if(b&1)
            s=s*a%c;
        a=a*a%c;
        b>>=1;
    }
    return s;
}

int main()
{
    int a,b,i,s;
    while(~scanf("%d%d",&a,&b))
    {
        printf("%d\n",pow_mod(a,b,10));
    }

    return 0;
}

快速幂按位取模，存代码

同余定理：所谓的同余，顾名思义，就是许多的数被一个数d去除，有相同的余数。d数学上的称谓为模。如a=6,b=1,d=5,则我们说a和b是模d同余的。因为他们都有相同的余数1。

(a*b)mod c= ((a mod c)*(b mod c)) mod c