GDUFE ACM-1010

题目：http://acm.gdufe.edu.cn/Problem/read/id/1010

 

A hard puzzle（简单题）

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

        GOJ gives a hard puzzle to C_Shit_Hu, super13, TestN,LOP: gave a and b,how to know the a^b.Everybody objects to this BT problem,so GOJ 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

思路：找规律，0 1 5 6每次的末位数都是自己本身，4 9是两次一循环，剩下的都是四次一循环，我直接用b%4看看应该用哪个末位数……超级暴力==

难度：如果像我这样取巧。。。还是比较简单的，但是。。还是用正常办法比较好，使用快速幂，我还没有学会，有时间再去看看。。

代码：

#include<stdio.h>
int main()
{
    long long int a,b,c,i;
    while(~scanf("%lld %lld",&a,&b))
    {
        c=1;
        if(b%4==0)
            c=a*a*a*a;
        else for(i=0;i<(b%4);i++)
        {
            c=c*a;
        }
        c=c%10;
        printf("%lld\n",c);
    }
    return 0;
}