杭电OJ 1097 A hard puzzle（我最后的博客水题报告）

A hard puzzle
Problem Description
lcy gives a hard puzzle tofeng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybodyobjects 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.Buteverybody is too lazy to slove this problem,so they remit to you who is wise.
 
 
Input
There are mutiple test cases. Each test casesconsists of two numbers a and b(0<a,b<=2^30)
 
 
Output
For each test case, you should output thea^b's last digit number.
 
 
Sample Input
7 66
8 800
 
 
Sample Output
9
6
 

这是我最后一次在博客上发水题的解题报告了。我虽然还会在杭电上刷水题，但是不会再做（过于简单的）水题的解题报告了。

当然，如果遇到困住我的水题（~~脸红），我还是会写出详细的解题报告，只不过不会把这道题放在水题的分类中，而是放在经验总结中。

就把这道无耻打表AC题，作为水题报告的终点吧。

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<pre name="code" class="cpp">/*规律十分明显，但由于取余的关系，对周期T取余的值域为{0, 1, 2,...,T-1}，
末尾数字i的周期T的最后一个数在ans[i][0]中，而不是ans[i
4000
][T]中*/
#include <stdio.h>
int ans[10][10] = {{0},
{1},
{6, 2, 4, 8},
{1, 3, 9, 7},
{6, 4},
{5},
{6},
{1, 7, 9, 3},
{6, 8, 4, 2},
{1, 9}};
int T[10] = {1, 1, 4, 4, 2, 1, 1, 4, 4, 2};
/*末尾数字i 对应的周期就是 T[i]*/
int main()
{
int a, b;
while (scanf("%d%d", &a, &b) != EOF) {
a = a % 10;
b = b % T[a];
printf("%d\n", ans[a][b]);
}

return 0;
}