HDU ACM 1097 Hard Puzzle

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19595    Accepted Submission(s): 7003

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

Author

eddy

Recommend

JGShining

 

 AC代码：

#include<stdio.h>
#include<string.h>
int temp[100];
int main()
{
    long i, n = 1, m;
    long long a, c;
    while(scanf("%lld%lld", &a, &c) != EOF)
    {
        memset(temp, 0, sizeof(temp));
        temp[0] = a%10;
        for(i=1; i<100; i++)
        {
            if((temp[i] = (temp[0]*temp[i-1])%10) == temp[0]) break;

        }
        printf("%d\n", temp[(c-1)%i]);
    }
    return 0;
}

结题报告：

实话实说，很久没编程的感觉就连ACM Step中简单的题目都忘了如何做了，似乎以前做过的题今天拿出来好似未曾相见。WA了差不多一个晚上，弄坏了心情最后什么都做不了！

本是不属于这一路的人，当初的选择没说过要后悔！