学步园

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13791    Accepted Submission(s): 4822

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

/**循环节找规律， 用if..else 去讨论太麻烦了，
 * 直接用循环找循环节
 * 再取模
*/ 
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int dp[201];

void solve( int a, int b)
{ 
int i, j, t, k;
    k = 2;
    dp[1] = a;
for(i = 2; i <=  b; i++) {
        dp[i] = t = dp[i-1] * a % 10;
for(j = 0; j < i; j++)
if (dp[j] == t) {
            b = b % (i -1);
if ( b == 0)
            printf("%d\n",dp[i-1]);
else
            printf("%d\n",dp[b]);
return ;
        }
    }
    printf("%d\n",dp[b]);
}



int main( )
{
int a, b;
while (scanf("%d%d",&a, &b) != EOF) {
        memset(dp, 0, sizeof(dp));
        solve(a % 10 , b); 

    }
return 0;
}