A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 34106 Accepted Submission(s): 12255



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

[align=left]Output[/align]
For each test case, you should output the a^b's last digit number.

[align=left]Sample Input[/align]

7 66
8 800

[align=left]Sample Output[/align]

9
6

[align=left] [/align]
#include<stdio.h>

#include<math.h>

main()

{

int a,b,i,j,c,x[1000],h;

while(scanf("%d%d",&a,&b)!=EOF)

{

c=a%10;

j=1;

x[1]=a%10;

for(i=2;i<=100;i++)

{

c=c*x[1];

c=c%10;

j++;

//printf("%d\n",c);

x[j]=c;

if(x[1]==x[j])

break;

}

h=b%(j-1);

if(h==0)

h=j-1;

printf("%d\n",x[h]);

}

return 0;

}