杭电oj1097

最新推荐文章于 2020-07-05 15:31:39 发布

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最新推荐文章于 2020-07-05 15:31:39 发布

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分类专栏： acm 文章标签： C 杭电

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A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23294 Accepted Submission(s): 8186

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

#include<stdio.h>
int main()
{
int a,b;
int map[9][9]={
  {1},
  {2,4,8,6},
  {3,9,7,1},
  {4,6},
  {5},
  {6},
  {7,9,3,1},
  {8,4,2,6},
  {9,1}
};
while(scanf("%d%d",&a,&b)==2)
{
  int n;
  a=a-10*(a/10);
  if( (a==1) || (a==5) || (a==6))                        { printf("%d\n",a); continue;}
  else if( (a==2) || (a==3) || (a==7) || (a==8) )          n=4;
  else if( (a==4) || (a==9) ) n=2;
  else if(a==0)     { printf("%d\n",0); continue;}
  b=b%n;
  if(b==0)
   printf("%d\n",map[a-1][n-1]);
  else
   printf("%d\n",map[a-1][b-1]);
}
return 0;
}

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