hdoj 1097 A hard puzzle

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本文提供了解决HDU在线评测平台问题1097的方法，即如何快速计算两个整数a和b的幂运算结果的末尾数字。通过字符串操作和循环迭代，实现对大范围数值的有效处理。

http://acm.hdu.edu.cn/showproblem.php?pid=1097

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28803 Accepted Submission(s): 10307

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
  
  
    
  
  
   #include<stdio.h>
#include<string.h>
int main()
{
    char a[20],c,s_[10];//2的30次方大概是10几位
    int b,n,s[200],i,j;
    while(scanf("%s %d",a,&b)!=EOF)
    {          
          n=strlen(a);
          c=a[n-1];
          memset(s,0,sizeof(s));
          for(i=1,s[0]=c-'0';;i++)//将末位数放到s[]中，每i+1个一循环 
          {  
               s[i]=(s[i-1]*(c-'0'))%10;        
               if(s[i]==c-'0')break;
          }
          b=b%i-1;
          if(b==-1)
          printf("%d\n",s[i-1]);
          else
          printf("%d\n",s[b]);
          //printf("%d@\n",b);
//          for(j=0;j<i;j++)
//          printf("%d#\n",s[j]);
////          
    } 
    //while(1);
    return 0;
}