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  • (HDOJ 1097)A hard puzzle

    A hard puzzle
    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
    6
     

    Author
    eddy
     

    Recommend
    JGShining
     

    AC code:

    #include<stdio.h>
    #include
    <math.h>
    int main()
    {
      
    long int a,b;
      
    int m,n,p;
      
    while(scanf("%ld%ld",&a,&b)!=EOF)
      {
          a
    =a%10;
          m
    =(a*a)%10;
          n
    =(m*a)%10;
          p
    =(n*a)%10;
          
    if(b%4==1)
          {
            printf(
    "%d\n",a);
          }
          
    else if(b%4==2)
          {
            printf(
    "%d\n",m);
          }
          
    else if(b%4==3)
          {
            printf(
    "%d\n",n);
          }
          
    else if(b%4==0)
          {
            printf(
    "%d\n",p);
          }
      } 
       
    return 0;
    } 

    作者:cpoint
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利.
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  • 原文地址:https://www.cnblogs.com/cpoint/p/2015289.html
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