杭电acm1097 A hard puzzle

最新推荐文章于 2019-01-03 11:00:48 发布

原创最新推荐文章于 2019-01-03 11:00:48 发布 · 919 阅读

0 ·

CC 4.0 BY-SA版权

杭电acm 专栏收录该内容

20 篇文章

订阅专栏

本文深入解析ACM难题中的末位数字循环规律，通过实例演示如何简化计算，快速找到任意数的幂次方的末位数字。采用循环理论与简单算法，巧妙解决复杂的数学问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

杭电acm1097

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18812 Accepted Submission(s): 6707

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

7 668 800

Sample Output

Author

eddy

Recommend

JGShining

思想：

尾数为0~9的数字的n次方都可以看做是四个数的循环，如：a为一的尾数，都是1。。a的最后一位为2时，当n次方时，始终都是2,4,8,6的循环，其他同理，所以可以

把b求余4来算，但防止b除以4后，余数为零，所以至少保证有a的4次方存在。。。

实现代码如下：

#include <iostream>
using namespace std;
int main()
{
int a,b,i,y;
while(cin>>a>>b)
{
  if(a<0 || b<0)
   break;
  a%=10;
  b%=4;
  y = (a*a*a*a);//防止b求余后为0；没进入for循环，至少保证怎样都有值
  for(i=0;i<b;i++)
  {
   y*=a;
   y%=10;
  }
  cout<<y%10<<endl;
}
return 0;
}