hdu1097 A hard puzzle

最新推荐文章于 2021-04-01 22:58:00 发布

kingsirre

最新推荐文章于 2021-04-01 22:58:00 发布

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本文介绍了一种快速计算a^b末位数字的方法，并提供了两种实现思路的C++代码示例。通过观察数字循环规律并利用数组存储，或者简化运算过程来提高效率。

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

解答：

这是求a^b最后一位的，可以求出0~9循环多少次又回到原来的值，比如2的次方是2 4 8 16 32 周期是4
然后把这些数存起来就是数组
int c[10][5]={
              {0},{1},{2,4,8,6},{3,9,7,1},{4,6},{5},
              {6},{7,9,3,1},{8,4,2,6},{9,1}
             };

#include<iostream>
#include<cmath>
using namespace std;
const
int c[10][5]={
              {0},{1},{2,4,8,6},{3,9,7,1},{4,6},{5},
              {6},{7,9,3,1},{8,4,2,6},{9,1}
             };
int main()
{
    int a,b;
    int sum,i,k;
    while(cin>>a>>b)
    {
     k=1;
     if(a>=10)
     a=a%10;
     sum=a;
     for(i=1;i<=k;i++)//最多循环5次
     {
      sum*=a;
      if(sum>=10)
      {
       sum%=10;
      }
      if(sum!=a)
      {k=k+1;}
     }
     if(b%(i-1)==0) cout<<c[a][b%(i-1)+(i-1)-1]<<endl;//注意输出，原来我没有
      else cout<<c[a][b%(i-1)-1]<<endl;
    }
    return 0;
}

另一种：

#include <iostream>
using namespace std;
int main()
{
	int a, b, n;
	while (cin>>a>>b) {
		a %= 10;
		b %= 4;
		switch(b) {
		case 1:
			n = a%10;
			break;
		case 2:
			n = a*a%10;
			break;
		case 3:
			n = a*a*a%10;
			break;
		case 0:
			n = a*a*a*a%10;
			break;
		default:
			break;
		}
		cout<<n<<endl;
	}
	return 0;
}