A hard puzzle

最新推荐文章于 2021-04-01 22:58:00 发布

森-Js

最新推荐文章于 2021-04-01 22:58:00 发布

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分类专栏： ACM. 数论算法文章标签：快速幂

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博客围绕快速幂展开，虽未给出具体内容，但可推测涉及快速幂的原理、应用等信息技术领域知识。快速幂是一种高效计算幂运算的算法，在编程和算法设计中有重要作用。

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A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49940 Accepted Submission(s): 18357

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

Author

eddy

#include<cstdio>
#include<algorithm>
using namespace std;  
long long shu(long long  n,long long m)       //快速幂。注意变量定义为long long 型
{
	long long result=1,a=n,b=m,i;
	while(b)
	{
		if(b%2!=0)
		result=result*a%10;
		a=a*a%10;
		b=b/2;
	}
	return result;
}
int main()
{
      long long n,m;
	while(~scanf("%lld%lld",&n,&m))
	{
		printf("%lld\n",shu(n,m));
	}
}