HDU 1097 A hard puzzle

最新推荐文章于 2021-04-01 22:58:00 发布

shyoldman

最新推荐文章于 2021-04-01 22:58:00 发布

阅读量289

点赞数

CC 4.0 BY-SA版权

分类专栏：数论

本文链接：https://blog.youkuaiyun.com/qq_37943488/article/details/78266008

数论专栏收录该内容

12 篇文章

订阅专栏

本文介绍了一种解决特定数学问题的快速幂取模算法，该算法能够高效地计算两个大数相乘后的结果的最后一位数字。通过使用位运算和循环结构，有效地降低了计算复杂度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45464 Accepted Submission(s): 16675

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

Author

eddy

一个普通的快速幂取模算法~

#include<iostream>
using namespace std;
typedef long long ll;
ll q_mod(ll a,ll b){
	ll ans=1;
	ll base=a%10;
	while(b){
		if(b&1) ans=ans*base%10;
		base=base*base%10;
		b>>=1;
	}
	return ans;
}
int main(){
	ll a,b;
	while(cin>>a>>b){
		cout<<q_mod(a,b)<<endl;
	}
}