HDU--1097--快速幂运算

最新推荐文章于 2021-08-14 10:29:24 发布

原创最新推荐文章于 2021-08-14 10:29:24 发布 · 214 阅读

0 ·

CC 4.0 BY-SA版权

HDU题目专栏收录该内容

102 篇文章

订阅专栏

本文介绍了一个编程问题，即如何高效地计算a的b次方的最后一位数字。通过使用快速幂运算模板，可以将时间复杂度降低到O(logn)，并提供了完整的C++代码实现。

题目链接：https://vjudge.net/problem/10781/origin

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

快速幂运算模板

//pow(x,n)%mod
typedef long long ll;
ll mod_pow(ll x,ll n,ll mod){//时间复杂度O(logn) 
	ll ans=1;
	while(n>0){
		if(n&1)
			ans=ans*x%mod;
			x=x*x%mod;
			n>>=1;
	}
	return ans;
}

求pow(a,b)%10;

#include<iostream>
using namespace std;
//pow(x,n)%mod
typedef long long ll;
ll mod_pow(ll x,ll n,ll mod){//时间复杂度O(logn) 
	ll ans=1;
	while(n>0){
		if(n&1)
			ans=ans*x%mod;
			x=x*x%mod;
			n>>=1;
	}
	return ans;
} 
int main(){
	int a,b;
	while(scanf("%d%d",&a,&b)!=EOF)
		cout<<mod_pow(a,b,10)<<endl;
}
/*Sample Input
7 66
8 800
Sample Output
9
6*/