Hello 2018-A. Modular Exponentiation

最新推荐文章于 2024-06-07 09:37:17 发布

原创最新推荐文章于 2024-06-07 09:37:17 发布 · 389 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces #913 #A

OI 专栏收录该内容

222 篇文章

订阅专栏

本文探讨了模幂运算的“逆”问题，即给定整数n和m，求解特定条件下的模运算结果。文章提供了一种高效算法，并通过实例解释了计算过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The following problem is well-known: given integers n and m, calculate

$\text{[math]}$ ,

where 2n = 2·2·...·2 (n factors), and $\text{[math]}$ denotes the remainder of division of x by y.

You are asked to solve the "reverse" problem. Given integers n and m, calculate

$\text{[math]}$ .

Input

The first line contains a single integer n (1 ≤ n ≤ 108).

The second line contains a single integer m (1 ≤ m ≤ 108).

Output

Output a single integer — the value of $\text{[math]}$ .

Examples
input
4
42
output
10
input
1
58
output
0
input
98765432
23456789
output
23456789

Note

In the first example, the remainder of division of 42 by 24 = 16 is equal to 10.

In the second example, 58 is divisible by 21 = 2 without remainder, and the answer is 0.

虽然n很大，但我们发现当n>26时2^26已经超过了数据范围。所以只要输出m就行了，当n<=26时直接判断。

Code：

#include<bits/stdc++.h>
using namespace std;
int mod,k;
int power(int p,int k)
{
	int ans=1,t=1;
	while(k)
	{
		if(t==1)t=p;else t=t*t;
		if(k%2==1)ans=ans*t;
		k>>=1;
	}
	return ans;
}
int main()
{
    scanf("%d%d",&k,&mod);
    if(k>26)printf("%d",mod);else
        printf("%d",mod%power(2,k));
    return 0;
}