codeforces 913A A. Modular Exponentiation

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给定整数n和m，求解2^n除以m的余数的逆问题。输入包含n和m，输出2的n次方除以m的商的余数。题目提供两个示例解释了计算过程。

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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 
    
      Copy
    
4
42

     output
   
10

     input 
    
      Copy
    
1
58

     output
   
0

     input 
    
      Copy
    
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.

很久没刷题了，我个人还是很喜欢codeforces的，水题

#include<bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
const int inf = 0x3f3f3f3f;  
int main()   
{  
    ll n,m;
    cin>>n>>m;
    ll now=2;
    int i;
    for(i=1;i<n;i++) 
    {
    	now*=2;
    	if(now>m)
    		break;
    }
    if(i<n)
    	cout<<m<<endl;
    else
    	cout<<m%now<<endl; 
    return 0;  
}