Codeforces 862C. Mahmoud and Ehab and the xor

最新推荐文章于 2021-08-01 01:32:11 发布

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最新推荐文章于 2021-08-01 01:32:11 发布

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本文探讨了一个有趣的问题：如何选择n个不同的非负整数，使得这些整数的连续异或操作结果恰好等于给定的目标值x。文章提供了一种高效的解决方案，并附带AC代码实现。

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Mahmoud and Ehab are on the third stage of their adventures now. As you know, Dr. Evil likes sets. This time he won't show them any set from his large collection, but will ask them to create a new set to replenish his beautiful collection of sets.

Dr. Evil has his favorite evil integer x. He asks Mahmoud and Ehab to find a set of n distinct non-negative integers such the bitwise-xor sum of the integers in it is exactly x. Dr. Evil doesn't like big numbers, so any number in the set shouldn't be greater than 10⁶.

Input

The only line contains two integers n and x (1 ≤ n ≤ 10⁵, 0 ≤ x ≤ 10⁵) — the number of elements in the set and the desired bitwise-xor, respectively.

Output

If there is no such set, print "NO" (without quotes).

Otherwise, on the first line print "YES" (without quotes) and on the second line print n distinct integers, denoting the elements in the set is any order. If there are multiple solutions you can print any of them.

Examples

input

5 5

output

YES
1 2 4 5 7

input

3 6

output

YES
1 2 5

Note

You can read more about the bitwise-xor operation here: https://en.wikipedia.org/wiki/Bitwise_operation#XOR

For the first sample $\text{[math]}$ .

For the second sample $\text{[math]}$ .

题意：让你选择n个不同的非负数，然后让他们连续异或，最后得到的结果等于x

思路：特判一下x == 0 && n == 2是NO，其他都是YES，然后看代码理解理解就懂啦，嘿嘿

AC代码：

#include<bits/stdc++.h>

using namespace std;

int vis[1000007];
void init(int cnt){
    int xx = 1;
    for(int i = 0; i < cnt; i++){
        while(!(vis[xx] == 0 && ((xx ^ 131072) < 1000000) && vis[(xx ^ 131072)] == 0)){
            xx++;
        }
        vis[xx] = 1;
        vis[(xx ^ 131072)] = 1;
    }
}
int main(){
    int n, m;
    scanf("%d%d", &n, &m);
    if(n == 2 && m == 0){
        printf("NO\n");
        return 0;
    }
    int k = (n - 1) / 4;
    printf("YES\n");
    if(n % 4 == 0){
        vis[m] = 1;
        vis[131072] = 1;
        init(k * 2 + 1);
    }
    else if(n % 4 == 1){
        vis[m] = 1;
        init(k * 2);
    }
    else if(n % 4 == 2){
        if(m == 0){
            vis[1] = vis[2] = vis[3] = 1;
            vis[6] = vis[15] = vis[9] = 1;
            n -= 6;
            init(n/2);
        }
        else{
            vis[m + 131072] = 1;
            vis[131072] = 1;
            init(k * 2);
        }
    }
    else{
        vis[m + 131072] = 1;
        init(k * 2 + 1);
    }
    for(int i = 0; i <= 1000000; i++){
        if(vis[i]){
            printf("%d ", i);
        }
    }
}