time limit per test 2 seconds
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
题意：构造一个含有n个元素的序列，使得每个值不同，并且所有的数的异或和等于x。
思路：直接从1开始构造。然后对最后一个和第一个进行特殊考虑。
#include "bits/stdc++.h" using namespace std; const int maxn = 1e6 + 10; int ans[maxn]; int main(int argc, char const *argv[]) { int N, K; scanf("%d%d", &N, &K); int res = 0; if (N == 1) return printf("YES %d ", K), 0; if (N == 2 && K == 0) return printf("NO "), 0; for (int i = 1; i < N; i++) { ans[i] = i; res ^= i; } ans[N] = res^K; if (ans[N] < N) { if (ans[N] == 1) ans[N - 1] |= (1<<19); else ans[1] |= (1<<19); ans[N] |= (1<<19); } printf("YES %d", ans[1]); for (int i = 2; i <= N; i++) { printf(" %d", ans[i]); } putchar(' '); return 0; }