CodeForces 909F

题意略。

思路：

第一问：

递归地来写，找对称，发现关于(1<<y) - 1和(1<<y)对称的数字做 & 结果为0。

第二问：

6，7特殊考虑。循环左移(1<<y) ~ (1<<(y + 1) - 1)，可以保证这个区间内的值与下标做 & 结果不为0。然而如果在这个段内只有1<<y这一个数，那么

我们无法循环左移，此时要特判为NO。

详见代码：

#include<bits/stdc++.h>
#define maxn 50
#define maxn2 100005
using namespace std;

int one[maxn],tail;
int ans1[maxn2],ans2[maxn2];
int six[] = {,,,,,};
int seven[] = {,,,,,,};

void init(){
    one[] = ;
    tail += ;
    for(int i = ;(<<i) -  <= ;++i){
        one[i] = (<<i) - ;
        tail += ;
    }
}
int cnt1(int x){
    int ret = ;
    while(x){
        x = x & (x - );
        ++ret;
    }
    return ret;
}
void construct(int x){
    if(x <= ) return;
    int pos = lower_bound(one,one + tail,x) - one - ;
    int mid = one[pos] + ;
    for(int i = mid;i <= x;++i) swap(ans2[i],ans2[mid - (i - mid) - ]);
    x = mid - (x - mid) -  - ;
    construct(x);
}

int main(){
    init();
    int n;
    scanf("%d",&n);
    if(n & ) printf("NO\n");
    else{
        printf("YES\n");
        for(int i = ;i <= n;++i) ans2[i] = i;
        construct(n);
        printf("%d",ans2[]);
        for(int i = ;i <= n;++i) printf(" %d",ans2[i]);
        printf("\n");
    }

    if(n <=  || cnt1(n) == ) printf("NO\n");
    else if(n == ){
        printf("YES\n");
        printf("%d",six[]);
        for(int i = ;i < ;++i) printf(" %d",six[i]);
        printf("\n");
    }
    else if(n == ){
        printf("YES\n");
        printf("%d",seven[]);
        for(int i = ;i < ;++i) printf(" %d",seven[i]);
        printf("\n");
    }
    else{
        printf("YES\n");
        for(int i = ;i <= ;++i) ans1[i] = seven[i - ];
        int lft = ,rht = min((lft<<) - ,n);
        while(lft < n){
            for(int i = lft;i <= rht;++i){
                if(i < rht) ans1[i] = i + ;
                else ans1[i] = lft;
            }
            lft = lft<<;
            rht = min((lft<<) - ,n);
        }
        printf("%d",ans1[]);
        for(int i = ;i <= n;++i) printf(" %d",ans1[i]);
        printf("\n");
    }
    return ;
}