题目链接：点击打开链接

题意：

构造一个n长的序列，m个限制：

每个限制[l, r] q

序列要满足 区间[l,r]的所有数 & 起来结果是q

思路：

直接构造，然后判可行就好了。。

#include <stdio.h>
#include <cstring>
#include <iostream>
#include <map>
template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) {
        putchar('-');
        x = -x;
    }
    if(x>9) pt(x/10);
    putchar(x%10+'0');
}
using namespace std;
typedef unsigned long long ull;
#define L(x) tree[x].l
#define R(x) tree[x].r
#define Lson(x) (x<<1)
#define Rson(x) (x<<1|1)
#define V(x) tree[x].val
const int N = 100005;
int hehe;
int haha;
struct node{
    int l, r, val;
}tree[N<<2];
int a[N];
void push_up(int id){
    V(id) = V(Lson(id)) & V(Rson(id));
}
void build(int l, int r, int id){
    L(id) = l ; R(id) = r;
    if(l == r){
        V(id) = a[l];
        return ;
    }
    int mid = (l+r)>>1;
    build(l, mid, Lson(id)); build(mid+1, r, Rson(id));
    push_up(id);
}
int query(int l, int r, int id){
    if(l == L(id) && R(id) == r){
        return V(id);
    }
    int mid = (L(id) + R(id)) >> 1;
    if(mid < l)
        return query(l, r, Rson(id));
    else if(r <= mid)
        return query(l, r, Lson(id));
    else return query(l, mid, Lson(id)) & query(mid+1, r, Rson(id));
}
int n, m;
struct Q{
    int l, r, q;
}q[N];
int o[N][32], t[N][32];

void put(){
    for(int i = 1; i <= n; i++)
    {
        pt(a[i]);
        i == n ? puts("") : putchar(' ');
    }
}
bool work(){
    int now = 0;
    int b[35] = {0};
    for(int i = 1; i <= n; i++)
    {
        for(int j = 0; j < 31; j++)
            if(o[i][j])
            {
                now |= (1<<j);
                b[j]+=o[i][j];
            }
        a[i] = now
        for(int j = 0; j < 31; j++)
            if(t[i][j])
            {
                b[j]+=t[i][j];
                if(b[j]==0)
                now &= ~(1<<j);
            }
    }
    build(1, n, 1);
    for(int i = 1; i <= m; i++) {
        if(query(q[i].l, q[i].r, 1) != q[i].q)return false;
    }
    return true;
}
bool input(){
    memset(o, 0, sizeof o);
    memset(t, 0, sizeof t);
    for(int i = 1; i <= m; i++){
        scanf("%d %d %d", &q[i].l, &q[i].r, &q[i].q);
        for(int j = 0; j < 31; j++)
            if(q[i].q&(1<<j))
            {
                o[q[i].l][j] ++;

                t[q[i].r][j] --;
            }
    }
    return true;
}
int main(){
    while(cin>>n>>m){
        input();
        if(work()){
            puts("YES");
            put();
        }
        else puts("NO");
    }
    return 0;
}

Codeforces 482B Interesting Array 构造+线段树判可行