题面

题目链接

大秦 为你打开 题目传送门

备用题面

构造一个长度为n的非负整数序列 $a_1,a_2,a_3,…,a_n$ 。

使得其满足 $m$ 个限制条件，限制条件的形式为 $l,r,q$ ，表示 $a_l\ \&\ a_l+1\ \&\ …\ \&\ a_{r−1}\ \&\ a_r = q$ 。

思路

首先要保证这些数字和起来等于 $q$ ，那么在做修改操作的时候得用或，而查询的时候要求按位和操作，所以合并用按位和。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
// #pragma GCC optimize(2)
// #pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef __int128 int128;

namespace FastIO 
{
    // char buf[1 << 20], *p1, *p2;
    // #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
    template<typename T> inline T read(T& x) {
        x = 0;
        int f = 1;
        char ch;
        while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
        x *= f;
        return x;
    }
    template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
        read(x);
        read(x_...);
        return;
    }
    inline ll read() {
        ll x;
        read(x);
        return x;
    }
};
using namespace FastIO;

const int N = 1e5 + 10;

class SegTree {
    private :
        struct Node {
            int L, R;
            int sum, lazy;
        }node[N << 2];
    public :
        void Add(int p, int v) {
            node[p].sum |= v;
            node[p].lazy |= v;
        }
        void PushDown(int p) {
            if(!node[p].lazy) return ;
            Add(p << 1, node[p].lazy);
            Add(p << 1 | 1, node[p].lazy);
            node[p].lazy = 0;
        }
        void Build(int p, int L, int R) {
            node[p].L = L, node[p].R = R;
            node[p].sum = 0, node[p].lazy = 0;
            if(L == R) return ;
            int mid = (L + R) >> 1;
            Build(p << 1, L, mid);
            Build(p << 1 | 1, mid + 1, R);
        }
        void Modify(int p, int L, int R, int v) {
            if(L <= node[p].L && node[p].R <= R) {
                Add(p, v);
                return ;
            }
            PushDown(p);
            int mid = (node[p].L + node[p].R) >> 1;
            if(L <= mid) Modify(p << 1, L, R, v);
            if(mid < R) Modify(p << 1 | 1, L, R, v);
            node[p].sum = node[p << 1].sum & node[p << 1 | 1].sum;
        }
        int Query(int p, int L, int R) {
            if(L <= node[p].L && node[p].R <= R) {
                return node[p].sum;
            }
            PushDown(p);
            int mid = (node[p].L + node[p].R) >> 1;
            int ret = INT_MAX;
            if(L <= mid) ret &= Query(p << 1, L, R);
            if(mid < R) ret &= Query(p << 1 | 1, L, R);
            return ret;
        }
};

int n, m;
int l[N], r[N], q[N];

SegTree tree;

void Input() {
    read(n, m);
    tree.Build(1, 1, n);
    for(int i = 1; i <= m; i++) {
        read(l[i], r[i], q[i]);
        tree.Modify(1, l[i], r[i], q[i]);
    }
}

void Work() {
    for(int i = 1; i <= m; i++) {
        if(tree.Query(1, l[i], r[i]) != q[i]) {
            printf("NO\n");
            return ;
        }
    }
    printf("YES\n");
    for(int i = 1; i <= n; i++) {
        printf("%d ", tree.Query(1, i, i));
    }
}

int main() {
    int T = 1;
    while(T--) {
        Input();
        Work();
    }
    return 0;
}