题目链接：Codeforces 482B Interesting Array

题目大意：给定一个长度为N的数组，如今有M个限制，每一个限制有l，r，q，表示从a[l]~a[r]取且后的数一定为q，问是

否有满足的数列。

解题思路：线段树维护。每条限制等于是对l~r之间的数或上q(取且的性质，对应二进制位一定为1)。那么处理全然部的

限制。在进行查询。查询相应每一个l~r之间的数取且是否还等于q。所以用线段树维护取且和。改动为或操作。

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 1e5 + 5;

const int INF = (1<<30)-1;

#define lson(x) ((x)<<1)

#define rson(x) (((x)<<1)|1)

int lc[maxn << 2], rc[maxn << 2], set[maxn << 2], val[maxn << 2];

inline void maintain (int u, int w) {

    val[u] |= w;

    set[u] |= w;

}

inline void pushup(int u) {

    val[u] = val[lson(u)] & val[rson(u)];

}

inline void pushdown(int u) {

    if (set[u]) {

        maintain(lson(u), set[u]);

        maintain(rson(u), set[u]);

        set[u] = 0;

    }

}

void build (int u, int l, int r) {

    lc[u] = l;

    rc[u] = r;

    set[u] = val[u] = 0;

    if (l == r)

        return;

    int mid = (lc[u] + rc[u]) >> 1;

    build(lson(u), l, mid);

    build(rson(u), mid + 1, r);

    pushup(u);

}

void modify(int u, int l, int r, int w) {

    if (l <= lc[u] && rc[u] <= r) {

        maintain(u, w);

        return;

    }

    pushdown(u);

    int mid = (lc[u] + rc[u]) >> 1;

    if (l <= mid)

        modify(lson(u), l, r, w);

    if (r > mid)

        modify(rson(u), l, r, w);

    pushup(u);

}

int query(int u, int l, int r) {

    if (l <= lc[u] && rc[u] <= r)

        return val[u];

    pushdown(u);

    int mid = (lc[u] + rc[u]) >> 1, ret = INF;

    if (l <= mid)

        ret &= query(lson(u), l, r);

    if (r > mid)

        ret &= query(rson(u), l, r);

    pushup(u);

    return ret;

}

int N, M, L[maxn], R[maxn], Q[maxn];

bool judge() {

    for (int i = 0; i < M; i++)

        if (query(1, L[i], R[i]) != Q[i])

            return true;

    printf("YES\n");

    for (int i = 1; i <= N; i++)

        printf("%d%c", query(1, i, i), i == N ?

'\n' : ' ');

    return false;

}

int main () {

    scanf("%d%d", &N, &M);

    build(1, 1, N);

    for (int i = 0; i < M; i++) {

        scanf("%d%d%d", &L[i], &R[i], &Q[i]);

        modify(1, L[i], R[i], Q[i]);

    }

    if (judge())

        printf("NO\n");

    return 0;

}