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 
     
      #include 
      
       #include 
       
        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; }