CF482B Interesting Array(线段树+位运算)

限制是区间按位与的结果必须为x，则根据按位与的性质，区间内的每一个数都至少“有”x，因此我们可以给这个区间的所有数或上x。因此我们先根据m个操作进行m个区间或操作。然后m次查询区间与是否满足要求。因此我们还要维护区间与。线段树即可。

#include <bits/stdc++.h>using namespace std;#define ll long long#define N 100010int const inf=(1<<30)-1;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}int n,m,a[N];struct qu{    int x,y,val;}q[N];struct node{    int x,lazy;}tree[N<<2];inline void pushup(int p){    tree[p].x=tree[p<<1].x&tree[p<<1|1].x;}inline void pushdown(int p){    if(!tree[p].lazy) return;int val=tree[p].lazy;tree[p].lazy=0;    tree[p<<1].lazy|=val;tree[p<<1|1].lazy|=val;    tree[p<<1].x|=val;tree[p<<1|1].x|=val;}void build(int p,int l,int r){    if(l==r) return;    int mid=l+r>>1;    build(p<<1,l,mid);build(p<<1|1,mid+1,r);}int ask(int p,int l,int r,int x,int y){    if(x<=l&&r<=y) return tree[p].x;    int mid=l+r>>1,res=inf;pushdown(p);    if(x<=mid) res&=ask(p<<1,l,mid,x,y);    if(y>mid) res&=ask(p<<1|1,mid+1,r,x,y);    return res;}void update(int p,int l,int r,int x,int y,int val){    if(x<=l&&r<=y){tree[p].lazy|=val;tree[p].x|=val;return;}    int mid=l+r>>1;pushdown(p);    if(x<=mid) update(p<<1,l,mid,x,y,val);    if(y>mid) update(p<<1|1,mid+1,r,x,y,val);    pushup(p);}int main(){//  freopen("a.in","r",stdin);    n=read();m=read();build(1,1,n);    for(int i=1;i<=m;++i){        q[i].x=read(),q[i].y=read(),q[i].val=read();        update(1,1,n,q[i].x,q[i].y,q[i].val);    }    for(int i=1;i<=m;++i)        if(ask(1,1,n,q[i].x,q[i].y)!=q[i].val){puts("NO");return 0;}    puts("YES");printf("%d",ask(1,1,n,1,1));    for(int i=2;i<=n;++i) printf(" %d",ask(1,1,n,i,i));    return 0;}