# 思路：

\(0\)和或\(1\)相当于重置区间，异或\(1\)相当于翻转区间，那么设出两个\(lazy\)搞一下。注意父区间\(pushdown\)重置标记时，子区间的翻转标记要清空。

# 代码：

``````#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e6 + 5;
const int MAXM = 3e6;
const ll MOD = 1e9 + 7;
const ull seed = 131;
const int INF = 0x3f3f3f3f;

#define lson (rt << 1)
#define rson (rt << 1 | 1)
int sum[5][maxn << 2];
int lazy[5][maxn << 2], rev[5][maxn << 2];
int a[maxn];
void pushup(int rt, int bit){
sum[bit][rt] = sum[bit][lson] + sum[bit][rson];
}
void pushdown(int rt, int bit, int l, int r){
int m = (l + r) >> 1;
if(lazy[bit][rt] != -1){
sum[bit][lson] = lazy[bit][rt] * (m - l + 1);
sum[bit][rson] = lazy[bit][rt] * (r - m);
lazy[bit][lson] = lazy[bit][rson] = lazy[bit][rt];
rev[bit][lson] = rev[bit][rson] = 0;    //!!!!!
lazy[bit][rt] = -1;
}
if(rev[bit][rt]){
sum[bit][lson] = m - l + 1 - sum[bit][lson];
sum[bit][rson] = r - m - sum[bit][rson];
rev[bit][lson] ^= 1;
rev[bit][rson] ^= 1;
rev[bit][rt] = 0;
}
}
void build(int l, int r, int bit, int rt){
lazy[bit][rt] = -1;
rev[bit][rt] = 0;
if(l == r){
sum[bit][rt] = (a[l] >> bit) & 1;
return;
}
int m = (l + r) >> 1;
build(l, m, bit, lson);
build(m + 1, r, bit, rson);
pushup(rt, bit);
}
void update(int L, int R, int l, int r, int op, int bit, int rt){
if(L <= l && R >= r){
if(op == 1){    //&0
lazy[bit][rt] = 0;
rev[bit][rt] = 0;
sum[bit][rt] = 0;
}
else if(op == 2){   //|1
lazy[bit][rt] = 1;
rev[bit][rt] = 0;
sum[bit][rt] = r - l + 1;
}
else{   //^1
rev[bit][rt] ^= 1;
sum[bit][rt] = r - l + 1 - sum[bit][rt];
}
return;
}
pushdown(rt, bit, l, r);
int m = (l + r) >> 1;
if(L <= m)
update(L, R, l, m, op, bit, lson);
if(R > m)
update(L, R, m + 1, r, op, bit, rson);
pushup(rt, bit);
}
int query(int L, int R, int l, int r, int bit, int rt){
if(L <= l && R >= r){
return sum[bit][rt];
}
pushdown(rt, bit, l, r);
int m = (l + r) >> 1, ret = 0;
if(L <= m)
ret += query(L, R, l, m, bit, lson);
if(R > m)
ret += query(L, R, m + 1, r, bit, rson);
return ret;
}
int main(){
int n, m, T;
scanf("%d", &T);
while(T--){
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 0; i < 5; i++) build(1, n, i, 1);
while(m--){
int op, l, r, x;
char ope[10];
scanf("%s", &ope);
if(ope[0] == 'A') op = 1;
else if(ope[0] == 'O') op = 2;
else if(ope[0] == 'X') op = 3;
else op = 4;
if(op == 4){
scanf("%d%d", &l, &r);
l++, r++;
ll ans = 0;
for(int i = 0; i < 5; i++){
x = query(l, r, 1, n, i, 1);
ans += 1LL * (1LL << i) * x;
}
printf("%lld\n", ans);
}
else{
scanf("%d%d%d", &x, &l, &r);
l++, r++;
if(op == 2 || op == 3){
for(int i = 0; i < 5; i++){
if((x >> i) & 1)
update(l, r, 1, n, op, i, 1);
}
}
else{
for(int i = 0; i < 5; i++){
if(!((x >> i) & 1))
update(l, r, 1, n, op, i, 1);
}
}
}
}
}
return 0;
}

``````

FZU2105 Digits Count（按位建线段树）题解

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