CF242E XOR on Segment
codeforces
洛谷
关于异或，无法运用懒标记实现区间异或；
可以像trie树一样拆位，将每个值拆成二进制数，对此建相应个数的线段树。
0 1与 0异或 数字不变
0 1与 1异或 数字翻转
由此，对于一个01串的每一个字符都与1异或 则 1的个数 = 串长 - 现在1的个数
查询：对所有的线段树[l,r]查询1的个数，在乘上对应的二进制位权，他们之和就是查询的答案。
Code
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
const int M = 20;
int n, now;
vector<vector<int>> a;
class segtree {
public:
struct node {
int tag = 0;
int64_t sum = 0;
void apply(int l, int r) {
// make v become node(tag,data) in modify
sum = r - l + 1 - sum;
tag ^= 1;
}
void init(int v) {
// in build_tree init
sum = v;
tag = 0;
}
};
node unite(const node &a, const node &b) const {
node res;
res.sum = a.sum + b.sum;
return res;
}
// about x: left son is x+1 , right son is x+((mid-l+1)<<1) ;
inline void push_down(int x, int l, int r) {
int y = (l + r) >> 1;
int z = x + ((y - l + 1) << 1);
// push from x into (x + 1) and z
if (tree[x].tag) {
tree[x + 1].apply(l, y);
tree[z].apply(y + 1, r);
tree[x].tag = 0;
}
}
int n;
vector<node> tree;
inline void push_up(int x, int z) { tree[x].sum = unite(tree[x + 1], tree[z]).sum; }
void build(int x, int l, int r) {
if (l == r) {
tree[x].init(a[l][now]);
return;
}
int y = (l + r) >> 1;
int z = x + ((y - l + 1) << 1);
build(x + 1, l, y);
build(z, y + 1, r);
push_up(x, z);
}
node get(int x, int l, int r, int ll, int rr) {
if (ll <= l && r <= rr) {
return tree[x];
}
int y = (l + r) >> 1;
int z = x + ((y - l + 1) << 1);
push_down(x, l, r);
node res{};
if (rr <= y)
res = get(x + 1, l, y, ll, rr);
else {
if (ll > y)
res = get(z, y + 1, r, ll, rr);
else
res = unite(get(x + 1, l, y, ll, rr), get(z, y + 1, r, ll, rr));
}
push_up(x, z);
return res;
}
void modify(int x, int l, int r, int ll, int rr) {
if (ll <= l && r <= rr) {
tree[x].apply(l, r);
return;
}
int y = (l + r) >> 1;
int z = x + ((y - l + 1) << 1);
push_down(x, l, r);
if (ll <= y) modify(x + 1, l, y, ll, rr);
if (rr > y) modify(z, y + 1, r, ll, rr);
push_up(x, z);
}
segtree(int _n = ::n) : n(_n) {
assert(n > 0);
tree.resize(2 * n - 1);
}
node get(int ll, int rr) {
assert(0 <= ll && ll <= rr && rr <= n - 1);
return get(0, 0, n - 1, ll, rr);
}
void modify(int ll, int rr) {
assert(0 <= ll && ll <= rr && rr <= n - 1);
modify(0, 0, n - 1, ll, rr);
}
}; // root's idx is 0 and the begin of vector is also 0;
// don't forget idx is from 0 to n-1 (equal [--x,--y]) when ask;
signed main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr), std::cout.tie(nullptr);
int q;
cin >> n;
segtree t[M];
a.resize(n, vector<int>(M));
for (int i = 0, x; i < n; ++i) {
cin >> x;
for (int j = 0; j < M; ++j) a[i][j] = x >> j & 1;
}
for (int i = 0; i < M; ++i) now = i, t[i].build(0, 0, n - 1);
cin >> q;
while (q--) {
int op, l, r;
cin >> op >> l >> r;
--l, --r;
if (op == 1) {
int64_t ans = 0, p = 1;
for (int i = 0; i < M; ++i, p <<= 1) ans += p * t[i].get(l, r).sum;
cout << ans << endl;
} else {
int64_t k;
cin >> k;
for (int i = 0; i < M; ++i)
if (k >> i & 1) t[i].modify(l, r);
}
}
return 0;
}
