洛谷 4883.mzf 的考验

看看 11 操作和 33 操作,这不一平衡树?
但是发现 22 操作没法简单地维护。

异或的话,可以维护子树每个数二进制每一位 11 的个数。
那么做一次异或就是把要异或的这个 dd 转为二进制,如果哪一位是 11 就把对应的个数反过来(即修改为数的个数减去这个数)。

然后据说 O(n2)O(n^2) 暴力过掉了?于是卡了时限……
被逼吸氧……

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
#pragma GCC optimize (2)
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#define ls(p) tree[p].lson
#define rs(p) tree[p].rson
using namespace std;

const int BUFF_SIZE = 1 << 20;
char BUFF[BUFF_SIZE],*BB,*BE;
#define gc() (BB == BE ? (BE = (BB = BUFF) + fread(BUFF,1,BUFF_SIZE,stdin),BB == BE ? EOF : *BB++) : *BB++)
template<class T>
inline void read(T &x)
{
x = 0;
char ch = 0,w = 0;
while(ch < '0' || ch > '9')
w |= ch == '-',ch = gc();
while(ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + (ch ^ '0'),ch = gc();
w ? x = -x : x;
}

const int N = 1e5;
const int LG = 20;
int n,m;
int p;
struct node
{
int val,rnd,sz;
int x,rev;
int cnt[LG + 5];
long long sum;
int lson,rson;
} tree[N + 10];
inline void up(int p)
{
tree[p].sz = tree[ls(p)].sz + 1 + tree[rs(p)].sz;
tree[p].sum = tree[ls(p)].sum + tree[p].val + tree[rs(p)].sum;
for(register int i = 1;i <= LG;++i)
tree[p].cnt[i] = tree[ls(p)].cnt[i] + ((tree[p].val >> i - 1) & 1) + tree[rs(p)].cnt[i];
}
inline void down(int p)
{
if(tree[p].rev)
{
swap(ls(p),rs(p));
if(ls(p))
tree[ls(p)].rev ^= 1;
if(rs(p))
tree[rs(p)].rev ^= 1;
tree[p].rev = 0;
}
if(tree[p].x)
{
int temp[LG + 5];
for(register int i = 1;i <= LG;++i)
temp[i] = (tree[p].x >> i - 1) & 1;
if(ls(p))
{
tree[ls(p)].val ^= tree[p].x,tree[ls(p)].x ^= tree[p].x;
tree[ls(p)].sum = 0;
for(register int i = 1;i <= LG;++i)
{
if(temp[i])
tree[ls(p)].cnt[i] = tree[ls(p)].sz - tree[ls(p)].cnt[i];
tree[ls(p)].sum += (1LL << i - 1) * tree[ls(p)].cnt[i];
}
}
if(rs(p))
{
tree[rs(p)].val ^= tree[p].x,tree[rs(p)].x ^= tree[p].x;
tree[rs(p)].sum = 0;
for(register int i = 1;i <= LG;++i)
{
if(temp[i])
tree[rs(p)].cnt[i] = tree[rs(p)].sz - tree[rs(p)].cnt[i];
tree[rs(p)].sum += (1LL << i - 1) * tree[rs(p)].cnt[i];
}
}
tree[p].x = 0;
}
}
void split(int p,int k,int &x,int &y)
{
if(!p)
{
x = y = 0;
return ;
}
down(p);
if(tree[ls(p)].sz < k)
x = p,split(rs(p),k - tree[ls(p)].sz - 1,rs(p),y);
else
y = p,split(ls(p),k,x,ls(p));
up(p);
}
int merge(int x,int y)
{
if(!x || !y)
return x | y;
down(x),down(y);
if(tree[x].rnd < tree[y].rnd)
{
rs(x) = merge(rs(x),y);
up(x);
return x;
}
else
{
ls(y) = merge(x,ls(y));
up(y);
return y;
}
}
int main()
{
srand(19260817);
read(n),read(m);
for(register int i = 1;i <= n;++i)
{
read(tree[i].val);
tree[i].sz = 1;
tree[i].rnd = rand();
tree[i].sum = tree[i].val;
for(register int j = 1;j <= LG;++j)
tree[i].cnt[j] = (tree[i].val >> j - 1) & 1;
p = merge(p,i);
}
int op,l,r,d,x,y,z;
while(m--)
{
read(op),read(l),read(r);
if(op == 1)
{
split(p,r,x,z);
split(x,l - 1,x,y);
tree[y].rev ^= 1;
p = merge(merge(x,y),z);
}
else if(op == 2)
{
read(d);
split(p,r,x,z);
split(x,l - 1,x,y);
tree[y].val ^= d,tree[y].x ^= d;
int temp[LG + 5];
for(register int i = 1;i <= LG;++i)
temp[i] = (d >> i - 1) & 1;
tree[y].sum = 0;
for(register int i = 1;i <= LG;++i)
{
if(temp[i])
tree[y].cnt[i] = tree[y].sz - tree[y].cnt[i];
tree[y].sum += (1LL << i - 1) * tree[y].cnt[i];
}
p = merge(merge(x,y),z);
}
else
{
split(p,r,x,z);
split(x,l - 1,x,y);
printf("%lld\n",tree[y].sum);
p = merge(merge(x,y),z);
}
}
}