Codeforces Round #326 (Div. 1) E. Duff as a Queen

题意：

给出一个长度为n的数组，要求支持一下两种操作：

1.对于l <= i <= r，a[i] ^= k

2.询问[l,r]的分数

区间[l,r]的分数定义为，能使用[l,r]内的数字异或出的数字的个数，一个都不取就是0



solution：

引理1：对于[l,r]中的数字，如果能构造出b1,b2,b3,...,bk，那么，一定能构造出b1 ^ b2 ^ b3 ^ ... ^ bk

把每个bi的构造方案看成是一个n位二进制数码，引理1得证。

构造数组{bn}，使得bi = ai ^ ai-1，b1 = a1，那么，有ai = b1 ^ b2 ^ b3 ^ ... ^ bi

引理2：score{an} == score{bn}

因为{bn}中的每个数都可以由{an}构造出来，由引理1，{bn}的任意组合也能由{an}构造。

因此，有score{bn} <= score{an}。同理可得，score{an} <= score{bn}

因此，[b]score{an} == score{bn}，引理2得证。[/b]



[b]最后，对于原数列{an}，构造出{bn}，区间修改就能转变为单点修改，询问的话就用线段树维护每段区间的xor线性基，复杂度O(nlog^3n)[/b]#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
#include<ext/pb_ds/priority_queue.hpp>
using namespace std;

const int maxn = 2E5 + 20;
const int T = 4;
const int N = 30;

struct data{
int a
; data(){memset(a,0,sizeof(a));}
data operator + (const data &b)
{
data c;
for (int i = 0; i < N; i++) c.a[i] = a[i];
for (int i = 0; i < N; i++)
{
if (!b.a[i]) continue; int x = b.a[i];
for (int j = i; j < N; j++)
if (x & (1<<j))
{
if (c.a[j]) x ^= c.a[j];
else {c.a[j] = x; break;}
}
}
return c;
}
}c[maxn*T],ret;

int n,m,mi[N + 1],A[maxn],sum[maxn];

int getint()
{
char ch = getchar(); int ret = 0;
while (ch < '0' || '9' < ch) ch = getchar();
while ('0' <= ch && ch <= '9')
ret = ret*10 + ch - '0',ch = getchar();
return ret;
}

void Build(int o,int l,int r)
{
if (l == r)
{
for (int i = l; i <= n; i += i&-i) sum[i] ^= A[l];
for (int i = 0; i < N; i++)
if (A[l] & (1<<i)) {c[o].a[i] = A[l]; break;}
return;
}
int mid = (l + r) >> 1;
Build(o<<1,l,mid); Build(o<<1|1,mid+1,r);
c[o] = c[o<<1] + c[o<<1|1];
}

void Modify(int o,int l,int r,int pos,int t)
{
if (l == r)
{
memset(c[o].a,0,sizeof(c[o].a)); A[l] ^= t;
for (int i = l; i <= n; i += i&-i) sum[i] ^= t;
for (int i = 0; i < N; i++)
if (A[l] & (1<<i)) {c[o].a[i] = A[l]; break;}
return;
}
int mid = (l + r) >> 1;
if (pos <= mid) Modify(o<<1,l,mid,pos,t);
else Modify(o<<1|1,mid+1,r,pos,t);
c[o] = c[o<<1] + c[o<<1|1];
}

void Query(int o,int l,int r,int ql,int qr)
{
if (ql <= l && r <= qr) {ret = ret + c[o]; return;}
int mid = (l + r) >> 1;
if (ql <= mid) Query(o<<1,l,mid,ql,qr);
if (qr > mid) Query(o<<1|1,mid+1,r,ql,qr);
}

int Sum(int pos)
{
int ret = 0;
for (int i = pos; i > 0; i -= i&-i) ret ^= sum[i];
return ret;
}

int main()
{

n = getint(); m = getint();
for (int i = 1; i <= n; i++) A[i] = getint();
for (int i = n; i > 1; i--) A[i] = A[i] ^ A[i - 1];
Build(1,1,n); mi[0] = 1;
for (int i = 1; i <= N; i++) mi[i] = mi[i-1] << 1;

while (m--)
{
int typ = getint(),l,r;
l = getint(); r = getint();
if (typ == 1)
{
int k = getint(); Modify(1,1,n,l,k);
if (r < n) Modify(1,1,n,r + 1,k);
}
else
{
memset(ret.a,0,sizeof(ret.a));
int tot = Sum(l),Ans = 0;
for (int i = 0; i < N; i++)
if (tot & (1<<i)) {ret.a[i] = tot; break;}
if (l < r) Query(1,1,n,l + 1,r);
for (int i = 0; i < N; i++) if (ret.a[i]) ++Ans;
printf("%d\n",mi[Ans]);
}
}
return 0;
}