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，score{an} == score{bn}，引理2得证。

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)

#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[N]; 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;}