codeforces 242E XOR on Segment

E. XOR on Segment

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You've got an array a, consisting of n integers a₁, a₂, ..., a_n. You are allowed to perform two operations on this array:

Calculate the sum of current array elements on the segment [l, r], that is, count value a_l + a_l + 1 + ... + a_r.
Apply the xor operation with a given number x to each array element on the segment [l, r], that is, execute . This operation changes exactly r - l + 1 array elements.

Expression means applying bitwise xor operation to numbers x and y. The given operation exists in all modern programming languages, for example in language C++ and Java it is marked as "^", in Pascal — as "xor".

You've got a list of m operations of the indicated type. Your task is to perform all given operations, for each sum query you should print the result you get.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the size of the array. The second line contains space-separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁶) — the original array.

The third line contains integer m (1 ≤ m ≤ 5·10⁴) — the number of operations with the array. The i-th of the following m lines first contains an integer t_i (1 ≤ t_i ≤ 2) — the type of the i-th query. If t_i = 1, then this is the query of the sum, if t_i = 2, then this is the query to change array elements. If the i-th operation is of type 1, then next follow two integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n). If the i-th operation is of type 2, then next follow three integers l_i, r_i, x_i (1 ≤ l_i ≤ r_i ≤ n, 1 ≤ x_i ≤ 10⁶). The numbers on the lines are separated by single spaces.

Output

For each query of type 1 print in a single line the sum of numbers on the given segment. Print the answers to the queries in the order in which the queries go in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, coutstreams, or the %I64d specifier.

Sample test(s)

input

5
4 10 3 13 7
8
1 2 4
2 1 3 3
1 2 4
1 3 3
2 2 5 5
1 1 5
2 1 2 10
1 2 3

output

input

6
4 7 4 0 7 3
5
2 2 3 8
1 1 5
2 3 5 1
2 4 5 6
1 2 3

output

38
28

比赛的时候我用一维线段搞，超时了。因为每次查询的时候都查询到最地下的子树了。后面看了别人几乎都是用二维线段树过的。

这题二维线段树的思路就是。每一维线段树分别存表示二进制的一个位。。那么任意一个区间的【l,r】的和，其实就是统计这个区间1的个数然后乘以这个线段所表示的所在位数。然后实现就是线段树的基本上传，下方了。

#include<cstdio>
using namespace std;
const int maxn = 100000+100;
typedef long long LL;
int a[maxn];
int T[22][maxn<<2];
bool lazy[22][maxn<<2];
void build(int k,int id,int l,int r){if(l==r){ T[k][id]=(a[l]>>k)&1; return ;}int m=(l+r)>>1;build(k,id<<1,l,m);  build(k,id<<1|1,m+1,r);T[k][id]=T[k][id<<1]+T[k][id<<1|1];
}
void update(int k,int id,int l,int r,int L,int R){if(L==l&&R==r){T[k][id]=r-l+1-T[k][id];lazy[k][id]^=1;return ;}int m=(L+R)>>1;if(m>=r)  update(k,id<<1,l,r,L,m);else if(l>m)  update(k,id<<1|1,l,r,m+1,R);else{update(k,id<<1,l,m,L,m);update(k,id<<1|1,m+1,r,m+1,R);}T[k][id]=T[k][id<<1]+T[k][id<<1|1];if(lazy[k][id])  T[k][id]=R-L+1-T[k][id];
}
LL query(int k,int id,int l,int r,int L,int R){if(L==l&&R==r){return T[k][id];}int m=(L+R)>>1,s;if(m>=r)   s=query(k,id<<1,l,r,L,m);else  if(l>m)  s=query(k,id<<1|1,l,r,m+1,R);else{s=query(k,id<<1,l,m,L,m)+query(k,id<<1|1,m+1,r,m+1,R);}if(lazy[k][id]) return r-l+1-s;else  return s;
}
int main(){int n,m;scanf("%d",&n);for(int i=1;i<=n;i++)  scanf("%d",&a[i]);for(int i=0;i<20;i++)  build(i,1,1,n);scanf("%d",&m);int id,l,r,x;while(m--){scanf("%d",&id);if(id==1){scanf("%d%d",&l,&r);long long sum=0;for(int i=0;i<20;i++) sum+=query(i,1,l,r,1,n)<<i;printf("%I64d\n",sum);}else{scanf("%d%d%d",&l,&r,&x);for(int i=0;i<20;i++){if(x>>i&1)update(i,1,l,r,1,n);}}}return 0;}

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