• the meaning of problems
To give you a sequence comprising a number n a, the operation on two defined sequences:
（1）$1,l,r\ :\ ans=\sum_{i=l}^{r}a_i$;
（2）$2,l,r,x\ :\ \forall\ i \in[l,r],a_i = a_i\ xor\ x$;
Output Operation (1) corresponding to the value ans;
•answer
Because the operation is an exclusive OR, and XOR operations are designed to be in binary;
So consider the establishment of tree line in accordance with the binary bits;
For each leaf node, the number corresponding to each bit is converted into binary information stored in the node;
Then is the update interval, the interval query;
•Code
• CF242E upgraded version
Sources with the title [ 2019ICPC Asian regional tournament Yinchuan Cup qualifiers station network A. Simple Data Structures ]
CF242E compared to this question, the operation of the two sequences increases:
（1）$1,l,r\ :\ ans=\sum_{i=l}^{r}a_i$;
（2）$2,l,r,x\ :\ \forall\ i \in[l,r],a_i = a_i\ xor\ x$;
（3）$2,l,r,x\ :\ \forall\ i \in[l,r],a_i = a_i\ |\ x$
（4）$2,l,r,x\ :\ \forall\ i \in[l,r],a_i = a_i\ \&\ x$
• ideas
And similar ideas on the topic, taking into account the opening 20 above tree line maintenance operations;
Difficulty is how lazy, and how pushDown ();
Taking into account ^, |, & characteristics;
For the i-th bit of a binary number, it is assumed to be 'bit [i];
易得 bit[i] ^ 0 = bit[i] , bit[i] | 0 = bit[i] , bit[i] &1 = bit[i];
bit [i] ^ 1: corresponds to the reverse bit [i];
bit[i] | 1 = 1;
bit[i] & 0 = 0;
That is only after three operations to be useful;
Considering the bit [i] may be done sequentially operate than before;
But one thing is certain, encountered bit [i] | 1, bit [i] = 1, no matter what kind of operation it has done before;
Similarly, encountered bit [i] & 0, bit [i] = 0;
If 0 & encountered first, in the face ^ 1, corresponding to the bit [i] value of 1;
Conversely, if the first encounter | 1, in the face ^ 1, is equivalent to bit [i] assigned to 0;
In this case, the definition of lazy there are four values:
$lazy=\begin{cases} -1\ ,\ no\ operate\\ \ 0\ \ ,\ become\ 0 \\ \ 1\ \ ,\ become\ 1 \\ \ 2\ \ ,\ reverse\end{cases}$;
lazy lazy and passed down a child node labeled lazy time, in conjunction with the parent node is updated;
•Code