题目链接：
http://codeforces.com/problemset/problem/242/E
E. XOR on Segment
time limit per test 4 seconds
memory limit per test 256 megabytes
#### 问题描述
> You've got an array a, consisting of n integers a1, a2, ..., an. 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 al + al + 1 + ... + ar.
> 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.
#### 输入
> The first line contains integer n (1 ≤ n ≤ 105) — the size of the array. The second line contains space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 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 ti (1 ≤ ti ≤ 2) — the type of the i-th query. If ti = 1, then this is the query of the sum, if ti = 2, then this is the query to change array elements. If the i-th operation is of type 1, then next follow two integers li, ri (1 ≤ li ≤ ri ≤ n). If the i-th operation is of type 2, then next follow three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106). The numbers on the lines are separated by single spaces.
#### 输出
> 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, cout streams, or the %I64d specifier.
#### 样例
> **sample 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
>
> **sample output**
> 26
> 22
> 0
> 34
> 11
## 题意
> 给你n个数，q个操作：
> 查询：返回l到r的和
> 更新：l到r的每个数都异或上x。
题解
对没个数字按二进制展开，存放在20颗线段树里面。
然后就转化成了一个经典的区间更新区间查询的线段树问题了。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#define lson (o<<1)
#define rson ((o<<1)|1)
#define M (l+(r-l)/2)
using namespace std;
const int maxn=1e5+10;
const int maxm=22;
typedef __int64 LL;
int n;
int sumv[maxm][maxn<<2];
int setv[maxn<<2];
void maintain(int o) {
for(int i=0; i<maxm; i++) {
sumv[i][o]=sumv[i][lson]+sumv[i][rson];
}
}
void pushdown(int o,int l,int r) {
for(int i=0; i<maxm; i++) {
if(setv[o]&(1<<i)) {
sumv[i][lson]=M-l+1-sumv[i][lson];
sumv[i][rson]=r-M-sumv[i][rson];
}
}
setv[lson]^=setv[o];
setv[rson]^=setv[o];
setv[o]=0;
}
void build(int o,int l,int r) {
if(l==r) {
int x;
scanf("%d",&x);
for(int i=0; i<maxm; i++) {
if(x&(1<<i)) sumv[i][o]=1;
else sumv[i][o]=0;
}
} else {
build(lson,l,M);
build(rson,M+1,r);
maintain(o);
}
}
int ql,qr,_v;
void update(int o,int l,int r) {
if(ql<=l&&r<=qr) {
for(int i=0; i<maxm; i++) {
if(_v&(1<<i)) {
sumv[i][o]=r-l+1-sumv[i][o];
}
}
setv[o]^=_v;
} else {
pushdown(o,l,r);
if(ql<=M) update(lson,l,M);
if(qr>M) update(rson,M+1,r);
maintain(o);
}
}
LL _sum;
void query(int o,int l,int r){
if(ql<=l&&r<=qr){
for(int i=0;i<maxm;i++){
_sum+=sumv[i][o]*((1LL)<<i);
}
}else{
pushdown(o,l,r);
if(ql<=M) query(lson,l,M);
if(qr>M) query(rson,M+1,r);
maintain(o);
}
}
int main() {
scanf("%d",&n);
build(1,1,n);
int q;
scanf("%d",&q);
while(q--) {
int cmd;
scanf("%d",&cmd);
scanf("%d%d",&ql,&qr);
if(cmd==1) {
_sum=0;
query(1,1,n);
printf("%I64d\n",_sum);
} else {
scanf("%d",&_v);
update(1,1,n);
}
}
return 0;
}
Notes
其实和XOR有关的题目大部分都是套路！！！！二进制展开，你会发现一个新的世界！！！！
