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.
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.
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.
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
26
22
0
34
11
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
38
28
一个用二维线段树操作的异或题，建立20棵线段数 tree[i]表示这n个数第i为在二进制下的情况 更新的时候将每个数 每个位更新 并用延迟符号 求和的时候按位加起来。
1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 #include <string.h> 5 #include <math.h> 6 #include <vector> 7 #include <stack> 8 using namespace std; 9 #define ll long long int 10 ll a[400000][20]; 11 ll b[400000]; 12 void build(int l,int r,int t) 13 { 14 int i; 15 if(l==r) 16 { 17 int x; 18 scanf("%d",&x); 19 i=0; 20 while(x) 21 { 22 a[t][i++]=x%2; 23 x/=2; 24 } 25 return ; 26 } 27 int m=(l+r)>>1; 28 build(l,m,t<<1); 29 build(m+1,r,t<<1|1); 30 for(i=0; i<20; i++) 31 a[t][i]=a[t<<1][i]+a[t<<1|1][i]; 32 } 33 void fun(int l,int r,int t) 34 { 35 int z=b[t]; 36 int m=(l+r)>>1; 37 b[t<<1]^=b[t]; 38 b[t<<1|1]^=b[t]; 39 int i=0; 40 while(z) 41 { 42 if(z%2){ 43 a[t<<1][i]=m-l+1-a[t<<1][i]; 44 a[t<<1|1][i]=r-m-a[t<<1|1][i]; 45 } 46 i++; 47 z/=2; 48 } 49 b[t]=0; 50 } 51 void update(int l,int r,int t,int x,int y,int z) 52 { 53 int i; 54 if(l>=x&&r<=y) 55 { 56 i=0; 57 b[t]^=z; 58 while(z) 59 { 60 if(z%2) 61 a[t][i]=r-l+1-a[t][i]; 62 i++; 63 z/=2; 64 } 65 return ; 66 } 67 if(b[t]) 68 fun(l,r,t); 69 int m=(l+r)>>1; 70 if(x<=m)update(l,m,t<<1,x,y,z); 71 if(y>m)update(m+1,r,t<<1|1,x,y,z); 72 for(i=0; i<20; i++) 73 a[t][i]=a[t<<1][i]+a[t<<1|1][i]; 74 } 75 ll query(int l,int r,int t,int x,int y) 76 { 77 if(l>=x&&r<=y) 78 { 79 ll sum=0; 80 for(int i=0;i<20;i++) 81 sum+=a[t][i]*(1<<i); 82 return sum; 83 } 84 if(b[t]) 85 fun(l,r,t); 86 int m=(l+r)>>1; 87 ll sum=0; 88 if(x<=m)sum+=query(l,m,t<<1,x,y); 89 if(y>m)sum+=query(m+1,r,t<<1|1,x,y); 90 for(int i=0; i<20; i++) 91 a[t][i]=a[t<<1][i]+a[t<<1|1][i]; 92 return sum; 93 } 94 int main() 95 { 96 memset(a,0,sizeof(a)); 97 memset(b,0,sizeof(b)); 98 int n; 99 cin>>n; 100 build(1,n,1); 101 int m; 102 cin>>m; 103 int i,s,x,y,z; 104 for(i=0; i<m; i++) 105 { 106 scanf("%d",&s); 107 if(s==1) 108 { 109 scanf("%d%d",&x,&y); 110 cout<<query(1,n,1,x,y)<<endl; 111 } 112 else 113 { 114 scanf("%d%d%d",&x,&y,&z); 115 update(1,n,1,x,y,z); 116 } 117 } 118 }