題目描述：

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

1. Calculate the sum of current array elements on the segment [l, r], that is, count value a_l + a_l + 1 + ... + a_r.
2. 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

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.

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

題目思路：

題意：N個點，M個操作，1是查詢區間和，2是區間內每個數異或x。

思路：異或只跟每個二進位制位有關，於是線段樹儲存區間內二進位制各個位1的數量，異或一個數時，若某位是1，那麼用區間長度減去原來的1數量就是異或後的1數量了。

真滴傻逼！，這個位運算子的優先順序非常低，必須要加括號，真滴傻逼！

還有就是爆int的數再強制轉換為long long的話這個數就會飆到long long的最大值，所以必須在爆int之前強制轉換為long long

程式碼：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#define ri(n) scanf("%d",&n)
#define oi(n) printf("%d\n",n)
#define rl(n) scanf("%lld",&n)
#define ol(n) printf("%lld\n",n)
#define rep(i,l,r) for(i=l;i<=r;i++)
#define rep1(i,l,r) for(i=l;i<r;i++)
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int epg=10-8;
const int maxn=1e5+10;
int val[maxn*4],a[maxn*4][25];
int x;
ll ans;
void uppush(int id)
{
    for(int i=0;i<22;i++)
        a[id][i]=a[id<<1][i]+a[id<<1|1][i];
}
void updown(int id,int sum)
{
    if(val[id])
    {
        val[id<<1]^=val[id];
        val[id<<1|1]^=val[id];
        for(int i=0;i<22;i++)
        {
            if((val[id]>>i)&1)
            {
                a[id<<1][i]=sum-(sum>>1)-a[id<<1][i];
                a[id<<1|1][i]=(sum>>1)-a[id<<1|1][i];
            }
        }
        val[id]=0;
    }
}
void build(int l,int r,int id)
{
    if(l==r)
    {
        scanf("%d",&x);
        for(int i=0;i<22;i++)
        {
            if((x>>i)&1)
                a[id][i]=1;
            else
                a[id][i]=0;
        }
        return ;
    }
    int m=(l+r)>>1;
    build(l,m,id<<1);
    build(m+1,r,id<<1|1);
    uppush(id);
}
void updata(int L,int R,int ad,int l,int r,int id)
{
    if(L<=l&&R>=r)
    {
        val[id]^=ad;
        for(int i=0;i<22;i++)
        {
            if((ad>>i)&1)
                a[id][i]=r-l+1-a[id][i];
        }
        return ;
    }
    updown(id,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)
        updata(L,R,ad,l,m,id<<1);
    if(R>m)
        updata(L,R,ad,m+1,r,id<<1|1);
    //updown(id,r-l+1);
    uppush(id);
}
void query(int L,int R,int l,int r,int id)
{
    if(L<=l&&R>=r)
    {
        for(int i=0;i<22;i++)
        {
             ans+=((ll)a[id][i]<<i);//真滴傻逼
        }
        return ;
    }
    updown(id,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)
        query(L,R,l,m,id<<1);
    if(R>m)
        query(L,R,m+1,r,id<<1|1);
    //updown(id,r-l+1);
}
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        memset(a,0,sizeof(a));
        memset(val,0,sizeof(val));
        build(1,n,1);
        int m;
        scanf("%d",&m);
        int y,L,R,ad;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&y);
            if(y==1)
            {
                scanf("%d%d",&L,&R);
                ans=0;
                query(L,R,1,n,1);
                printf("%lld\n",ans);
            }
            else
            {
                scanf("%d%d%d",&L,&R,&ad);
                updata(L,R,ad,1,n,1);
            }
        }
    }
    return 0;
}