线段树+区间xor+区间求和

题目描述：

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 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, cout streams, or the %I64d specifier.

Example

Input

54 10 3 13 781 2 42 1 3 31 2 41 3 32 2 5 51 1 52 1 2 101 2 3

Output

262203411

Input

64 7 4 0 7 352 2 3 81 1 52 3 5 12 4 5 61 2 3

Output

3828

题目思路：

题意：N个点，M个操作，1是查询区间和，2是区间内每个数异或x。

思路：异或只跟每个二进制位有关，于是线段树储存区间内二进制各个位1的数量，异或一个数时，若某位是1，那么用区间长度减去原来的1数量就是异或后的1数量了。

真滴傻逼！，这个位运算符的优先级非常低，必须要加括号，真滴傻逼！

还有就是爆int的数再强制转换为long long的话这个数就会飙到long long的最大值，所以必须在爆int之前强制转换为long long

代码：

#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;}