[CodeForces242E]XOR on Segment-线段树

XOR on Segment

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.

Input

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.

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.

Examples

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

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水题一眼切……
然而调tag调了半天……

(╯‵□′)╯︵┻━┻

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题意:
维护一个数据结构，支持两个操作:
1 l r ：输出[l,r]区间的和
2 l r x ：将[l,r]区间的数异或上x

思路:
看到异或显然是要拆位了~
然后显然是要用线段树。
在每个节点维护一个桶，分别存每个二进制位的0和1分别出现了多少个即可~
异或操作相当于打个tag，将对应有1的二进制位处的0和1的个数交换即可~

是不是很水然而咱就是调了将近1h.jpg

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;const int N=1e5+9;struct node{    int bit[22][2];    node(){memset(bit,0,sizeof(bit));}    inline void operator = (int x)    {        for(int i=20;i>=0;i--)            bit[i][(x>>i)&1]++;    }    inline ll get()    {        ll ans=0;        for(int i=20;i>=0;i--)            ans=(ans<<1)+(ll)bit[i][1];        return ans;    }    inline node operator + (node o)    {        node ret;        for(int i=20;i>=0;i--)        {            ret.bit[i][0]=bit[i][0]+o.bit[i][0];            ret.bit[i][1]=bit[i][1]+o.bit[i][1];        }        return ret;    }};node t[N<<2];int n,m,tag[N<<2],a[N];inline int read(){    int x=0;char ch=getchar();    while(ch<'0' || '9'<ch)ch=getchar();    while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();    return x;}inline void flush(int x,int val){    for(int i=20;i>=0;i--)        if((val>>i)&1)            swap(t[x].bit[i][0],t[x].bit[i][1]);}inline void push(int x,int l,int r){    if(l==r)    {        tag[x]=0;        return;    }    if(tag[x])    {        int mid=l+r>>1;        flush(x<<1,tag[x]);        flush(x<<1|1,tag[x]);        tag[x<<1]^=tag[x];        tag[x<<1|1]^=tag[x];        tag[x]=0;    }}inline void update(int x){    t[x]=t[x<<1]+t[x<<1|1];}void build(int x,int l,int r){    tag[x]=0;    if(l==r)    {        t[x]=a[l]=read();        return;    }    int mid=l+r>>1;    build(x<<1,l,mid);    build(x<<1|1,mid+1,r);    update(x);}void modify(int x,int l,int r,int dl,int dr,int val){    push(x,l,r);    if(l==dl && r==dr)    {        tag[x]^=val;        flush(x,tag[x]);        return;    }    int mid=l+r>>1;    if(dr<=mid)        modify(x<<1,l,mid,dl,dr,val);    else if(mid<dl)        modify(x<<1|1,mid+1,r,dl,dr,val);    else    {        modify(x<<1,l,mid,dl,mid,val);        modify(x<<1|1,mid+1,r,mid+1,dr,val);    }    update(x);}ll query(int x,int l,int r,int dl,int dr){    push(x,l,r);    if(l==dl && r==dr)        return t[x].get();    int mid=l+r>>1;    if(dr<=mid)        return query(x<<1,l,mid,dl,dr);    if(mid<dl)        return query(x<<1|1,mid+1,r,dl,dr);    return query(x<<1,l,mid,dl,mid)+query(x<<1|1,mid+1,r,mid+1,dr);}int main(){    n=read();    build(1,1,n);    m=read();    for(int i=1,op,l,r;i<=m;i++)    {        op=read();        l=read();        r=read();        if(op==1)            printf("%I64d\n",query(1,1,n,l,r));        else            modify(1,1,n,l,r,read());    }    return 0;}