CF 242E XOR on Segment（二维线段树）

最新推荐文章于 2022-09-06 11:03:04 发布

KinFungYu

最新推荐文章于 2022-09-06 11:03:04 发布

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分类专栏：树状数组线段树 RMQ 文章标签：二维线段树

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XOR on Segment

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You've got an array a, consisting of n integers a₁, a₂, ..., a_n. 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 a_l + a_l + 1 + ... + a_r.
Apply the xor operation with a given number x to each array element on the segment [l, r], that is, execute $\text{[math]}$ . This operation changes exactly r - l + 1 array elements.

Expression $\text{[math]}$ 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.

Sample test(s)

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

Output

Input

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

Output

38
28

题意：有两个操作：1.区间求和：求给定区间元素的和；区间更新：将给定区间里的所有数与一个给定的数异或。

思路：异或不满足分配律。每个数化成二进制时，每个位上的异或是独立的，这样我们就可以在线段树的节点上保存一定区间内的所有数的第i(0<i<20)位上的1的个数总和，求和时只要统计每位上1的个数，再化为十进制即可。用到二维线段树。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdlib>
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
#define ll long long
using namespace std;

const int maxn=100005;
struct node
{
    int l,r;
    ll sum,xo;
}tree[22][4*maxn];
int num[maxn];
int n;
void build(int id,int l,int r,int rt)
{
    tree[id][rt].l=l;
    tree[id][rt].r=r;
    tree[id][rt].xo=0;
    if(l==r)
    {
        tree[id][rt].sum=(num[l]>>id)&1;
        return;
    }
    int mid=(l+r)>>1;
    build(id,l,mid,L(rt));
    build(id,mid+1,r,R(rt));
    tree[id][rt].sum=tree[id][L(rt)].sum+tree[id][R(rt)].sum;
}
void update(int id,int l,int r,int rt)
{
    if(tree[id][rt].l==l&&tree[id][rt].r==r)
    {
        tree[id][rt].sum=r-l+1-tree[id][rt].sum;
        tree[id][rt].xo^=1;
        return;
    }
    if(tree[id][rt].xo)
    {
        tree[id][L(rt)].sum=tree[id][L(rt)].r-tree[id][L(rt)].l+1-tree[id][L(rt)].sum;
        tree[id][L(rt)].xo^=1;
        tree[id][R(rt)].sum=tree[id][R(rt)].r-tree[id][R(rt)].l+1-tree[id][R(rt)].sum;
        tree[id][R(rt)].xo^=1;
        tree[id][rt].xo=0;
    }
    if(r<=tree[id][L(rt)].r) update(id,l,r,L(rt));
    else if(l>=tree[id][R(rt)].l) update(id,l,r,R(rt));
    else
    {
        update(id,l,tree[id][L(rt)].r,L(rt));
        update(id,tree[id][R(rt)].l,r,R(rt));
    }
    tree[id][rt].sum=tree[id][L(rt)].sum+tree[id][R(rt)].sum;
}
ll query(int id,int l,int r,int rt)
{
    if(tree[id][rt].l==l&&tree[id][rt].r==r)
    return tree[id][rt].sum;
    if(tree[id][rt].xo)
    {
        tree[id][L(rt)].sum=tree[id][L(rt)].r-tree[id][L(rt)].l+1-tree[id][L(rt)].sum;
        tree[id][L(rt)].xo^=1;
        tree[id][R(rt)].sum=tree[id][R(rt)].r-tree[id][R(rt)].l+1-tree[id][R(rt)].sum;
        tree[id][R(rt)].xo^=1;
        tree[id][rt].xo=0;
    }
    if(r<=tree[id][L(rt)].r) return query(id,l,r,L(rt));
    else if(l>=tree[id][R(rt)].l) return query(id,l,r,R(rt));
    else return query(id,l,tree[id][L(rt)].r,L(rt))+query(id,tree[id][R(rt)].l,r,R(rt));
}
int main()
{
    //freopen("1.txt","r",stdin);
    int a,b,op,m;
    ll x;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
            cin>>num[i];

        for(int i=0;i<=20;i++)
        build(i,1,n,1);

        cin>>m;
        while(m--)
        {
            cin>>op;
            if(op==1)
            {
                cin>>a>>b;
                ll ans=0;
                for(int i=0;i<=20;i++)
                ans+=query(i,a,b,1)<<i;
                cout<<ans<<endl;
            }
            else
            {
                cin>>a>>b>>x;
                for(int i=0;i<=20;i++)
                if(x>>i&1)
                update(i,a,b,1);
            }
        }
    }
    return 0;
}