Codeforces 703D Mishka and Interesting sum (树状数组求区间内不同的数的异或和)

http://www.codeforces.com/contest/703/problem/D

Little Mishka enjoys programming. Since her birthday has just passed, her friends decided to present her with array of non-negative integers a1, a2, …, an of n elements!

Mishka loved the array and she instantly decided to determine its beauty value, but she is too little and can’t process large arrays. Right because of that she invited you to visit her and asked you to process m queries.

Each query is processed in the following way:

Two integers l and r (1 ≤ l ≤ r ≤ n) are specified — bounds of query segment.
Integers, presented in array segment [l,  r] (in sequence of integers al, al + 1, ..., ar) even number of times, are written down.
XOR-sum of written down integers is calculated, and this value is the answer for a query. Formally, if integers written down in point 2 are x1, x2, ..., xk, then Mishka wants to know the value , where  — operator of exclusive bitwise OR. 

Since only the little bears know the definition of array beauty, all you are to do is to answer each of queries presented.
Input

The first line of the input contains single integer n (1 ≤ n ≤ 1 000 000) — the number of elements in the array.

The second line of the input contains n integers a1, a2, …, an (1 ≤ ai ≤ 109) — array elements.

The third line of the input contains single integer m (1 ≤ m ≤ 1 000 000) — the number of queries.

Each of the next m lines describes corresponding query by a pair of integers l and r (1 ≤ l ≤ r ≤ n) — the bounds of query segment.
Output

Print m non-negative integers — the answers for the queries in the order they appear in the input.

Examples
Input

3
3 7 8
1
1 3

Output

0

Input

7
1 2 1 3 3 2 3
5
4 7
4 5
1 3
1 7
1 5

Output

0
3
1
3
2


题意:求区间内出现偶数次的数的所有异或值。
解题思路:

区间内所有数的异或值 = 区间内出现奇数次的数的所有异或值

区间内出现偶数次的数的所有异或值 = 区间内所有数的异或值 ^ 区间内不同的数的异或值。

然而区间内所有数的异或值可以用前缀异或数组得到,问题转化成求区间内不同的数的异或值。一开始用莫队算法,O(n*sqrt(n)) TLE了。标答是 O(nlog(n))。

标答是O(nlog(n)): 维护一个last[]数组,记录的是当前i位置的数在它前面出现的位置。然后按照询问的r从小到大排序。枚举n,插入该位置的数,当这个数前面出现过,即last[i]不等于0,那么先将这个数删掉,然后再插入当前位置的的数。这样就保证了每个数在树状数组里面仅出现过一次。
这种做法要记住。可活用。


先贴一份TLE的莫队算法:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <math.h>
using namespace std;

typedef long long LL;
const int N = 1e6 + 100;
LL arr[N],XOR[N],ANS[N];
int size;
map<LL,int> Hash;
struct Node
{
        int l,r,id;
        friend bool operator < (const Node &a, const Node &b)
        {
                if(a.l/size == b.l/size) return a.r < b.r;
                else return a.l/size < b.l/size;
        }
}q[N];
void work(int k)
{
        LL now = 0;
        int curL = q[1].l, curR = q[1].r;
        for(int i=curL; i<=curR ;i++)
        {
               now ^=  (Hash[arr[i]] ? 0 : arr[i]);
               Hash[arr[i]]++;
        }
        ANS[q[1].id] =  now ^ (XOR[q[1].l-1] ^ XOR[q[1].r]);

        for(int i=2;i<=k;i++)
        {
                while(q[i].l > curL)
                {
                        if(Hash[arr[curL]]==1) now ^= arr[curL];
                        Hash[arr[curL]]--;
                        curL++;
                }
                while(q[i].l < curL)
                {
                       curL--;
                       if(Hash[arr[curL]]==0) now ^= arr[curL];
                       Hash[arr[curL]]++;
                }
                while(q[i].r < curR)
                {
                        if(Hash[arr[curR]]==1) now^= arr[curR];
                        Hash[arr[curR]]--;
                        curR--;
                }
                while(q[i].r > curR)
                {
                        curR++;
                        if(Hash[arr[curR]]==0) now^= arr[curR];
                        Hash[arr[curR]]++;
                }
                ANS[q[i].id] = now ^ XOR[q[i].l-1] ^ XOR[q[i].r];
        }
}
int main()
{
        int n;
        scanf("%d",&n);
        size = (int)sqrt(n);
        for(int i=1;i<=n;i++) scanf("%I64d",&arr[i]), XOR[i] = XOR[i-1]^arr[i];
        int k;
        scanf("%d",&k);
        for(int i=1;i<=k;i++)
        {
                scanf("%d%d",&q[i].l,&q[i].r);
                q[i].id = i;
        }
        sort(q+1,q+1+k);
        work(k);
        for(int i=1;i<=k;i++) printf("%I64d\n",ANS[i]);
        return 0;
}

再贴一份AC代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;

const int N = 1e6 + 100;
int arr[N],n,XOR[N],last[N],tr[N],ans[N];
struct Node
{
    int l,r,id;
    friend bool operator < (const Node &a,const Node &b)
    {
        return a.r < b.r;
    }
}Q[N];
map<int,int>Hash;
int lowbit(int x) { return x&(-x); }
void add(int i,int val)
{
    for(;i<=n;i+=lowbit(i)) 
        tr[i] ^= val;
}
int query(int i)
{
    int res = 0;
    for(;i>=1;i-=lowbit(i))
        res ^= tr[i];
    return res;
}
int main()
{
    scanf("%d",&n);
    Hash.clear();
    memset(last,0,sizeof(last));
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&arr[i]);
        XOR[i] = XOR[i-1]^arr[i];
        last[i] = Hash[arr[i]];
        Hash[arr[i]] = i;
    }
    int m;
    scanf("%d",&m);
    for(int i=1;i<=m;i++) scanf("%d%d",&Q[i].l,&Q[i].r),Q[i].id = i;
    sort(Q+1,Q+1+m);

    int px = 1;
    for(int i=1;i<=n;i++)
    {
            if(last[i]) add(last[i],arr[i]);
            add(i,arr[i]);
            while( px <=m && Q[px].r == i  )
            {
                ans[Q[px].id] = query(Q[px].r) ^ query(Q[px].l-1);
                ans[Q[px].id] ^= XOR[Q[px].r] ^ XOR[Q[px].l-1];
                px++;
            }
    }
    for(int i=1;i<=m;i++) printf("%d\n",ans[i]);
    return 0;
}
### CodeForces 上关于树状数组的题目列表 在CodeForces平台上,存在多个涉及树状数组据结构问题。以下是部分精选题目及其简短描述: #### 1. **799C - Fountains** 此题要实现单点更新操作,在`A[]`组中修改某一位置的值并相应地更新辅助组`tree[]`。具体来说,当执行增加操作时,会通过循环不断更新受影响节点直到超出范围为止[^1]。 ```cpp void add(int k, int num) { while (k <= n) { tree[k] += num; k += k & (-k); } } ``` #### 2. **1042D - Array and Operations** 该挑战涉及到利用树状数组或线段树来高效处理特定类型的查询请。核心思路在于将原始问题转化为计算前缀中小于给定阈值的量统计问题,从而简化了解决方案的设计过程[^2]。 #### 3. **1635F - Closest Pair** 这道难题不仅考察了参赛者对树状数组的理解程度,还测试了其解决复杂逻辑的能力。题目背景设定在一个有序序列基础上,目标是在指定范围内找到具有最小化评估函的一对索引组合[^3]。 #### 4. **703D - Mishka and Interesting Contest** 本题聚焦于如何有效地区间内所有不同值的异或结果。采用的方法是预先按右端点排序各区间,并记录每种元素最后一次出现的位置;之后借助树状数组完成后续所需的快速查找与更新工作[^4]。 这些例子展示了树状数组作为一种强大工具的应用场景,能够显著提升算法效率特别是在面对大规模据集的情况下表现尤为突出。
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