CF&&CC百套计划2 CodeChef December Challenge 2017 Chef And Easy Xor Queries

https://www.codechef.com/DEC17/problems/CHEFEXQ

题意：

位置i的数改为k

询问区间[1,i]内有多少个前缀的异或和为k

分块

sum[i][j] 表示第i块内，有多少个前缀，他们的异或和为j

a[i] 表示位置i的数

位置i改为k：

若 g=x1^x2^x3……

把 x1 改为 k 后，那新的g=x1^x1^k^x2^x3……

所以修改可以看做整体异或修改后的值^原来的值

即

区间[i，n] 异或上a[i]^k

i所在块单个改，后面的块整体打标记

查询：

i所在块单个查

前面的块累加sum[][k^标记]

#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
 
using namespace std;
 
#define N 100001
#define S 318
const int K=1<<20;
 
int sum[S][K+1];
 
int tag[S];
 
int a[N],prexo[N];
 
int bl[N];
 
void read(int &x)
{
    x=0; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); }
}
 
int main()
{
    int n,q;
    read(n); read(q);
    for(int i=1;i<=n;++i)
    {
        read(a[i]);
        prexo[i]=prexo[i-1]^a[i];
    }
    int siz=sqrt(n);
    for(int i=1;i<=n;++i)
    {
        bl[i]=(i-1)/siz+1;
        sum[bl[i]][prexo[i]]++;
    }
    int tot=bl[n];
    int ty,x,k;
    int m,res,pos;
    int ans;
    while(q--)
    {
        read(ty); read(x); read(k);
        pos=bl[x];
        if(ty==1)
        {
            res=a[x]^k;
            a[x]=k;
            m=min(pos*siz,n);
            if(tag[pos])
            {
                for(int i=(pos-1)*siz+1;i<=m;++i)
                {
                    sum[pos][prexo[i]]--;
                    prexo[i]^=tag[pos];
                    sum[pos][prexo[i]]++;
                }
                tag[pos]=0;
            }
            for(int i=x;i<=m;++i)
            {
                sum[pos][prexo[i]]--;
                prexo[i]^=res;
                sum[pos][prexo[i]]++;
            }
            for(int i=pos+1;i<=tot;++i) tag[i]^=res;
        }
        else
        {
            ans=0;
            for(int i=(pos-1)*siz+1;i<=x;++i)
            {
                if((prexo[i]^tag[pos])==k) ans++;
            } 
            for(int i=1;i<pos;++i) ans+=sum[i][k^tag[i]];
            cout<<ans<<'\n';
        }
    }
}

Read problems statements in Mandarin chinese, Russian andVietnamese as well.

Chef always likes to play with arrays. He came up with a new term "magical subarray". A subarray is called magical if its starting index is 1 (1-based indexing). Now, Chef has an array of N elements and 2 types of queries:

type 1: Given two numbers i and x, the value at index i should be updated to x.
type 2: Given two numbers i and k, your program should output the total number ofmagical subarrays with the last index ≤ i in which the xor of all elements is equal tok.

Input

The first line of the input contains two integers N and Q denoting the number of elements in the array and the number of queries respectively.
The second line contains N space-separated integers A₁, A₂ ... A_N denoting the initial values of the array.
Each of the following Q lines describes an operation. If the first integer is 1, it means that the operation is of type 1 and it will be followed by two integers i and x. If the first integer is 2, it means that the operations is of type 2 and it will be followed by two integers i and k.

Output

For each operation of type 2, print the number of magical subarrays on a separate line.

Constraints

1 ≤ N, Q ≤ 100,000
1 ≤ A[i] ≤ 1,000,000
1 ≤ i ≤ N
1 ≤ x, k ≤ 1,000,000

Subtasks

Subtask #1 (20 points): 1 ≤ N, Q ≤ 1,000

Subtask #2 (30 points): 1 ≤ N, Q ≤ 10,000

Subtask #3 (50 points): original constraints

Example

Input:

5 3
1 1 1 1 1
2 5 1
1 3 2
2 5 1

Output:

3
1

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