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Commit 7b90b29

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‎DPandbitmasking/Countingstrings.cpp‎

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/*
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Name: Mehul Chaturvedi
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IIT-Guwahati
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*/
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/*
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PROBLEM STATEMENT
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Given a string 's' consisting of upper case alphabets, i.e. from 'A' to 'Z'. Your task is to find how many strings 't' with length equal to that of 's', also consisting of upper case alphabets are there satisfying the following conditions:
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-> String 't' is lexicographical larger than string 's'.
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-> When you write both 's' and 't' in the reverse order, 't' is still lexicographical larger than 's'.
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Find out number of such strings 't'. As the answer could be very large, take modulo 10^9 + 7.
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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typedef unordered_map<int, int> umapii;
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typedef unordered_map<int, bool> umapib;
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typedef unordered_map<string, int> umapsi;
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typedef unordered_map<string, string> umapss;
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typedef map<string, int> mapsi;
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typedef map<pair<int, int>, int> mappiii;
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typedef map<int, int> mapii;
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typedef pair<int, int> pii;
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typedef pair<long long, long long> pll;
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typedef unordered_set<int> useti;
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#define uset unordered_set
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#define it iterator
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#define mp make_pair
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#define pb push_back
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#define all(x) (x).begin(), (x).end()
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#define f first
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#define s second
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#define MOD 1000000007
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#include<bits/stdc++.h>
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using namespace std;
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long long ans[100001],mod=1e9+7;
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int go(string s){
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string str="";
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int n=s.length();
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for(int i=n-1;i>=1;i--)
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str.push_back(s.at(i));
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for(int i=0;i<n-1;i++)
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ans[i]=(ans[i-1]*26%mod+90-str.at(i))%mod;
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long long a=0;
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ans[-1]=0;
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for(int i=0;i<n;i++)
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{
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a=(a+(90-s.at(i))*(ans[n-i-2]+1)%mod)%mod;
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}
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return a;
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}
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int countStrings(char* s) {
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int i=0;
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string s1="";
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while(s[i]!='0円'){
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s1=s1+s[i];
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i++;
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}
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return go(s1);
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}
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int main( int argc , char ** argv )
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{
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ios_base::sync_with_stdio(false) ;
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cin.tie(NULL) ;
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char s[100005];
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cin>>s;
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int ans = countStrings(s);
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cout<<ans<<endl;
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return 0 ;
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}

‎Tries/Test.txt‎

Whitespace-only changes.

‎Tries/maximumxorsub.cpp‎

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/*
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Name: Mehul Chaturvedi
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IIT-Guwahati
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*/
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/*
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PROBLEM STATEMENT
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Given an array of n integers, find subarray whose xor is maximum.
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Input Format
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First line contains single integer n (1<=n<=1000000).
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Next line contains n space separated integers.
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Output Format
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Print xor of the subarray whose xor of all elements in subarray is maximum over all subarrays.
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Constraints
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1 <= n <= 1000000
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1 <= A[i] <=100000
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Sample Input
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4
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1 2 3 4
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Sample Output
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7
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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typedef unordered_map<int, int> umapii;
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typedef unordered_map<int, bool> umapib;
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typedef unordered_map<string, int> umapsi;
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typedef unordered_map<string, string> umapss;
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typedef map<string, int> mapsi;
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typedef map<pair<int, int>, int> mappiii;
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typedef map<int, int> mapii;
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typedef pair<int, int> pii;
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typedef pair<long long, long long> pll;
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typedef unordered_set<int> useti;
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#define uset unordered_set
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#define it iterator
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#define mp make_pair
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#define pb push_back
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#define all(x) (x).begin(), (x).end()
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#define f first
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#define s second
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#define MOD 1000000007
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#include<bits/stdc++.h>
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using namespace std;
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// Assumed int size
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#define INT_SIZE 32
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// A Trie Node
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struct TrieNode
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{
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int value; // Only used in leaf nodes
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TrieNode *arr[2];
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};
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// Utility function tp create a Trie node
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TrieNode *newNode()
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{
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TrieNode *temp = new TrieNode;
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temp->value = 0;
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temp->arr[0] = temp->arr[1] = NULL;
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return temp;
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}
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// Inserts pre_xor to trie with given root
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void insert(TrieNode *root, int pre_xor)
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{
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TrieNode *temp = root;
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// Start from the msb, insert all bits of
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// pre_xor into Trie
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for (int i=INT_SIZE-1; i>=0; i--)
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{
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// Find current bit in given prefix
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bool val = pre_xor & (1<<i);
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// Create a new node if needed
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if (temp->arr[val] == NULL)
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temp->arr[val] = newNode();
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temp = temp->arr[val];
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}
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// Store value at leaf node
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temp->value = pre_xor;
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}
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// Finds the maximum XOR ending with last number in
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// prefix XOR 'pre_xor' and returns the XOR of this maximum
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// with pre_xor which is maximum XOR ending with last element
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// of pre_xor.
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int query(TrieNode *root, int pre_xor)
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{
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TrieNode *temp = root;
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for (int i=INT_SIZE-1; i>=0; i--)
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{
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// Find current bit in given prefix
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bool val = pre_xor & (1<<i);
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// Traverse Trie, first look for a
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// prefix that has opposite bit
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if (temp->arr[1-val]!=NULL)
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temp = temp->arr[1-val];
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// If there is no prefix with opposite
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// bit, then look for same bit.
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else if (temp->arr[val] != NULL)
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temp = temp->arr[val];
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}
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return pre_xor^(temp->value);
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}
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// Returns maximum XOR value of a subarray in arr[0..n-1]
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int maxSubarrayXOR(int arr*, int n)
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{
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// Create a Trie and insert 0 into it
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TrieNode *root = newNode();
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insert(root, 0);
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// Initialize answer and xor of current prefix
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int result = INT_MIN, pre_xor =0;
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// Traverse all input array element
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for (int i=0; i<n; i++)
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{
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// update current prefix xor and insert it into Trie
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pre_xor = pre_xor^arr[i];
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insert(root, pre_xor);
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// Query for current prefix xor in Trie and update
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// result if required
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result = max(result, query(root, pre_xor));
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}
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return result;
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}
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int main( int argc , char ** argv )
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{
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ios_base::sync_with_stdio(false) ;
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cin.tie(NULL) ;
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int n;
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cin>>n;
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int* arr = new int[n];
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for (int i = 0; i < n; ++i)
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{
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cin>>arr[i];
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}
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cout << maxSubarrayXOR(arr, n) << '\n';
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return 0 ;
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}

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