Sansa and XOR
Problem Statement :
Sansa has an array. She wants to find the value obtained by XOR-ing the contiguous subarrays, followed by XOR-ing the values thus obtained. Determine this value. Example Subarray Operation Result 3 None 3 4 None 4 5 None 5 3,4 3 XOR 4 7 4,5 4 XOR 5 1 3,4,5 3 XOR 4 XOR 5 2 Now we take the resultant values and XOR them together: . Return . Function Description Complete the sansaXor function in the editor below. sansaXor has the following parameter(s): int arr[n]: an array of integers Returns int: the result of calculations Input Format The first line contains an integer , the number of the test cases. Each of the next pairs of lines is as follows: - The first line of each test case contains an integer , the number of elements in . - The second line of each test case contains space-separated integers .
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
#define s(k) scanf("%d",&k);
#define sll(k) scanf("%lld",&k);
#define p(k) printf("%d\n",k);
#define pll(k) printf("%lld\n",k);
#define f(i,N) for(i=0;i<N;i++)
#define f1(i,N) for(i=0;i<=N;i++)
#define f2(i,N) for(i=1;i<=N;i++)
typedef long long ll;
int main(){
int t,n,k,sum,i;
s(t)
while(t--){
s(n)
if(!(n%2)){
printf("0\n");
f(i,n)
s(k)
} else {
sum=0;
f(i,n){
s(k)
if(!(i%2))
sum^=k;
}
p(sum);
}
}
return 0;
}
Solution in C++ :
In C++ :
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <limits>
#include <cstring>
#include <string>
using namespace std;
#define pairii pair<int, int>
#define llong long long
#define pb push_back
#define sortall(x) sort((x).begin(), (x).end())
#define INFI numeric_limits<int>::max()
#define INFLL numeric_limits<llong>::max()
#define INFD numeric_limits<double>::max()
#define FOR(i,s,n) for (int (i) = (s); (i) < (n); (i)++)
#define FORZ(i,n) FOR((i),0,(n))
void solve() {
int n;
scanf("%d",&n);
int res = 0;
FORZ(i,n) {
int x;
scanf("%d",&x);
if ((i+1)%2 && (n-i)%2) {
res ^= x;
}
}
printf("%d\n",res);
}
int main() {
#ifdef DEBUG
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int t;
scanf("%d",&t);
FORZ(i,t) solve();
return 0;
}
Solution in Java :
In Java :
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
int num = Integer.parseInt(line);
for (int i = 0; i < num; i++) {
String count = br.readLine();
String[] ns = br.readLine().split(" ");
sansaXor(ns, Integer.valueOf(count));
}
}
public static void sansaXor(String[] strs, int count){
Map<Integer,Integer> map = new HashMap<Integer,Integer>();
long result = 0;
for(int i = 0; i < count; i++){
int numCount = (i + 1)*(count - i);
int tmp = Integer.valueOf(strs[i]);
if(map.containsKey(tmp)){
map.put(tmp, numCount+map.get(tmp));
}else{
map.put(tmp, numCount);
}
}
for(int k: map.keySet()){
int value = map.get(k);
if(value%2!=0){
result = result^k;
}
}
System.out.println(result);
}
}
Solution in Python :
In Python3 :
for _ in range(int(input())):
N = int(input())
A = tuple(map(int,input().split()))
X = 0
for i, x in enumerate(A):
if ((i+1) * (N-i)) % 2 == 1:
X ^= x
print(X)
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