**Cipher**

### Problem Statement :

Jack and Daniel are friends. They want to encrypt their conversations so that they can save themselves from interception by a detective agency so they invent a new cipher. Every message is encoded to its binary representation. Then it is written down times, shifted by bits. Each of the columns is XORed together to get the final encoded string. If and it looks like so: 1001011 shift 0 01001011 shift 1 001001011 shift 2 0001001011 shift 3 ---------- 1110101001 <- XORed/encoded string s Now we have to decode the message. We know that . The first digit in so our output string is going to start with . The next two digits are also , so they must have been XORed with . We know the first digit of our shifted string is a as well. Since the digit of is , we XOR that with our and now know there is a in the position of the original string. Continue with that logic until the end. Then the encoded message and the key are sent to Daniel. Jack is using this encoding algorithm and asks Daniel to implement a decoding algorithm. Can you help Daniel implement this? Function Description Complete the cipher function in the editor below. It should return the decoded string. cipher has the following parameter(s): k: an integer that represents the number of times the string is shifted s: an encoded string of binary digits Input Format The first line contains two integers and , the length of the original decoded string and the number of shifts. The second line contains the encoded string consisting of ones and zeros. Output Format Return the decoded message of length , consisting of ones and zeros.

### Solution :

` ````
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int n, k, i, j, pre, cur, a[1000000];
char input[2000000];
scanf("%d", &n);
scanf("%d", &k);
scanf("%s", input);
if (input[0] == '1')
pre = 1;
else
pre = 0;
printf("%d", pre);
a[0] = pre;
for (i = 1; i < n; i++) {
if (input[i] == '1') {
printf("%d", (pre ^ 1) & 1);
a[i] = (pre ^ 1) & 1;
} else {
printf("%d", (pre ^ 0) & 1);
a[i] = (pre ^ 0) & 1;
}
pre = pre ^ a[i];
if ((i - k + 2) > 0)
pre = pre ^ a[i - k + 1];
}
printf("\n");
return 0;
}
```

` ````
Solution in C++ :
In C++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int N,K;
cin>>N>>K;
char a[N+K-1];
cin>>a;
int arr[N],sum=0,i;
arr[0]=a[0]-48;
sum=arr[0];
for(i=1;i<N;i++){
if(i>=K)sum=(sum-arr[i-K]);
if(a[i]=='0') arr[i]=sum%2;
else arr[i]=(sum+1)%2;
sum=sum+arr[i];
}
// arr[N-1]=a[N+K-2]-48;
for(i=0;i<N;i++)
cout<<arr[i];
return 0;
}
```

` ````
Solution in Java :
In Java :
import java.io.*;
public class Solution {
public static void solve(Input in, PrintWriter out) throws IOException {
int n = in.nextInt();
int k = in.nextInt();
char[] c = in.next().toCharArray();
int x = 0;
char[] ans = new char[n];
for (int i = 0; i < n; ++i) {
ans[i] = (char) (((c[i] - '0') ^ x) + '0');
x ^= ans[i] - '0';
if (i >= k - 1) {
x ^= ans[i - k + 1] - '0';
}
}
out.println(ans);
}
public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
solve(new Input(new BufferedReader(new InputStreamReader(System.in))), out);
out.close();
}
static class Input {
BufferedReader in;
StringBuilder sb = new StringBuilder();
public Input(BufferedReader in) {
this.in = in;
}
public Input(String s) {
this.in = new BufferedReader(new StringReader(s));
}
public String next() throws IOException {
sb.setLength(0);
while (true) {
int c = in.read();
if (c == -1) {
return null;
}
if (" \n\r\t".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
int c = in.read();
if (c == -1 || " \n\r\t".indexOf(c) != -1) {
break;
}
sb.append((char)c);
}
return sb.toString();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}
```

` ````
Solution in Python :
In Python3 :
n, k = map(int, input().split())
s = input()
ans = []
pref = 0
for i in range(n):
val = pref ^ (ord(s[i]) - ord('0'))
ans.append(val)
pref ^= val
if i >= k - 1:
pref ^= ans[i - k + 1]
print(''.join(map(str, ans)))
```

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