Romantic Reversals CodeChef Solution

Problem -Romantic Reversals CodeChef Solution

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Romantic Reversals CodeChef Solution in C++17

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n,k;
        cin >> n >> k;
        string s;
        cin >> s;
        string ans;
        int first=0, last=k-1;
        int c=0;
        while(first<=last)
        {
            if(c%2==0){
                ans += s[first];
                first++;
            }
            else{
                ans+=s[last];
                last--; 
            }
            c++;
        }
        reverse(ans.begin(), ans.end());
        for(int i=k; i<n; ++i){
            ans += s[i];
        }
        cout << ans << endl;
    }
}

Romantic Reversals CodeChef Solution in C++14

#include <bits/stdc++.h>
using namespace std;



void solve()
{
    int n,k;
    cin>>n>>k;
    string s;
    cin>>s;
    string ans="";
    int  it1=(k/2),it2=(k/2)-1;
    if(k%2==1)
    {
        cout<<s[it1];
        it1++;
    }
    for(int i=0;i<k/2;i++)
    {
         cout<<s[it1];
        cout<<s[it2];
        it2--;
        it1++;
    }
    cout<<ans;
    for(int i=k;i<n;i++)
    {
        cout<<s[i];
    }
    cout<<"\n";
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

Romantic Reversals CodeChef Solution in PYTH 3

# cook your dish here
t=int(input())
for _ in range(t):
    n,k=map(int,input().split())
    s=input()
    x=s[:k]
    y=s[k:]
    r=""
    for i in range(len(x)//2):
        r+=x[i]
        r+=x[-(i+1)]
    if(len(x)%2):
        r+=x[len(x)//2]
    r=r[::-1]
    print(r+y)

Romantic Reversals CodeChef Solution in C

#include <stdio.h>

void solve(int n, int k)
{
    char str[n];
    char ans[n];
    char dir = 1;
    int i, j = 0, s = k - 1, p = k;
    char temp;
    for (i = 0; i < n; i++)
        scanf("%c", &str[i]);
    i = 0;
    while (s >= j)
    {

        if (dir)
        {
            ans[i++] = str[j++];
            dir = 0;
        }
        else
        {
            ans[i++] = str[s--];
            dir = 1;
        }
    }
    for (i--; i >= 0; i--)
        printf("%c", ans[i]);
    for (i = p; i < n; i++)
        printf("%c", str[i]);
    printf("\n");
}
int main()
{
    int t, n, k;

    scanf("%d", &t);
    while (t > 0)
    {
        t--;
        scanf("%d%d\n", &n, &k);
        solve(n, k);
    }
    return 0;
}

Romantic Reversals CodeChef Solution in JAVA

import java.io.*;
import java.lang.*;
import java.util.*;

//Atg
class Codechef {

  public static void main(String[] args) throws java.lang.Exception {
    Scanner sc = new Scanner(System.in);
    int t = 1;
    t = sc.nextInt();
    while (t-- > 0) {
      int n = sc.nextInt();
      int k = sc.nextInt();
      String a = sc.next();
      //   char[] brr = new char[n + 1];
      String str = a;
      StringBuilder brr = new StringBuilder(str);

      int ind = 0;
      int tmp = 1;
      if (k % 2 == 1) {
        tmp = 2;
      }
      int kk = k;

      for (int i = k + 1; i < n + 1; i++) {
        brr.setCharAt(i-1, a.charAt(i - 1));
      }

      while (k > 0) {
        brr.setCharAt(k-1, a.charAt(ind++));
        k -= 2;
      }

      while (tmp < kk) {
        brr.setCharAt(tmp-1, a.charAt(ind++));
        tmp += 2;
      }

      System.out.print(brr);
      System.out.println();
    }
  }
}

Romantic Reversals CodeChef Solution in PYPY 3

T = int(input())

for _ in range(T):
    N, K = map(int, input().split())
    
    S = input()
    lst = list(S)
    s = ''
    i = 0
    j = K-1
    l = K-1
    while i < j:
        lst[l] = S[i]
        l -= 1
        lst[l] = S[j]
        l-= 1
        j -= 1
        i += 1
    
    if i == j:
        lst[l] = S[i]
    print(''.join(lst))

Romantic Reversals CodeChef Solution in PYTH

t = int(raw_input())
for i in range(t):
	st = raw_input().split()
	N = int(st[0])
	K = int(st[1])
	S = raw_input().strip()
	p = K/2
	st = S[p]
	if K%2 == 0:
		m = -1
	else:
		m = 1
	# endif
	for k in range(1,K):
		p += m*k
		m = -1*m
		st += S[p]
	# endfor k
	for k in range(K,N):
		st += S[k]
	# endfor k
	print st
# endfor i

Romantic Reversals CodeChef Solution in C#

// Problem: RMNTREV - Romantic Reversals
// Author: Gusztav Szmolik

using System;
using System.Text;

public class RomanticReversals
    {
    public static int Main ()
        {
        ushort t = UInt16.Parse (Console.ReadLine ());
        string[] parts = null;
        for (ushort i = 0; i < t; i++)
            {
            parts = Console.ReadLine ().Split ();
            uint n = UInt32.Parse (parts[0]);
            uint k = UInt32.Parse (parts[1]);
            string s = Console.ReadLine ();
            StringBuilder ans = new StringBuilder ();
            bool oddLength = (k%2 == 1);
            if (oddLength)
                ans.Append (s[Convert.ToInt32 ((k-1)/2)]);
            for (uint j = (oddLength ? (k+1)/2 : k/2); j < k; j++)
                {
                ans.Append (s[(int)j]);
                ans.Append (s[Convert.ToInt32 (k-1-j)]);
                }
            ans.Append (s.Substring ((int)k));
            Console.WriteLine (ans);
            }
        return 0;
        }
    }

Romantic Reversals CodeChef Solution in GO

package main

import (
	"bufio"
	"bytes"
	"fmt"
	"os"
)

func main() {
	reader := bufio.NewReader(os.Stdin)

	tc := readNum(reader)
	var buf bytes.Buffer
	for tc > 0 {
		tc--
		n, k := readTwoNums(reader)
		S, _ := reader.ReadString('\n')
		res := solve(n, k, S)
		buf.WriteString(fmt.Sprintf("%s\n", res))
	}
	fmt.Println(buf.String())
}

func readInt(bytes []byte, from int, val *int) int {
	i := from
	sign := 1
	if bytes[i] == '-' {
		sign = -1
		i++
	}
	tmp := 0
	for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
		tmp = tmp*10 + int(bytes[i]-'0')
		i++
	}
	*val = tmp * sign
	return i
}

func readNum(reader *bufio.Reader) (a int) {
	bs, _ := reader.ReadBytes('\n')
	readInt(bs, 0, &a)
	return
}

func readTwoNums(reader *bufio.Reader) (a int, b int) {
	res := readNNums(reader, 2)
	a, b = res[0], res[1]
	return
}

func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
	res := readNNums(reader, 3)
	a, b, c = res[0], res[1], res[2]
	return
}

func readNNums(reader *bufio.Reader, n int) []int {
	res := make([]int, n)
	x := 0
	bs, _ := reader.ReadBytes('\n')
	for i := 0; i < n; i++ {
		for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
			x++
		}
		x = readInt(bs, x, &res[i])
	}
	return res
}

func readUint64(bytes []byte, from int, val *uint64) int {
	i := from

	var tmp uint64
	for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
		tmp = tmp*10 + uint64(bytes[i]-'0')
		i++
	}
	*val = tmp

	return i
}
func solve(n, k int, X string) string {
	buf := make([]byte, n)
	copy(buf[k:], X[k:])
	l, r := 0, k-1
	pos := k - 1
	for l <= r {
		buf[pos] = X[l]
		if pos > 0 {
			buf[pos-1] = X[r]
		}
		pos -= 2
		l++
		r--
	}

	return string(buf)
}

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