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Problem Statement :
Johnny is playing with a large binary number, . The number is so large that it needs to be compressed into an array of integers, , where the values in even indices () represent some number of consecutive bits and the values in odd indices () represent some number of consecutive bits in alternating substrings of . For example, suppose we have array . represents , represents , represents , represents , and represents . The number of consecutive binary characters in the substring of corresponds to integer , as shown in this diagram: When we assemble the sequential alternating sequences of 's and 's, we get . We define setCount() to be the number of 's in a binary number, . Johnny wants to find a binary number, , that is the smallest binary number where setCount() = setCount(). He then wants to compress into an array of integers, (in the same way that integer array contains the compressed form of binary string ). Johnny isn't sure how to solve the problem. Given array , find integer array and print its length on a new line. Then print the elements of array as a single line of space-separated integers. Input Format The first line contains a single positive integer, , denoting the number of test cases. Each of the subsequent lines describes a test case over lines: The first line contains a single positive integer, , denoting the length of array . The second line contains positive space-separated integers describing the respective elements in integer array (i.e., ). Output Format For each test case, print the following lines: Print the length of integer array (the array representing the compressed form of binary integer ) on a new line. Print each element of as a single line of space-separated integers. It is guaranteed that a solution exists.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
int t;
scanf("%d", &t);
for(int num_test_cases = 0; num_test_cases < t; num_test_cases++)
{
int n;
scanf("%d", &n);
unsigned long long a[n];
for(int i = 0; i < n; i++)
{
scanf("%llu", &a[i]);
}
if((n == 1) && (a[0] == 1))
{
printf("2\n1 1\n");
continue;
}
int size = n + 2;
unsigned long long b[size];
if(n % 2) //odd
{
size = n + 2;
for(int i = 0; i < n - 2; i++)
{
b[i] = a[i];
}
b[n - 2] = a[n - 2] - 1;
b[n - 1] = 1;
b[n] = 1;
b[n + 1] = a[n - 1] - 1;
} else { //even
size = n + 1;
for(int i = 0; i < n - 3; i++)
{
b[i] = a[i];
}
b[n - 3] = a[n - 3] - 1;
b[n - 2] = 1;
b[n - 1] = a[n - 1] + 1;
b[n] = a[n - 2] - 1;
}
int zero_index = -1;
//check for 0's
if(b[size - 1] == 0)
{
size--;
}
for(int i = 0; i < size; i++)
{
if(b[i] == 0)
{
zero_index = i; //note where that 0 is
size = size - 2;
break;
}
}
printf("%d\n", size);
for(int i = 0; i < size; i++)
{
if(zero_index == -1)
{
printf("%llu", b[i]);
} else {
if(i == zero_index - 1)
{
printf("%llu", b[i] + 1);
} else if((i == zero_index) || (i == zero_index + 1)) {
size++;
continue;
} else {
printf("%llu", b[i]);
}
}
if(i < size - 1)
{
printf(" ");
}
}
printf("\n");
}
return 0;
}
Solution in C++ :
In C++ :
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
int T;
scanf("%d", &T);
for (int testcase = 0; testcase < T; testcase++)
{
int n;
scanf("%d", &n);
vector<long long> dat(n);
for (int i = 0; i < n; i++) {
scanf("%lld", &dat[i]);
}
vector<long long> ans;
if (n == 1) {
ans.push_back(1);
ans.push_back(1);
ans.push_back(dat[0] - 1);
}
else if (n == 2) {
ans.push_back(1);
ans.push_back(1 + dat[1]);
ans.push_back(dat[0] - 1);
}
else if (n % 2 == 0) {
for (int i = 0; i < n - 3; i++) {
ans.push_back(dat[i]);
}
ans.push_back(dat[n - 3] - 1);
ans.push_back(1);
ans.push_back(1 + dat[n - 1]);
ans.push_back(dat[n - 2] - 1);
}
else {
for (int i = 0; i < n - 2; i++) {
ans.push_back(dat[i]);
}
ans.push_back(dat[n - 2] - 1);
ans.push_back(1);
ans.push_back(1);
ans.push_back(dat[n - 1] - 1);
}
for (;;)
{
vector<long long> res;
for (int i = 0; i < ans.size(); i++)
{
long long v = ans[i];
if (v == 0) {
if (i + 1 < ans.size()) {
res.back() += ans[i + 1];
i++;
}
continue;
}
res.push_back(v);
}
if (res == ans)
break;
ans = res;
}
printf("%d\n", (int)ans.size());
for (int i = 0; i < ans.size(); i++) {
printf("%lld%c", ans[i], " \n"[i + 1 == ans.size()]);
}
}
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
int n = sc.nextInt();
long[] a = new long[n];
for (int j = 0; j < n; j++) {
a[j] = sc.nextLong();
}
if (n==1) {
if (a[0]==1) {
System.out.println(2);
System.out.println("1 1");
} else {
System.out.println(3);
System.out.println("1 1 "+(a[0]-1));
}
continue;
} else if (n==2) {
if (a[0]==1) {
System.out.println(2);
System.out.println("1 "+(a[1]+1));
} else {
System.out.println(3);
System.out.println("1 "+(a[1]+1)+" "+(a[0]-1));
}
continue;
}
int last = (n-1)/2*2;
ArrayList<Long> al = new ArrayList<Long>();
for (int j = 0; j < last-1; j++) {
al.add(a[j]);
}
al.add(a[last-1]-1);
al.add(1l);
if (last < n-1)
al.add(1l+a[n-1]);
else
al.add(1l);
al.add(a[last]-1);
for (int j = 0; j < al.size(); j++) {
if (al.get(j)==0) {
long repnum = al.remove(j-1);
al.remove(j-1);
if (j-1<al.size())
repnum += al.remove(j-1);
al.add(j-1,repnum);
}
}
System.out.println(al.size());
for (int j = 0; j < al.size(); j++) {
if (j > 0)
System.out.print(" ");
System.out.print(al.get(j));
}
System.out.println();
}
}
}
Solution in Python :
In Python3 :
T = int(input())
for t in range(T):
N = int(input())
A = [int(x) for x in input().split()]
if N == 1:
A = [1, 1, A[0] - 1]
elif N == 2:
A = [1, A[1] + 1, A[0] - 1]
elif N & 1:
A.append(1)
A.append(A[-2] - 1)
A[-4] -= 1
A[-3] = 1
else:
A.append(A[-2] - 1)
A[-4] -= 1
A[-3] = 1
A[-2] += 1
while A[-1] == 0: A.pop()
for i in range(len(A)-2, 0, -1):
if A[i] == 0:
A[i-1] += A[i+1]
del A[i:i+2]
print(len(A))
print(' '.join(map(str, A)))
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