I wrote a program to print a N x N square pattern with alternate 0's and 1's. For eg. A 5 x 5 square would looks like this:
I used the following code-
#include<iostream.h>
int main()
{
int i, n;
cin >> n; //number of rows (and columns) in the n x n matrix
for(i = 1; i <= n*n; i++)
{
cout << " " << i%2;
if(i%n == 0)
cout << "\n";
}
fflush(stdin);
getchar();
return 0;
}
This code works fine for odd numbers but for even numbers it prints the same thing in each new line and not alternate pattern.For 4 it prints this-
Where am I going wrong?
In my opinion the best way to iterate over matrix is using loop in another loop.
I think this code will be helpful for you:
for(i = 0; i < n; i++) {
for (j = 1; j <= n; j++) {
cout<<" "<< (j + i) % 2;
}
cout<<"\n";
}
where n is number of rows, i and j are ints.
Try to understand why and how it works.
If you're a beginner programmer, then I suggest (no offence) not trying to be too clever with your methodology; the main reason why your code is not working is (apart from various syntax errors) a logic error - as pointed out by blauerschluessel.
Just use two loops, one for rows and one for columns:
for (int row = 1; row <= n; row++)
{
for (int col = 0; col < n; col++)
cout << " " << ((row % 2) ^ (col % 2));
cout << "\n";
}
EDIT: since you wanted a one-loop solution, a good way to do so would be to set a flip flag which handles the difference between even and odd n:
bool flip = false;
int nsq = n * n;
for (int i = 1; i <= nsq; i++)
{
cout << " " << (flip ^ (i % 2));
if (i % n == 0) {
if (n % 2 == 0) flip = !flip;
cout << "\n";
}
}
The reason that it isn't working and creating is because of your logic. To fix this you need to change what the code does. The easiest way to handle that is to think of what it does and compare that to what you want it to do. This sounds like it is for an assignment so we could give you the answer but then you would get nothing from our help so I've writen this answer to guide you to the logic of solving it yourself.
Lets start with what it does.
Currently it is going to print 0 or 1 n*n times. You have a counter named i that will increment every time starting from 0 and going to (n*n)-1. If you were to print this number i you would get the following table for n=5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
Now you currently check if the value i is odd or even i%2 and this makes the value 0 or 1. Giving you the following table
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
Now in the case of n=4 your counter i would print out to give you a table
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Now if you print out the odd or even pattern you get
1 0 1 0
1 0 1 0
1 0 1 0
1 0 1 0
This pattern diffrence is because of the changing pattern of printed numbers due to the shape of the square or more accurately matrix you are printing. To fix this you need to adjust the logic of how you determine which number to print because it will only work for matrixes that have odd widths.
You just need to add one more parameter to print the value. Below mentioned code has the updated for loop which you are using:
int num = 0;
for(i = 1; i <= n*n; i++)
{
num = !num;
std::cout << " " << num;
if(i%n == 0) {
std::cout << "\n";
num = n%2 ? num : !num;
}
}
The complete compiled code :
#include <iostream>
#include <stdio.h>
int main()
{
int i, n, num = 0;
std::cin >> n; //number of rows (and columns) in the n x n matrix
for(i = 1; i <= n*n; i++)
{
num = !num;
std::cout << " " << num;
if(i%n == 0) {
std::cout << "\n";
num = n%2 ? num : !num;
}
}
fflush(stdin);
getchar();
return 0;
}
Related
I don't think you'll need to know the context of the problem to answer this question, but I'll give it just in case.
-In the past N weeks, we've measured the amount of rainfall every day, and noted it down for each day of the week. Return the number of the first week of the two week period where there were the most days without rain.
The code gives no warnings or errors, and if I try to print dryestweeks inside the second for loop, then it returns the correct answer. However, all of the code after the second for loop seems to be getting ignored, and I'm getting Process returned -1073741819 (0xC0000005). The issue has to lie in the 2nd for loop, because if I comment it out then both "test2" and dryestweeks get printed, and the program returns 0.
#include <iostream>
#include <vector>
#include <bits/stdc++.h>
using namespace std;
int main() {
int weeks;
cin >> weeks;
vector<int> v[weeks];
for (int i = 0;i < weeks; i++) {
int a, b, c, d, e, f, g;
cin >> a >> b >> c >> d >> e >> f >> g;
v[i].push_back(a);
v[i].push_back(b);
v[i].push_back(c);
v[i].push_back(d);
v[i].push_back(e);
v[i].push_back(f);
v[i].push_back(g);
}
int mostdrydays = 0;
int dryestweeks = 0;
for (int i = 0; i < weeks; i++) {
int weeklydrydays = count(v[i].begin(), v[i].end(), 0);
int nextweekdrydays = count(v[i+1].begin(), v[i+1].end(), 0);
int biweeklydrydays=weeklydrydays+nextweekdrydays;
if (biweeklydrydays > mostdrydays) {
mostdrydays = biweeklydrydays;
dryestweeks = i + 1;
}
}
cout << "test2" << endl;
cout << dryestweeks << endl;
return 0;
}
An example of an input would be:
6
5 10 15 20 25 30 35
0 2 0 0 0 0 0
0 0 0 1 0 3 0
0 1 2 3 4 5 6
5 1 0 0 2 1 0
0 0 0 0 0 0 0
The program should print "2" with the above input.
The second loop has an overflow.
You first defined v[weeks] and then the second loop goes from [0, weeks[ but you are retrieving the next week with v[i + 1]. I don't know exactly what are you are trying to achieve, but if you do
for(int i = 0; i < weeks - 1; i++)
{
...
}
it executes properly.
For the given example of input, in the last iteration (i = 5) of the second loop, index i + 1(=6) will be out of the bound for v[i + 1] (legal indices for v will be from 0 to 5).
The second loop is iterating one more time than required.
I am getting a SIGSEV fault while executing the program. If the numbers were listed in increasing order of their index then the group A would have every other one starting with the first number in the list and the group B would have every other one starting with the second number on in the list. However, certain numbers, as specified in the second line of each case are to be skipped. Then the numbers of the two groups are to be printed out.
The first line contains a single integer T ≤ 50 indicating the number of test cases to follow. Each test case consists of two lines. The first line contains two numbers n,m satisfying 0 ≤ m ≤ n ≤ 1000. Here, n is the highest number till which we run and m is the number the values to be skipped. The second line contains a list of m distinct integers between 1 and n. These are the number to be skipped. Consecutive integers are separated by a single space.
Eg:-
Input:
3
6 3
2 4 1
3 2
3 2
8 2
3 8
Output:
3 6
5
1
1 4 6
2 5 7
My code runs this case perfectly, and other cases where it accesses the end indexes too, but the website is not accepting the solution, with the specified fault appearing.
Please note that here I am trying to solve a practice coding site's question, hence the lack of checks, and that I am a beginner, so I don't have information regarding concepts like vectors, etc.
The compiler is GCC14
Thanks in advance if you read up this huge question and do answer!
#include <iostream>
using namespace std;
int main()
{
long cases, tasks, dones, t_done, i, k = 1;
cin >> cases;
for( int j = 0 ; j < cases ; j++ )
{
k = 1;
cin >> tasks >> dones;
int task[tasks] = {0} ;
for( i = 0 ; i < dones ; i++ )
{
cin >> t_done ;
task[ t_done - 1 ] = 1 ;
}
while(k <= tasks)
{
for( ; k <= tasks; k++)
{
if( task[k-1] != 1)
{
task[k-1] = 2;
break ;
}
}
k++;
for( ; k <= tasks; k++)
{
if( task[k-1] != 1)
{
task[k-1] = 3;
break ;
}
}
k++;
}
for( i = 1 ; i <= tasks ; i++)
{
if(task[i-1] == 2)
cout << i << " ";
}
cout << "\n" ;
for( i = 1 ; i <= tasks ; i++)
{
if(task[i-1] == 3)
cout << i << " ";
}
cout << "\n" ;
}
return 0 ;
}
So I N - numbers I have to input, and I got M - numbers of places for those numbers and I need to find all combinations with repetition of given numbers.
Here is example:
Let's say that N is 3(I Have to input 3 numbers), and M is 4.
For example let's input numbers: 6 11 and 533.
This should be result
6,6,6,6
6,6,6,11
6,6,6,533
6,6,11,6
...
533,533,533,533
I know how to do that manualy when I know how much is N and M:
In example where N is 3 and M is 4:
int main()
{
int N = 3;
int M = 4;
int *numbers = new int[N + 1];
for (int i = 0; i < N; i++)
cin >> numbers[i];
for (int a = 0; a < N; a++)
for (int b = 0; b < N; b++)
for (int c = 0; c < N; c++)
for (int d = 0; d < N; d++)
{
cout << numbers[a] << " " << numbers[b] << " " << numbers[c] << " " << numbers[d] << endl;
}
return 0;
}
But how can I make algorithm so I can enter N and M via std::cin and I get correct resut?
Thanks.
First one short tip: don't use "new" or C-style arrays in C++ when we have RAII and much faster data structures.
For the solution to your problem I would suggest making separate function with recursion. You said you know how to do it manually so the first step in making it into algorithm is to tear down you manual solution step by step. For this problem when you solve it by hand you basically start with array of all first numbers and then for last position you just loop through available numbers. Then you go to the second last position and again loop through available numbers just now with the difference that for every number there you must also repeat the last spot number loop. Here is the recursion. For every "n"th position you must loop through available numbers and for every call the same function for "n+1"th number.
Here is a simplified solution, leaving out the input handling and exact print to keep code shorter and more focused on the problem:
#include <vector>
#include <iostream>
void printCombinations(const std::vector<int>& numbers, unsigned size, std::vector<int>& line) {
for (unsigned i = 0; i < numbers.size(); i++) {
line.push_back(numbers[i]);
if (size <= 1) { // Condition that prevents infinite loop in recursion
for (const auto& j : line)
std::cout << j << ","; // Simplified print to keep code shorter
std::cout << std::endl;
line.erase(line.end() - 1);
} else {
printCombinations(numbers, size - 1, line); // Recursion happens here
line.erase(line.end() - 1);
}
}
}
int main() {
std::vector<int> numbers = {6, 11, 533};
unsigned size = 4;
std::vector<int> line;
printCombinations(numbers, size, line);
return 0;
}
If you have any questions feel free to ask.
Totally there is no need for recursion here. This is a typical job for dynamic programming. Just get the first solution right for n = 1 (1 slot is available) which means the answer is [[6],[11],[533]] and then move on one by one by relying on the one previously memoized solution.
Sorry that i am not fluent in C, yet in JS this is the solution. I hope it helps.
function combosOfN(a,n){
var res = {};
for(var i = 1; i <= n; i++) res[i] = res[i-1] ? res[i-1].reduce((r,e) => r.concat(a.map(n => e.concat(n))),[])
: a.map(e => [e]);
return res[n];
}
var arr = [6,11,533],
n = 4;
console.log(JSON.stringify(combosOfN(arr,n)));
Normally the easiest way to do dynamic nested for loops is to create your own stack and use recursion.
#include <iostream>
#include <vector>
void printCombinations(int sampleCount, const std::vector<int>& options, std::vector<int>& numbersToPrint) {
if (numbersToPrint.size() == sampleCount) {
// got all the numbers we need, print them.
for (int number : numbersToPrint) {
std::cout << number << " ";
}
std::cout << "\n";
}
else {
// Add a new number, iterate over all possibilities
numbersToPrint.push_back(0);
for (int number : options) {
numbersToPrint.back() = number;
printCombinations(sampleCount, options, numbersToPrint);
}
numbersToPrint.pop_back();
}
}
void printCombinations(int sampleCount, const std::vector<int>& options) {
std::vector<int> stack;
printCombinations(sampleCount, options, stack);
}
int main()
{
printCombinations(3, {1,2,3});
}
output
1 1 1
1 1 2
1 1 3
1 2 1
1 2 2
1 2 3
1 3 1
1 3 2
1 3 3
2 1 1
2 1 2
2 1 3
2 2 1
2 2 2
2 2 3
2 3 1
2 3 2
2 3 3
3 1 1
3 1 2
3 1 3
3 2 1
3 2 2
3 2 3
3 3 1
3 3 2
3 3 3
Here is an algorithm to solve this, that does't use recursion.
Let's say n=2 and m=3. Consider the following sequence that corresponds to these values:
000
001
010
011
100
101
110
111
The meaning of this is that when you see a 0 you take the first number, and when you see a 1 you take the second number. So given the input numbers [5, 7], then 000 = 555, 001=557, 010=575 etc.
The sequence above looks identical to representing numbers from 0 to 7 in base 2. Basically, if you go from 0 to 7 and represent the numbers in base 2, you have the sequence above.
If you take n=3, m=4 then you need to work in base 3:
0000
0001
0002
0010
0011
0012
....
So you go over all the numbers from 0 to 63 (4^3-1), represent them in base 3 and follow the coding: 0 = first number, 1 = second number, 2 = third number and 3 = fourth number.
For the general case, you go from 0 to M^N-1, represent each number in base N, and apply the coding 0 = first number, etc.
Here is some sample code:
#include <stdio.h>
#include <math.h>
void convert_to_base(int number, char result[], int base, int number_of_digits) {
for (int i = number_of_digits - 1; i >= 0; i--) {
int remainder = number % base;
number = number / base;
result[i] = '0' + remainder;
}
}
int main() {
int n = 2, m = 3;
int num = pow(n, m) - 1;
for (int i = 0; i <= num; i++) {
char str[33];
convert_to_base(i, str, n, m);
printf("%s\n", str);
}
return 0;
}
Output:
000
001
010
011
100
101
110
111
I want to ask your help, I want to display a two dimensional array in c++, whose every second two rows are changing periodically between 0 and 1.
For example n = 5, m = 4 the program will write this:
0 0 0 0
0 0 0 0
1 1 1 1
1 1 1 1
0 0 0 0
I'm still a beginner in C++. I tried it myself to see if I could do it.
The only thing you need to know is, when to assign 1 to your row and when to assign 0.
These are the the rows where you have to give every number the value 0:
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
again:
0,1,4,5,8,9,12,13,16,17
If you look at them like this, they don't make much sense. But if you divide them into two lists like this:
0,4,8,12,16
and
1,5,9,13,17
you can see that the first list consist of numbers, that are the result of n * 4, and the second list is the result of n * 4 + 1
So you just have to make an if-statement where the requirement is
(i%4 == 0 || i%4 == 1)
Here is my complete code. It's probably not the best solution:
#include <iostream>
using namespace std;
int main()
{
const int n = 20, m = 4;
int array[n][m];
int i = 0, j = 0;
while ( i < n)
{
if (j < 4)
{
if (i%4 == 0 || i%4 == 1)
array[i][j] = 0;
else
array[i][j] = 1;
cout << array[i][j] << ", ";
j++;
}
else {
j = 0;
i++;
cout << endl;
}
}
}
I tried solving this interview question. My code runs for test cases but fails for all the real input test cases. I tried hard to find the mistake but unable to do so. Please find my code below the question
Bob loves sorting very much. He is always thinking of new ways to sort an array.His friend Ram gives him a challenging task. He gives Bob an array and an integer K. The challenge is to produce the lexicographical minimal array after at most K-swaps. Only consecutive pairs of elements can be swapped. Help Bob in returning the lexicographical minimal array possible after at most K-swaps.
Input: The first line contains an integer T i.e. the number of Test cases. T test cases follow. Each test case has 2 lines. The first line contains N(number of elements in array) and K(number of swaps). The second line contains n integers of the array.
Output: Print the lexicographical minimal array.
Constraints:
1 <= T <= l0
1 <= N,K <= 1000
1 <= A[i] <= 1000000
Sample Input (Plaintext Link)
2
3 2
5 3 1
5 3
8 9 11 2 1
Sample Output (Plaintext Link)
1 5 3
2 8 9 11 1
Explanation
After swap 1:
5 1 3
After swap 2:
1 5 3
{1,5,3} is lexicographically minimal than {5,1,3}
Example 2:
Swap 1: 8 9 2 11 1
Swap 2: 8 2 9 11 1
Swap 3: 2 8 9 11 1
#include <iostream>
using namespace std;
void trySwap(int a[], int start, int end, int *rem)
{
//cout << start << " " << end << " " << *rem << endl;
while (*rem != 0 && end > start)
{
if (a[end - 1] > a[end])
{
swap(a[end - 1], a[end]);
(*rem)--;
end--;
}
else end--;
}
}
void getMinimalLexicographicArray(int a[], int size, int k)
{
int start , rem = k, window = k;
for (start = 0; start < size; start++)
{
window = rem;
if (rem == 0)
return;
else
{
//cout << start << " " << rem << endl;
int end = start + window;
if (end >= size)
{
end = size - 1;
}
trySwap(a, start, end, &rem);
}
}
}
int main()
{
int T, N, K;
int a[1000];
int i, j;
cin >> T;
for (i = 0; i < T; i++)
{
cin >> N;
cin >> K;
for (j = 0; j < N; j++)
{
cin >> a[j];
}
getMinimalLexicographicArray(a, N, K);
for (j = 0; j < N; j++)
cout << a[j] << " ";
cout << endl;
}
return 0;
}
Python solution can be easily translated to C++:
def findMinArray(arr, k):
i = 0
n = len(arr)
while k > 0 and i < n:
min_idx = i
hi = min(n, i + k + 1)
for j in range(i, hi):
if arr[j] < arr[min_idx]:
min_idx = j
for j in range(min_idx, i, -1):
arr[j - 1], arr[j] = arr[j], arr[j - 1]
k -= min_idx - i
i += 1
return arr
Here are two failed test cases.
2
2 2
2 1
5 3
3 2 1 5 4
In the first, your code makes no swaps, because K >= N. In the second, your code swaps 5 and 4 when it should spend its third swap on 3 and 2.
EDIT: the new version is still too greedy. The correct output for
1
10 10
5 4 3 2 1 10 9 8 7 6
is
1 2 3 4 5 10 9 8 7 6
.