Suppose I have a loop:
for(int i=1; i<=1024; i++)
I want to fill a file with 128 columns (not rows!), so the first column contains numbers from 1 to 8, second from 9 to 16 and so on and so forth.
The items for each columns are just 8i to 8i+7. You can write mutiple loops.for(int i = 0; i< 128;i++)
for(int j = 0; j <8;j++)
int k = 8* i+ j;
void write_in_file( ofstream &fout, int start){
for(int i = 1; i <= 128; i++){
fout<<start <<"\t";
start+=8;
}
fout<<"\n";
}
int main(){
ofstream fout;
fout.open("out.txt");
for(int i=1;i<=8;i++){
write_in_file(fout,i);
}
}
Explanation:
As we need 8 rows so we call function write_in_file 8 times. Each time function write_in_file writes 128 entries in the file.
The simplest way is to make two loops - the first by lines and nested by columns - and then calculate numbers with easy math expression.
E.g.:
for(int line = 0; line < 8; line++)
{
for( int col = 0; col < 128; col++)
{
cout << setw(5) << line + col * 8 + 1;
}
cout << endl;
}
setw() has parameter 5 to make identical width for colums with numbers from 1 to 1024
EDITED:
If you want to use only one for loop and exact as you give in the question, you can use more complicated math expressions.
First, let's calculate number of line as (i-1) / 128 (numbers start from 0) and number of column as (i-1) % 128 (numbers from 0 to 127). And now you can make the following loop with additional new-line-conditional output:
for(int i=1; i<=1024; i++)
{
cout << setw(5) << ( 1 + 8 * ((i-1) % 128) ) + ( (i-1) / 128 );
if( i % 128 == 0 ) // new line once per 128 numbers
cout << endl;
}
Of course if you want to make file, you should do something with the output - redirect standard output to file or change cout to other stream.
Related
Okay so I'm tryna create a program that:
(1) swaps my array
(2) performs caesar cipher substitution on the swapped array
(3) convert the array from (2) that is in decimal form into 8-bit binary form
And so far I've successfully done the first 2 parts but I'm facing problem with converting the array from decimal to binary form.
And this is my coding of what I've tried
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
void swapfrontback(int a[], int n);
int main()
{
int a[10], i, n;
cout << "enter size" << endl;
cin >> n;
if (n == 0)
{
cout << "Array is empty!\n";
}
else
{
cout << "p = " << endl;
for (i = 0; i < n; i++)
{
cin >> a[i];
}
}
swapfrontback(a,n);
//caesar cipher
int shift = 0;
cout << "input shift: ";
cin >> shift;
int modulus = 0;
cout << "input modulus: ";
cin >> modulus;
cout << "p''=" << endl;
for (i = 0; i < n; i++)
{
a[i] = (a[i] + shift) % modulus;
cout << a[i] << endl;
}
// Function that convert Decimal to binary
int b;
b = 8;
cout<< "p'''=" << endl;
for (i = 0; i < n; i++)
{
for(int i=b-1;i>=0;i--)
{
if( a[i] & ( 1 << i ) ) cout<<1;
else cout<<0;
}
}
return 0;
}
void swapfrontback(int a[], int n)
{
int i, temp;
for (i = 0; i < n / 2; i++)
{
temp = a[i];
a[i] = a[n - i-1];
a[n - i-1] = temp;
}
cout << "p' = '" << endl;
for (i = 0; i < n; i++)
{
cout << a[i] << endl;
}
}
the problem is that instead of converting the array of decimal from the 2nd part which is the caesar cipher into its binary form, I'm getting 000000010000000100000001 .
My initial array is
3
18
25
Shift 8 and modulo 26. If anyone knows how to fix this please do help me.
Well, there seems to be something that may be an issue in the future (like the n being larger than 10, but, regarding your question, this nested for sentence is wrong.
for (i = 0; i < n; i++)
{
for(int i=b-1;i>=0;i--) //here you are using the variable 'i' twice
{
if( a[i] & ( 1 << i ) ) cout<<1; //i starts at 7, which binary representation in 4 bits is 0111
else cout<<0;
}
}
When you're using nested for sentences, it is a good idea to not repeat their iterating variables' names since they can affect each other and create nasty things like infinite loops or something like that. Try to use a different variable name instead to avoid confusion and issues:
for(int j=b-1;j>=0;j--) //this is an example
Finally, the idea behind transforming a base 10 number to its binary representation (is to use the & operator with the number 1 to know if a given bit position is a 1 (true) or 0 (false)) for example, imagine that you want to convert 14 to its binary form (00001110), the idea is to start making the & operation with the number 1, an continue with powers of 2 (since them will always be a number with a single 1 and trailing 0s) 1-1 2-10 4-100 8-1000, etc.
So you start with j = 1 and you apply the & operation between it and your number (14 in this case) so: 00000001 & 00001110 is 0 because there is not a given index in which both numbers have a '1' bit in common, so the first bit of 14 is 0, then you either multiply j by two (j*=2), or shift their bits to the left once (j = 1<<j) to move your bit one position to the left, now j = 2 (00000010), and 2 & 14 is 2 because they both have the second bit at '1', so, since the result is not 0, we know that the second bit of 14 is '1', the algorithm is something like:
int j = 128; 128 because this is the number with a '1' in the 8th bit (your 8 bit limit)
int mynumber = 14;
while(j){ // when the j value is 0, it will be the same as false
if(mynumber & j) cout<<1;
else cout<<0;
j=j>>1;
}
Hope you understand, please ensure that your numbers fit in 8 bits (255 max).
I am trying to add elements to my file using a method I have already used and was proven to be successful, however now when I do it I get the numbers I want as well as a bunch of other numbers that aren't in my file and don't make any sense
const int MAX_SIZE = 21;
int readSquare(int square[MAX_SIZE][MAX_SIZE], string inputFileName){ //reads file into an array
int value;
ifstream inFile;
inFile.open(inputFileName);
if (inFile) //if the input file to be read open successfully then goes on
{
int temp;
inFile >> temp;
if (temp>21) {
temp=21;
}
for (int i = 0; i < MAX_SIZE; i++)
{
for(int j = 0; j < MAX_SIZE; j++)
{
inFile >> square[i][j];
}
}
} else {
inFile.close();
return 0; //returns 0 if couldnt open file
}
inFile.close();
cout << "Magic square" << endl;
for(int i=0;i<MAX_SIZE;i++)
{
for(int j=0;j<MAX_SIZE;j++)
{
cout << square[i][j] << " ";
}
cout<<endl;
}
return 1;
}
This is the file I am using on my code
3
4 9 2
3 5 7
8 1 6
And this is the result I get(goes on for a while but I only took the top portion)
4 9 2 3 5 7 8 1 6 16840768 6619136 6643024 23198772 0 1942212500 127 917504 6643024 786434 6643032 0
65536 30 0 31 0 13930549 30 593 6619744 6619744 -2 127 46 6420808 1997546816 -1759127226 -2 6420704 1997359545 4096 4104
0 6420680 6634144 6619136 6421232 4104 6619744 0 3 0 4096 6420732 1997535944 6420804 655612 655360 2 9 0 2 6420976
0 1997378284 6420976 663276 1952 229640288 663200 655360 0 1997377793 6421060 661336 9 16777596 0 13080 236 661336 2 16777596 -530786634
Hat tip to #melpomene for working through the details in the main comments.
Op, you're iterating over the entire range of the array regardless of the availability of input data. I suggest you do the following so the results are less random in appearance:
initialize the values in the 2D array to zero.
limit the input samples to the quantity you're expecting, and not to exceed the size of the array in either dimension.
In your post, you're showing the value of 3 in the first line of the input file. What does that mean -- 3 lines, 3 samples, or 3 samples for each of the 3 lines?
Since the input file has 3 samples per line, I'm guessing the initial value in the data file represents the samples per line where the values for each line are assigned to an individual inner array.
Without deviating too much from your post, consider the following:
// clear the array for easier diags
for (int n = 0; n < MAX_SIZE; n++)
for (int m = 0; m < MAX_SIZE; m++)
square[n][m] = 0;
int cols;
inFile >> cols; // first line of data file indicating the samples in each row
if (cols > MAX_SIZE) // don't exceed the size of the inner array
cols = MAX_SIZE;
for (int i = 0; i < MAX_SIZE; i++)
{
for(int j = 0; j < cols; j++)
{
if (!(inFile >> square[i][j])) // read until EOF
{
i = MAX_SIZE; // force outer loop to terminate since break only affects the inner loop.
break;
}
}
}
See How does ifstream's eof() work?
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;
}
#include <iostream>
using namespace std;
void bi(int a);
int main()
{
// here is the issue how do start a loop, where i want the answer from 16 to 31 numbers
int a=0;
cout<<"Baum-Sweet Sequence From 16 to 31 \n";
for(int j=a;j>16 && j<31;j++)
{
cout<<j;
}
bi(a);
system("Pause");
}
// Rest is working properly
void bi(int a)
{
int myArr[15],i=0,f=0,n=0;
for (int h = 0 ; h <= a; h++)
{
int num = h;
for (i = 0 ; i < 4 ; i++)
{
myArr[i] = num%2;
num = num/2;
}
for (int t = 0 ; t < 4 ; t++)
{
if (myArr[t]%2==0)
f++;
}
if (f%2==0)
cout << " = " << 1;
else
cout << " = " << 0;
cout <<endl;
}
}
i want to show the sequence from 16 to 31 decimal number but its not showing :\ could anyone help me out here
There is an error in the for loop.
The for loop has three parts separated by a semicolon.
for (INITIALIZATION; CONDITION; AFTERTHOUGHT)
{
// Source code for the for-loop's body
}
The first part initializes the variable (e.g. "int j = 16;" means that through the variable j you begin counting by 16);
The second part checks a condition and it quits the loop when false (e.g. j <=31 means that it quits the loop when j will have value 31);
The third one is performed once each time the loop ends and then repeats (e.g. j++ means that at each iteration of the loop j will be incremented by 1).
Each iteration will execute the code in the body of the for loop.
Considering that you want to call the bi function for each value from 16 to 31 your for loop body should include bi(j). Your main should be modified like the code below:
int main()
{
cout<<"Baum-Sweet Sequence From 16 to 31 \n";
for(int j=16;j<=31;j++)
{
cout<<j;
bi(j);
}
system("Pause");
return 0;
}
Your problem is that you set j to 0, but then make a condition for the loop that it will only execute if j (which is set to a), is bigger than 16.
Your first thing to do is to make the loop conditions this:
for (int j = 16; j <= 32; j++)
This is a part of my question.I tried many times but couldn't get the answer
Problem Statement
You are given a list of N people who are attending ACM-ICPC World Finals. Each of them are either well versed in a topic or they are not. Find out the maximum number of topics a 2-person team can know. And also find out how many teams can know that maximum number of topics.
Note Suppose a, b, and c are three different people, then (a,b) and (b,c) are counted as two different teams.
Input Format
The first line contains two integers, N and M, separated by a single space, where N represents the number of people, and M represents the number of topics. N lines follow.
Each line contains a binary string of length M. If the ith line's jth character is 1, then the ith person knows the jth topic; otherwise, he doesn't know the topic.
Constraints
2≤N≤500
1≤M≤500
Output Format
On the first line, print the maximum number of topics a 2-person team can know.
On the second line, print the number of 2-person teams that can know the maximum number of topics.
Sample Input
4 5
10101
11100
11010
00101
Sample Output
5
2
Explanation
(1, 3) and (3, 4) know all the 5 topics. So the maximal topics a 2-person team knows is 5, and only 2 teams can achieve this.
this is a a part of my work.Any clue how can i get this to work
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n, m, max = 0, max1 = 0, count = 0;
cin >> n >> m; //for input of N and M
int a[n][m];
for (int i = 0; i<n; i++) //for input of N integers of digit size M
for (int j = 0; j<m; j + >>
cin >> a[i][j];
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
max = 0;
for (int k = 0; k<m; k++)
{
if (a[i][k] == 1 || a[j][k] == 1) max++;
cout << k;
if (k = m - 1 && max>max1) max1 = max;
if (k == m - 1 && max == max1) count++;;
}
}
}
cout << max1 << endl << count;
return 0;
}
I think the way of taking my input logic is wrong.could you please help me out.I am stuck in this question from 5 days.
PLease only help me on how should i take input and how to read the digit of integer.
Don't have a compiler with me so there's probably a syntax boner or two in there, but the logic walks through on paper.
Builds the storage:
std::cin >> n >> m; //for input of N and M
std::vector<std::vector<bool>>list(n,std::vector<bool>(m, false));
Loads the storage:
char temp;
for (int i = 0; i < n; i++) //for input of N integers of digit size M
{
for (int j = 0; j < m; j++)
{
std::cin >> temp;
if (temp == 1)
{
list[i][j] = true;
}
}
}
Runs the algorithm
for (int a = 0; a < n; a++)
{
for (int b = a+1; b < n; b++)
{
int knowcount = 0;
for (int j = 0; j < m; j++)
{
if (list[a][j] | list[b][j])
{
knowcount ++;
}
}
if (knowcount > max)
{
groupcount = 1;
max = know;
}
else if(knowcount == max)
{
groupcount ++;
}
}
}
Your method of input is wrong. According to your method, the input will have to be given like this (with spaces between individual numbers):
1 0 1 0 1
1 1 1 0 0
1 1 0 1 0
0 0 1 0 1
Only then it makes sense to create a matrix. But since the format in the question does not contain any space between a number in the same row, thus this method will fail. Taking into consideration the test case, you might be tempted to store the 'N' numbers in a single dimensional integer array, but keep in mind the constraints ('M' can be as big as 500 and int or even unsigned long long int data type cannot store such a big number).