I want to know how the modulo operator is used here
Input and output are as follows:
Input : 6
1 2 3 4 5 6
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 0
6 7 8 9 0 1
#include <stdio.h>
int main(){
int i, j, n;
printf("Input n: ");
scanf("%d", &n);
for(i=0; i<n; i++)
{
for (j=1; j<=n; j++)
{
printf ("%d ", (i+j)%10);
}
printf("\n");
}
return 0;
}
The expression (i+j)%10 calculates i+j modulo 10, which means taking the last digit of the sum of i and j.
Related
The question is:
Given a 2-D array A of size M x N . Write a program in C to print A in spiral form starting from bottom right corner in anti-clockwise direction. There must be space after each element. Note the following
1<=M
N<=20
0<=Aij<=1000
Example:
Input
1 8
1 2 3 4 5 6 7 8
Output
8 7 6 5 4 3 2 1
Input
3 3
1 2 3
4 5 6
7 8 9
Output
9 6 3 2 1 4 7 8 5
I have tried it....can anyone help me with this.
#include<stdio.h>
int main()
{
int m,n;
scanf("%d%d",&m,&n) ;
int matrix[m][n] ;
int last_row = (m-1) ;
int last_column = (n-1) ;
int first_row = 0;
int first_column = 0;
int i=0,j=0,r=0,c=0;
while(i<m)
{
while(j<n)
{
scanf("%d",&matrix[i][j]) ;
++j;
} j=0;
++i;
}
while(first_column <=last_column && first_row <= last_row)
{ for( r=last_row ; r>=first_row ; --r)
{ c=last_column;
printf("%d ",matrix[r][c]) ;
} --last_column;
for (c=last_column; c>=first_column ; --c)
{
r=first_row;
printf("%d ",matrix[r][c]) ;
} ++first_row ;
for(r=first_row; r<=last_row; r++)
{
c=first_column;
printf("%d ",matrix[r][c]) ;
} ++first_column;
for(c=first_column ; c<=last_column ; c++)
{
r=last_row;
printf("%d ",matrix[r][c]) ;
} --last_row;
}
return 0;
}
The current output which I am getting is :
Input
8 1
1
2
3
4
5
6
7
8
Output
8 7 6 5 4 3 2 1 2 3 4 5 6 7 8
Input
1 8
1 2 3 4 5 6 7 8
Output
8 7 6 5 4 3 2 1 2 3 4 5 6 7
I just worked to my final exam with simple codes; when I try to sorting strings, I face annoying error. Why 2 is not smaller than 10 on my CodeBlocks IDE but is smaller than 10 on real and onlinegdb.com?
This is the annoying code:
#include <string.h>
#include <stdio.h>
#define STR_SIZ 20
int main()
{
char strArr[][STR_SIZ] = {"abc", "hdas", "sdfasf", "kakldf", "caksl", "casd", "keam", "cznjcx", "mnxzv", "jkalkds"};
char minStr[STR_SIZ];
strcpy(minStr, strArr[0]);
int N = sizeof(strArr)/sizeof(minStr);
// int N = 10;
for(int x = 0; x < N-1; x++)
{
printf("%d", x);
strcpy(minStr,strArr[x]);
int j;
for(j=1+x; j < 10; j++)
{
printf("%4d\n", j);
int cmp = strcmp(strArr[j], minStr);
if(cmp < 0)
strcpy(minStr,strArr[j]);
}
char temp[STR_SIZ];
strcpy(temp,strArr[x]);
strcpy(strArr[x], minStr);
strcpy(strArr[j], temp);
}
return 0;
}
Output on onlinegdb.com:
0 1
2
3
4
5
6
7
8
9
1 2
3
4
5
6
7
8
9
2 3
4
5
6
7
8
9
3 4
5
6
7
8
9
4 5
6
7
8
9
5 6
7
8
9
6 7
8
9
7 8
9
8 9
Output on CodeBlocks:
0 1
2
3
4
5
6
7
8
9
1 2
3
4
5
6
7
8
9
2
PS: I just have used Codeblock in the morning and it was okey with executing.
strArr has 10 elements. At the end of your loop, you call strcpy(strArr[j], temp);. This will write to strArr[10], which is out of bounds and will overwrite some unknown memory. Anything can happen after that.
You should save the j value when you copy a string into minStr.
FYI, your code above prints this as your final string order with onlinegdb:
abc
caksl
caksl
caksl
caksl
casd
cznjcx
cznjcx
jkalkds
jkalkds
So I think you have other problems as well.
try this
#include <string.h>
#include <stdio.h>
#define STR_SIZ 20
int main()
{
char strArr[][STR_SIZ] = {"abc", "hdas", "sdfasf", "kakldf", "caksl", "casd", "keam", "cznjcx", "mnxzv", "jkalkds"};
strcpy(minStr, strArr[0]);
// Calculate the number of elements this way.
const int N = sizeof(strArr)/sizeof(strArr[0]);
// int N = 10;
for(int x = 0; x < N-1; x++)
{
printf("%d", x);
int j;
for(j=1+x; j < N; j++) // Use N here too!
{
printf("%4d\n", j);
int cmp = strcmp(strArr[j], strArr[x]);
if(cmp < 0)
{
// Do the swaps only when needed.
char temp[STR_SIZ];
strcpy(temp,strArr[x]);
strcpy(strArr[x], strArr[j]);
strcpy(strArr[j], temp);
}
}
}
// Verify result
for(int x = 0; x < N; x++) printf("%s\n", strArr[x]);
return 0;
}
I moved your swap into your if check and got rid of your minStr as it was not needed. Notice how I calculate the N size too. Honestly, you were close, but you needed to verify your output.
Hello a beginner here who needs some of your help. My C program is good and does what it is supposed to do only that it is not supposed to use any kind of if statements. I wrote it that way as I saw it would be easier so that I can then replace the if statements. I have been trying to replace the if statements but am now stuck. What can I use instead of the if statement to still produce the same output.
The program is supposed to generate a sequence of thirty random integers between 0 and 9 and then print out the sequence both forward and backwards. Then print out a count of how many times each number between 0 and 9 appeared in the sequence.
This is the output
Here is a sequence of 30 random numbers between 0 and 9:
3 6 7 5 3 5 6 2 9 1 2 7 0 9 3 6 0 6 2 6 1 8 7 9 2 0 2 3 7 5
Printing them backwards, that's:
5 7 3 2 0 2 9 7 8 1 6 2 6 0 6 3 9 0 7 2 1 9 2 6 5 3 5 7 6 3
There were 3 0's
There were 2 1's
There were 5 2's
There were 4 3's
There were no 4's
There were 3 5's
There were 5 6's
There were 4 7's
There was only 1 8
There were 3 9's
This is my C program
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i, j, array[30]={0}, count=0,check;
srand(time(NULL));
for(i=0;i<30;i++)
array[i]=rand()%10;
for(i=0;i<30;i++)
printf("%d ",array[i]);
printf("\n\n");
for(i=29;i>=0;i--)
printf("%d ",array[i]);
printf("\n\n");
for(i=0;i<30;i++){
check=array[i];
if(array[i]!=-1)
array[i]=-1;
if(check == -1)
continue;
count =1;
for(j=0;j<30;j++){
if((i==j) || (array[j]==-1))
continue;
if(check==array[j]){
count++;
array[j]=-1;
}
}
printf("There were %d %d's\n",count,check);
}
return 0;
}
You'll understand the algorithm from comments:
#include <stdio.h>
#include <stdlib.h>
//time.h is needed for time()
#include <time.h>
int main()
{
int i, array[30] = {0};
srand(time(NULL));
//generate and print 30 random numbers
for(i = 0; i < 30; i++){
array[i] = rand() % 10;
printf("%d ", array[i]);
}
puts("\n\n");
//print these numbers backwards
for(i = 29; i >= 0; i--)
printf("%d ",array[i]);
puts("\n\n");
// print out a count of how many times each number
// between 0 and 9 appeared in the sequence.
int count[10] = {0};
for(i = 0; i < 30; i++)
count[array[i]]++;
//output the count for each number
for(i = 0; i < 10; i++)
printf("There were %d %d's\n",count[i], i);
return 0;
}
Output:
9 2 3 9 8 4 3 8 1 3 6 4 3 2 5 3 2 3 0 1 9 0 3 5 1 3 3 8 2 0
0 2 8 3 3 1 5 3 0 9 1 0 3 2 3 5 2 3 4 6 3 1 8 3 4 8 9 3 2 9
There were 3 0's
There were 3 1's
There were 4 2's
There were 9 3's
There were 2 4's
There were 2 5's
There were 1 6's
There were 0 7's
There were 3 8's
There were 3 9's
The following is a simplification of your original source while removing the if statements. There are implied if statements in several places where a logical expression is used as part of a source code statement.
For instance for(j = 0; j < 30 && match >= 0; j++) has several logical expressions but no if appears in this statement. The logical expressions are j < 30 and match >= 0 and the complete expression of j < 30 && match >= 0.
This example uses a logical expression and the evaluation short circuit behavior of the C compiler (see Short-circuit evaluation in Wikipedia) in the statement array[j] == match && ++count && (array[j] = -1); so that if the logical expression array[j] == match evaluates to false then the rest of the statement will not be executed.
We also depend on the preincrement operator with the ++count to increment count and then take the resulting value to check if it is false (zero) or true (non-zero). Since the variable count is initialized to zero and when incremented will always be non-zero then the next logical expression in the statement is evaluated the (array[j] = -1). We put the assignment statement within parenthesis to enforce the order of evaluation. We want the variable array[j] to be assigned the value of -1 and for the result to then be used in the logical statement. Since this is the last logical expression of the entire logical statement, whether it evaluates to false (zero) or true (non-zero) doesn't matter as what we want is the side effect of assigning the value of -1 to the array element.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int i, array[30] = {0};
srand(time(NULL));
for(i = 0; i < 30; i++)
array[i] = rand() % 10;
for(i = 0; i < 30; i++)
printf("%d ", array[i]);
printf("\n\n");
for(i = 29; i >= 0; i--)
printf("%d ",array[i]);
printf("\n\n");
for(i = 0; i < 30; i++){
int j;
int count = 0;
int match = array[i];
for(j = 0; j < 30 && match >= 0; j++){
array[j] == match && ++count && (array[j] = -1);
// replaces the following if as value of count is
// tested after it is incremented so will always be nonzero.
// if (array[j] == match) {
// count++; array[j] = -1;
// }
}
// if this is a valid array element value we are trying to match
// then print the count and the value being matched. printf()
// is a function that returns an int indicating number of character written.
match >= 0 && printf("There were %d %d's\n", count, match);
}
return 0;
}
An example output.
8 2 4 0 8 0 8 1 1 4 6 9 3 9 7 6 3 9 0 1 0 7 1 2 4 0 3 0 2 3
3 2 0 3 0 4 2 1 7 0 1 0 9 3 6 7 9 3 9 6 4 1 1 8 0 8 0 4 2 8
There were 3 8's
There were 3 2's
There were 3 4's
There were 6 0's
There were 4 1's
There were 2 6's
There were 3 9's
There were 4 3's
There were 2 7's
I am trying to make a program that takes 3 inputs from the user: row/columns count and number range start and number range end.
In my case and for example lets say 4 1 16 so it means 4 rows and columns print numbers from 1-16.
I have a problem accomplishing this.
#include <stdio.h>
#include <stdlib.h>
int num1,num2,count,i,y;
int main()
{
count = 4;
num1 = 1;
num2 = 16;
for(i=1; i<=count; i++){
for(y=num1; y<=num2; y++){
printf("%d ",y);
}
printf("\n");
}
return 0;
}
The output is
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Whereas I want my output to be:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
for(y=1; y<=count; y++){
printf("%d ",num1++);
}
That way you will ensure that all the numbers will be printed. you can use num1 variable that you have declared and initialized with 1.
Don't use unnecessary global variables. And keep the declarations close to where you use them. That way you won't have to look in the top of the page to check the type or if you initialized or not.
for(i=1; i<=count; i++){
printf("%d ",i);
if( i % 4 == 0)
printf("\n");
}
Print 4*(i-1)+y instead of y:
printf("%d ",4*(i-1)+y);
I am working on a pattern that prints the following code using for loop
1
2 6
3 7 10
4 8 11 13
5 9 12 14 15
code is as follows, which results in slightly wrong output.
main(){
int i,j,k,n,num;
printf("\n Enter no of rows: ");
scanf("%d",&num);
for(i=1;i<=num;i++,k=num){
for(j=1,n=i;j<=i;j++,n+=k){
printf("%d ",n);
}
printf("\n");
}
}
And the code gives me this .
which is wrong from the output i wanted
1
2 6
3 7 11
4 8 12 16
5 9 13 17 21
The key to this is how you update n in the inner loop. You need it to take into account not just num but also i as you descend through each iteration. I've fixed num at 5 here and placed the assignment to k at the start of the outer loop as this will always be constant:
#include <stdio.h>
int main(void) {
int i, j, k, n, num;
num = 5;
for(i = 1, k = num; i <= num; i++){
for(j = 1, n = i; j <= i; n += k - j, j++){
printf("%d ", n);
}
printf("\n");
}
}
Gives:
1
2 6
3 7 10
4 8 11 13
5 9 12 14 15