I have made a code where I have to
input
1 . no of questions
2. difficulty for every question
3 . no. of queries( query is to find median of difficulty questions between questions no. given in next line)
4 . question no.s to find median
loop goes on till no input is given .
#include<stdio.h>
int main()
{
int ques,count=0;
while(scanf("%d",&ques))
{
int i,diff,se,fi,j,query,arr[100];
for(i=0;i<ques;i++)
{
scanf("%d",&diff);
arr[i] = diff;
}
count++;
printf("Case %d:\n",count);
scanf("%d",&query);
for(i=0;i<query;i++)
{
int sum = 0;
scanf("%d %d",&fi,&se);
for(j=fi-1;j<se;j++)
{
sum = sum+ arr[j];
}
sum = sum/((se-fi)+1);
printf("%d\n",sum);
}
}
return 0;
}
Here i give two inputs
5
5 3 2 4 1
3
1 3
2 4
3 5
5
10 6 4 8 2
4
1 3
2 4
3 5
2 3
6
2 56 2 3 5 4
1
2 5
but my output should be upto case 3 only instead:
Case 1:
3
3
2
Case 2:
6
6
4
5
Case 3:
16
Case 4:
4
Case 5:
4
Case 6:
4
Case 7:
4
Case 8:
4
Case 9:
4
Case 10:
4
Case 11:
4
Case 12:
4
Case 13:
4
Case 14:
4
Case 15:
4
and it goes on and on : Tell me why is this happening:
You need to check the return value of scanf, if you give control-D, it will return -1, in your program that means you need to quit, but -1 will cause the while condition be true.
Whwnever you are ending , no more values in there in your input . it returns -1 ; but your code seems to take '0' as quit which is making -1 to be true and your code is running .
Related
I have to create a Sudoku grid with the following pattern in C:
1 2 3 4
3 4 1 2
2 3 4 1
4 1 2 3
The first number in the top left corner (here: 1) must be an editable variable for a start value. There is another variable to create the grid with by the square size, in this example the square size is 2 and the 1 2 3 4 are in one square. 3 4 1 2 are in another square and so on...
If the start value is e.g. 3, the grid looks like this:
3 4 1 2
1 2 3 4
4 1 2 3
2 3 4 1
I noticed that there is a pattern: If the row number is odd, the new start value of the next row is the second last one. If the row number is even, the new start value of the next row is the last one. I tried to do it in C, but the even rows are cloning themselves. Note that arrays and pointers are not allowed here, only loops and other control-structures.
I tried the following approach, but the even rows are cloning themselves:
#include <stdio.h>
const int intSquareSize = 2;
const int intFieldLength = intSquareSize * intSquareSize;
int intStartValue = 3;
int main() {
int a = 0;
int b = 0;
int m = 0;
for (int intRowCounter = 1; intRowCounter <= intFieldLength; intRowCounter++) {
m = intFieldLength - 1;
for (int intColumnCounter = 1; intColumnCounter <= intFieldLength; intColumnCounter++) {
a = intStartValue + (intColumnCounter - 1);
b = a;
if (a > intFieldLength) {
a = intFieldLength - m;
m--;
}
if (intRowCounter % 2 == 0 && intColumnCounter == intFieldLength) {
intStartValue = a;
} else if (intRowCounter % 2 == 1 && intColumnCounter == (intFieldLength - 1)) {
intStartValue = b;
}
printf("%d\t", a);
}
printf("\n");
}
return 0;
}
What did I wrong and how can I fix it?
If the row number is odd, the new start value of the next row is the second last one. If the row number is even, the new start value of the next row is the last one.
I don't think it is helpful to think in terms of odd and even. The involved numbers are just symbols, and one could even replace them with distinct colors (for example). Odd/even is not a significant thing here, and it would certainly not play the same role in other board sizes.
The pattern I see is this:
In the first intSquareSize rows, the values shift horizontally (compared to the previous row) by intSquareSize. For example with intSquareSize=3 the first three rows could be:
3 4 5 6 7 8 9 1 2
6 7 8 9 1 2 3 4 5
9 1 2 3 4 5 6 7 8
Notice the shift of 3 positions to the left at each next row.
Then the pattern for the next chunks of intSquareSize rows would be the same, but with one shift. So the complete 9x9 sudoku would look like this:
3 4 5 6 7 8 9 1 2
6 7 8 9 1 2 3 4 5
9 1 2 3 4 5 6 7 8
4 5 6 7 8 9 1 2 3
7 8 9 1 2 3 4 5 6
1 2 3 4 5 6 7 8 9
5 6 7 8 9 1 2 3 4
8 9 1 2 3 4 5 6 7
2 3 4 5 6 7 8 9 1
This is just one possible pattern you could follow. There could be others. But the following code will apply the above logic. Note I used your variables, but I prefer to use 0-based logic, so loop variables start at 0, and the value of a is also 0-based. Only at the time of printing 1 is added to that value, so it becomes 1-based:
int a = intStartValue - 1; // Move from 1-based to 0-based
for (int intBlockCounter = 0; intBlockCounter < intSquareSize; intBlockCounter++) {
for (int intRowCounter = 0; intRowCounter < intSquareSize; intRowCounter++) {
for (int intColumnCounter = 0; intColumnCounter < intFieldLength; intColumnCounter++) {
printf("%d\t", (a + 1)); // back to 1-based
a = (a + 1) % intFieldLength;
}
printf("\n");
a = (a + intSquareSize) % intFieldLength; // Shift within a block
}
a = (a + 1) % intFieldLength; // Shift between blocks
}
I have a file, with format:
Course - Grade Count - Grades
Programming 10 3 4 5 4 3 2 4 5 2 3
Mathematics 8 3 3 4 5 3 2 2 3
Physics 6 3 4 5 3 4 5
Design 6 5 4 5 3 2 4
Logistics 8 3 4 5 3 1 1 2 4
Ex: Course - Programming, Grade Count - 10 and Grades - 3 4 5 4 3 2 4 5 2 3
I already have
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define SIZE 70
int main(void)
{
char subject[SIZE];
int gradeCount;
int grades[SIZE];
FILE *fp = fopen("C:\\Project\\project.txt", "r"); //opening already created file
if (fp == NULL) {
perror("Error opening file");
return(-1);
}
for (int i = 0; i < SIZE; i++) {
fscanf(fp, "%s %d", &subject[i], &gradeCount);
printf("%s \n", &subject[i]);
//printf("%d \n", gradeCount);
for (int k = 0; k < gradeCount; k++)
{
fscanf(fp, "%d", &grades[k]);
// printf("%d \n" , grades[k]);
}
if (i == SIZE) {
break;
}
}
fclose(fp);
return 0;
}
I need to print out "Course", "Grade Count" and "Grades" without any problems, later on I need to make a search and so I need to separate them from each other, but that is not the case, now I will show you the outputs for all cases, when I output first "Subject/Course" then "Grade Count" and finally "Grades".
For Courses:
Programming
Mathematics
Physics
Design
Logistics
ogistics
gistics
istics
stics
tics
ics
cs
s
#
##
#
#
For Grade Count:
10
8
6
6
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
And for Grades:
3
4
5
4
3
2
4
5
2
3
3
3
4
5
3
2
2
3
3
4
5
3
4
5
5
4
5
3
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
In all cases, additional things are added to original stuff that should be printed out, I don't know where it comes from, I thought about pointers, but don't know much about them. Any suggestions?
Just need to print everything normally to normally search for everything (Courses, grade count and grades) later on.
You need to exit the loop early if it fails to read anything in. You can do that by checking the return value of fscanf. If the first call doesn't return 2, you know that it didn't read in 2 values and can break out of the loop.
You're also calling fscanf and printf incorrectly for dealing with a string. You are moving the starting point of where you read into/print from, which isn't needed and reduces the maximum space available to you.
Updated code looks something like this
for (int i = 0; i < SIZE; i++) {
if(fscanf(fp, "%s %d", subject, &gradeCount) != 2) {
break;
}
printf("%s ", subject);
//printf("%d \n", gradeCount);
for (int k = 0; k < gradeCount; k++)
{
fscanf(fp, "%d", &grades[k]);
// printf("%d" , grades[k]);
}
}
I am trying to construct a 2D array for an assignment. I've used a nested for loop to construct the 2D array using scanf():
int width;
int height;
scanf("%d %d",&width,&height);
int array[width][height];
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
scanf("%d",&array[i][j]);
}
}
However when I print the array, I can see that it has been constructed in a strange way, where all the numbers of the first line past a certain point are the first few numbers from the second line (instead of what they should be). The next lines after work fine.
Example:
Input:
6 2
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
The created array looks like this:
1 3 2 4 6 8 (<-- these last 4 numbers are the first 4 numbers of the second line)
2 4 6 8 0 2 (correct)
3 4 2 0 1 3 (correct)
Any ideas? Thanks a lot.
Your declaration of array
int array[width][height];
is wrong. The outer loop goes from 0 to height - 1, but array[i] can only go
from 0 to width - 1. The same applies for the inner loop. You swapped width
and height in the declaration of the array, it should be
int array[height][width];
Also note that for the matrix
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
the width is 6 and the height is 3, so the correct input should be
6 3
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
I compiled and run this code:
#include <stdio.h>
int main(void)
{
int width;
int height;
scanf("%d %d",&width,&height);
int array[height][width];
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
scanf("%d",&array[i][j]);
}
}
printf("----------------\n");
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
printf("%d ", array[i][j]);
}
printf("\n");
}
}
And the output is:
$ ./b
6 3
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
----------------
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
as you can see, now it's reading correctly. See https://ideone.com/OJjj0Y
if anyone could answer me why this works, it would be greatly appreciated. The exercise (chapter 4, ex 7 and 8) says that if you have the expression:
9 - ((total - 1) % 10)
then, you could be tempted to simplify it like this:
10 - (total % 10)
But this would not work. Instead he offers the alternative:
(10 - (total % 10)) % 10
Now, I understand how he got to the first simplification, but not why it's wrong, or why does the second one works.
Thanks in advance
x %m has a range of (-m, m) in most C implementations. Mathematically it is generally defined from (0, m). Hence by adding m the modulo again will convert the C to the mathematical one.
Consider the outputs for total = 10 to see that the second expression is not equivalent.
Note also that the third expression is not equivalent to the first expression unless total > 0 (because the behaviour of % is implementation-defined in pre-C99 C, and defined but not what you want in C99).
Assuming that total > 0, the first and third expressions are equivalent due to the following mathematical identity:
(a % b) == (((a + c) % b) - c) % b
To understand why, imagine doing the operations on a clock-face.
This is because modulo in C allows for negative numbers.
so -5 % 10 is -5 instead of 5.
In the first case, the 9 - ((total - 1) % 10) is always positive.
In the second case it can be negative if -10 < total < 0. In the 3rd case it is again wrapped around for negatives back into the positive range.
It is a common thing for modulo because generally you want it for positives only(not sure why they implemented it for negatives).
To show why 9-((total)%10) is wrong, use a contradiction.
Let total = 10.
Then 9-((10-1)%10) ==> 9-(9%10) ==> 9-9 = 0.
But, 10-(10%10) ==> 10 -0 = 10.
Thus, 10-((total)%10) is not equivalent to 9-((total-1)%10)
The alternative is not a simplification and neither expression is equvalent to the first so the premise is flawed from the start:
The following:
int total ;
for( total = -10; total <= 10; total++ )
{
printf( "%d:\t%d\t%d\t%d\n", total,
9 - ((total - 1) % 10),
10 - (total % 10),
(10 - (total % 10)) % 10 ) ;
}
Produces:
-10: 10 10 0
-9: 9 19 9
-8: 18 18 8
-7: 17 17 7
-6: 16 16 6
-5: 15 15 5
-4: 14 14 4
-3: 13 13 3
-2: 12 12 2
-1: 11 11 1
0: 10 10 0
1: 9 9 9
2: 8 8 8
3: 7 7 7
4: 6 6 6
5: 5 5 5
6: 4 4 4
7: 3 3 3
8: 2 2 2
9: 1 1 1
10: 0 10 0
The last is only equvalent for integers greater than zero.
I need to swap first n elements from two non repeating sequences(arrays), where n is a random integer.
Seq1: 1 4 5 6 9 8 2 3 7
Seq2: 3 9 1 2 8 7 4 5 6
If n = 4
Seq1: 3 9 1 2 | 9 8 2 3 7
Seq2: 1 4 5 6 | 8 7 4 5 6
Now i need to repair the sequence by replacing the repeated numbers after '|'.
How to do this?
This is my effort..
for(left1 = 0; left1<pivot; left1++)
{
for(right1 = pivot; right1 < no_jobs; right1++)
{
if(S1->sequence[left1] == S1->sequence[right1])
{
for(left2 = 0; left2<pivot; left2++)
{
for(right2 = pivot; right2<no_jobs; right2++)
{
if(S2->sequence[left2] == S2->sequence[right2])
{
swap_temp = S1->sequence[right1];
S1->sequence[right1] = S2->sequence[right2];
S2->sequence[right2] = swap_temp;
break;
}
}
}
}
}
}
Swapping the first n elements is straightforward using a single for loop.
for(int i = 0; i < n; i++){
int tmp = array1[i];
array1[i] = array2[i];
array2[i] = tmp;
}
Now you need to find what has changed in the arrays. You can do this by comparing the parts you swapped.
int m1 = 0, m2 = 0;
int missing_array1[n];
int missing_array2[n];
for(int i = 0; i < n; i++){
bool found = false;
for(int j = 0; j < n; j++){
if(array1[i] == array2[j]){
found = true;
break;
}
}
if(!found){
missing_array2[m2++] = array1[i];
}
}
for(int i = 0; i < n; i++){
bool found = false;
for(int j = 0; j < n; j++){
if(array2[i] == array1[j]){
found = true;
break;
}
}
if(!found){
missing_array1[m1++] = array2[i];
}
}
missing_array2 now contains the numbers that are missing from array2. These are all the numbers that will be duplicated in array1. The same goes for missing_array1. Next you need to scan both arrays and replace the duplicates with the missing numbers.
while(m1 >= 0){
int z = 0;
while(missing_array1[m1] != array2[n + z]){
z++;
}
array2[n + z] = missing_array2[m1--];
}
while(m2 >= 0){
int z = 0;
while(missing_array2[m2] != array1[n + z]){
z++;
}
array1[n + z] = missing_array1[m2--];
}
In summary, you compare the parts you swapped to find the values that will be missing from each array. These value are also the values that will be duplicated in the opposite array. Then you scan each of the arrays and replace the duplicate values with one of the missing values (I assume you don't care which of the missing values, as long as all the values are unique.
If the swapped portions of the sequences contain the same values, then there would be no repeats - performing the swap would just shuffle the first n elements. So the values you need to repair are the values which occur in one of the swapped sequences
Firstly, I'd create a histogram of the n swapped elements, with those from sequence 1 counting as bit 0, and those from sequence 2 as bit 1. If any members of the histogram are non-zero, then they occur in one or the other sequence only.
If there are values requiring repair, then you can construct a look-up table of the values which require rewriting. This should map i to i unless i is one of the asymmetric values in the histogram, in which case it needs to map to the another asymmetric value.
Seq1: 1 4 5 6 9 8 2 3 7
Seq2: 3 9 1 2 8 7 4 5 6
If n = 4
Seq1: 3 9 1 2 | 9 8 2 3 7
Seq2: 1 4 5 6 | 8 7 4 5 6
histogram
value 1 2 3 4 5 6 7 8 9
count 3 1 1 2 2 2 0 0 1
mapping for sequence 1 ( while histogram [S1[i]] & 1, replace[S1[i]] with S2[i] )
value 1 2 3 4 5 6 7 8 9
replace 1 6 5 4 5 6 7 8 4
apply mapping to sequence 1 for i > n
Seq1: 3 9 1 2 | 9 8 2 3 7
replace - - - - | 4 8 6 5 7
result 3 9 1 2 | 4 8 6 5 7
mapping for sequence 2 ( while histogram [S2[i]] & 2, replace[S2[i]] with S1[i] )
value 1 2 3 4 5 6 7 8 9
replace 1 2 3 9 3 2 7 8 9
apply mapping to sequence 1 for i > n
Seq2: 1 4 5 6 | 8 7 4 5 6
replace - - - - | 8 7 9 3 2
result 1 4 5 6 | 8 7 9 3 2
Alternatively, replace with the next value with the other bit set in the histogram (the iterated replace will also need to check for replacing a value with itself); I'm assuming it doesn't really matter what value is used as the replacement as long as the values in the result are unique.