Every past-paper program, built step by step

#include <stdio.h>

int main(void) {
    char c;
    char letters[100];          /* only the alphabets go here */
    int count = 0;              /* how many letters stored */
    int sum = 0;                /* sum of the digits */
    int i;
    char highest;

    printf("Enter digits and lowercase letters: ");
    scanf("%c", &c);
    while (c != '\n') {         /* Enter is character 10 = '\n' */
        if (c >= '0' && c <= '9')
            sum = sum + (c - '0');      /* '7' - '0' = 7 */
        else
            letters[count++] = c;
        scanf("%c", &c);
    }

    printf("Sum of digits: %d\n", sum);
    if (count == 0) {
        printf("No alphabets entered\n");
    } else {
        highest = letters[0];
        for (i = 1; i < count; i++)
            if (letters[i] > highest)
                highest = letters[i];
        printf("Highest alphabet: %c\n", highest);
    }
    return 0;
}

Runs, including an input with no letters:

Enter digits and lowercase letters: ab12cd34
Sum of digits: 10
Highest alphabet: d

Enter digits and lowercase letters: a1z2g3f9
Sum of digits: 15
Highest alphabet: z

Enter digits and lowercase letters: 2025
Sum of digits: 9
No alphabets entered

#include <stdio.h>
int main(){
    int n;  //Size of account number
    scanf("%d",&n);
    int last_accno[n]; //Scan last account number
    for(int i=0;i<n;i++)
        scanf("%d",&last_accno[i]);
    int new_accno[n];
    //Write logic for new account no.
    int i;
    for (i = 0; i < n; i++)          /* step 1: copy */
        new_accno[i] = last_accno[i];
    i = n - 1;                       /* step 2: start at the last digit */
    while (i >= 0 && new_accno[i] == 9) {
        new_accno[i] = 0;            /* 9 + 1 = 10: write 0, carry 1 */
        i--;
    }
    if (i >= 0)
        new_accno[i] = new_accno[i] + 1;   /* the carry lands here */
    for (i = 0; i < n; i++)
        printf("%d", new_accno[i]);
    printf("\n");
    return 0;
}

4 1 0 9 9
1100

3 1 2 3
124

5 3 0 9 9 9
31000

#include <stdio.h>

int main(void) {
    int n, t, d, m = 0;
    scanf("%d", &n);
    for (d = 0; d <= 9; d++)                 /* try digits smallest first */
        for (t = n; t > 0; t = t / 10)       /* peel n's digits one by one */
            if (t % 10 == d)
                m = m * 10 + d;              /* found d: append it */
    printf("Output = %d\n", m);
    return 0;
}

62371
Output = 12367

50314
Output = 1345

Note the second run: a 0 digit ends up first and disappears, giving 1345.

#include <stdio.h>

int main(void) {
    int n, t, d, m, small;
    scanf("%d", &n);                    /* a 5-digit number, digits distinct */

    small = 10;                         /* smallest NONZERO digit goes first */
    for (t = n; t > 0; t = t / 10)
        if (t % 10 != 0 && t % 10 < small)
            small = t % 10;

    m = small;
    for (d = 0; d <= 9; d++) {          /* then every other digit, smallest up */
        if (d == small)
            continue;
        for (t = n; t > 0; t = t / 10)
            if (t % 10 == d)
                m = m * 10 + d;         /* append d on the right */
    }
    printf("Output = %d\n", m);
    return 0;
}

62371
Output = 12367

50314
Output = 10345

98765
Output = 56789

The published model answer is about 30 lines (find the minimum, remove it, rebuild the number, repeat five times). It's correct for the sample, and outputs 1345 for 50314. Both readings are defensible if stated; an unstated choice is what risks marks.

#include <stdio.h>

int main(void) {
    int n, i, val, pos;

    printf("Number of elements: ");
    scanf("%d", &n);
    int a[n];                          /* the only array */
    printf("Enter %d elements: ", n);
    for (i = 0; i < n; i++)
        scanf("%d", &a[i]);
    printf("Element to insert: ");
    scanf("%d", &val);
    printf("Position (index 0 to %d): ", n - 1);
    scanf("%d", &pos);

    for (i = n - 1; i > pos; i--)      /* shift RIGHT-TO-LEFT; old a[n-1] is lost */
        a[i] = a[i - 1];
    a[pos] = val;

    printf("Resultant array:");
    for (i = 0; i < n; i++)
        printf(" %d", a[i]);
    printf("\n");
    return 0;
}

Number of elements: 5
Enter 5 elements: 2 5 6 4 9
Element to insert: 10
Position (index 0 to 4): 2
Resultant array: 2 5 10 6 4

Number of elements: 5
Enter 5 elements: 2 5 6 4 9
Element to insert: 10
Position (index 0 to 4): 4
Resultant array: 2 5 6 4 10

Number of elements: 5
Enter 5 elements: 2 5 6 4 9
Element to insert: 10
Position (index 0 to 4): 0
Resultant array: 10 2 5 6 4

And the classic wrong version, shifting left-to-right (for (i = pos + 1; i < n; i++) a[i] = a[i - 1];), compiled on the same sample:

 2 5 10 6 6

#include <stdio.h>

int main(void) {
    int a, b, c;
    printf("Enter three distinct integers: ");
    scanf("%d %d %d", &a, &b, &c);

    if (a > b) {                 /* a beat b: now only a and c can win */
        if (a > c)
            printf("Largest = %d\n", a);
        else
            printf("Largest = %d\n", c);
    } else {                     /* b beat a: now only b and c can win */
        if (b > c)
            printf("Largest = %d\n", b);
        else
            printf("Largest = %d\n", c);
    }
    return 0;
}

Enter three distinct integers: 3 7 5
Largest = 7

Enter three distinct integers: 9 2 4
Largest = 9

Enter three distinct integers: -1 -5 -3
Largest = -1

#include <stdio.h>
int main() {
    int n = 6;
    switch(n > 0 ? 1 : n < 0 ? -1 : 2) {
        case -1:
            printf("Negative");
            break;                 /* added */
        case 1:
            printf("Positive");
            break;                 /* added */
        default:
            printf("Zero");
    }
    return 0;
}

Positive

Changing the 6 to −4 prints Negative; to 0 prints Zero.

#include<stdio.h>
int main(){
    int price = 0;                          //for the currency note
    int deno1 = 0, deno2 = 0, deno3 = 0;    // for the denominations
    int ways = 0;                           // for the number of ways the change possible
    int d1, d2, d3;
    printf("Enter price: ");
    scanf("%d", &price);
    printf("Enter deno1 deno2 deno3 (largest first): ");
    scanf("%d %d %d", &deno1, &deno2, &deno3);

    for (d1 = 0; d1 <= price / deno1; d1++)
        for (d2 = 0; d2 <= price / deno2; d2++)
            for (d3 = 0; d3 <= price / deno3; d3++)
                if (d1 * deno1 + d2 * deno2 + d3 * deno3 == price) {
                    printf("<%d, %d, %d>\n", d1, d2, d3);
                    ways++;
                }

    if (ways > 0)
        printf("Number of ways = %d\n", ways);
    else
        printf("Not Possible\n");
    return 0;
}

Enter price: 110
Enter deno1 deno2 deno3 (largest first): 100 20 10
<0, 0, 11>
<0, 1, 9>
<0, 2, 7>
<0, 3, 5>
<0, 4, 3>
<0, 5, 1>
<1, 0, 1>
Number of ways = 7

Enter price: 25
Enter deno1 deno2 deno3 (largest first): 20 10 4
Not Possible

The second run: every note is even and 25 is odd, so no combination works.

#include <stdio.h>

int main(void) {
    int runs[10];
    int i, total = 0, highest, lowest, centuries = 0;
    float average;

    printf("Enter runs in 10 matches: ");
    for (i = 0; i < 10; i++) {
        scanf("%d", &runs[i]);
        if (runs[i] < 0 || runs[i] > 200) {
            printf("Error: runs must be between 0 and 200\n");
            return 0;                        /* terminate */
        }
    }

    highest = runs[0];
    lowest = runs[0];
    for (i = 0; i < 10; i++) {
        total = total + runs[i];
        if (runs[i] > highest) highest = runs[i];
        if (runs[i] < lowest)  lowest = runs[i];
        if (runs[i] >= 100)    centuries++;
    }
    average = total / 10.0;                  /* 10.0, not 10 */

    printf("Total Runs = %d\n", total);
    printf("Average Runs = %.2f\n", average);
    printf("Highest Score = %d\n", highest);
    printf("Lowest Score = %d\n", lowest);
    printf("Number of Centuries = %d\n", centuries);
    if (highest - lowest <= 50)
        printf("Performer Status = Consistent Performer\n");
    else
        printf("Performer Status = Inconsistent Performer\n");
    return 0;
}

Enter runs in 10 matches: 120 35 0 88 150 60 45 100 10 75
Total Runs = 683
Average Runs = 68.30
Highest Score = 150
Lowest Score = 0
Number of Centuries = 3
Performer Status = Inconsistent Performer

Enter runs in 10 matches: 120 35 250 88 150 60 45 100 10 75
Error: runs must be between 0 and 200

Enter runs in 10 matches: 60 70 55 90 80 75 65 85 88 95
Total Runs = 763
Average Runs = 76.30
Highest Score = 95
Lowest Score = 55
Number of Centuries = 0
Performer Status = Consistent Performer

#include <stdio.h>
#include <math.h>

int main(void) {
    int a, b, c, disc, real;
    double r1, r2, re, im;

    printf("Enter a b c (a not 0): ");
    scanf("%d %d %d", &a, &b, &c);
    disc = b * b - 4 * a * c;
    real = (disc >= 0);                  /* 1 if real roots, 0 if complex */

    switch (real) {
    case 1:
        r1 = (-b + sqrt(disc)) / (2.0 * a);
        r2 = (-b - sqrt(disc)) / (2.0 * a);
        printf("Roots are %f and %f\n", r1, r2);
        break;
    case 0:
        re = -b / (2.0 * a);
        im = sqrt(-disc) / (2.0 * a);
        printf("Roots are %f+%fi and %f-%fi\n", re, im, re, im);
        break;
    }
    return 0;
}

Enter a b c (a not 0): 1 4 5
Roots are -2.000000+1.000000i and -2.000000-1.000000i

Enter a b c (a not 0): 1 3 2
Roots are -1.000000 and -2.000000

Enter a b c (a not 0): 1 1 1
Roots are -0.500000+0.866025i and -0.500000-0.866025i

Enter a b c (a not 0): 2 3 1
Roots are -0.500000 and -1.000000