`luc035.c`¶

Problem Statement

According to a study, the approximate level of intelligence of a person can be calculated using the following formula. i = 2 + (y + 0.5x) write a program that will produce a table of values of i, y and x, where y varies from 1 to 6, and, for each value of y, x varies from 5.5 to 12.5 in steps of 0.5

Metadata¶

Property	Detail
Author	Amit Dutta (amitdutta4255@gmail.com)
License	MIT
Difficulty	Beginner (index: 1 / 10)

Concepts¶

Beta Feature

This concept detection system is still in beta and may occasionally show incorrect or incomplete results.

Pointers
Recursion
Sorting (possible)
Iteration

Actions¶

Raw View on GitHub

You can print or save this file by opening Raw and using your browser.

Source Code¶

#include <stdio.h>
int main()
{
    double y, x;
    printf("\n--- Approximate Intelligence ---\n");
    for (y = 1; y <= 6; y++)
    {
        printf("\tY = %d\n", y);
        for (x = 5.5; x <= 12.5; x += 0.5)
        {
            printf("Y: %g\t X: %.2g\t I: %g\n", y, x, 2 + (y + 0.5 * x));
        }
        printf("-----------------------\n");
    }
    return 0;
}

Explanation¶

Explain with AI

Copy the prompt below and paste it into any AI assistant.

    You are explaining a C programming code to a beginner.

    STRICT RULES:

    - Only use the given code. Do NOT assume anything not present.

    - Do NOT add extra examples.

    - Keep explanation clear and short.

    - If something is unclear, say "Not clear from code".

    - Follow the exact format below. Do NOT change headings.

    FORMAT:

    [START]

    ## What it does

    (Explain the overall purpose in 1-2 sentences)

    ## Step-by-step

    (Explain how the code works in steps, simple language)

    ## Key Concepts

    (List concepts like loop, condition, function, etc.)

    ## Notes

    (Mention any limitations, errors, or assumptions)

    [END]

    CODE (luc035.c):

    #include <stdio.h>
    int main()
    {
        double y, x;
        printf("\n--- Approximate Intelligence ---\n");
        for (y = 1; y <= 6; y++)
        {
            printf("\tY = %d\n", y);
            for (x = 5.5; x <= 12.5; x += 0.5)
            {
                printf("Y: %g\t X: %.2g\t I: %g\n", y, x, 2 + (y + 0.5 * x));
            }
            printf("-----------------------\n");
        }
        return 0;
    }

luc035.c¶