luc018-logic.c¶
Problem Statement
According to Gregorian calender, it was Monday on the date 01/01/01. Write a program to find out what is the day on 1st January of any input year.
Metadata¶
| Property | Detail |
|---|---|
| Author | Amit Dutta amitdutta4255@gmail.com |
| Date | 12 Dec 2025 |
| License | MIT License (See the LICENSE file for details) |
| Difficulty | Beginner (index: 2 / 10) |
Concepts¶
Beta Feature
This concept detection system is still in beta and may occasionally show incorrect or incomplete results.
- Pointers
- Array
- Sorting (possible)
- Iteration
Actions¶
You can print or save this file by opening Raw and using your browser.
Source Code¶
#include <stdio.h>
/**
* @brief Determines if a given year is a leap year.
* * The rule: A year is a leap year if it is divisible by 4, UNLESS it is
* divisible by 100 but NOT by 400.
* * @param year The year to check.
* @return 1 if it is a leap year, 0 otherwise.
*/
int is_leap(int year) {
// Check if divisible by 400 OR (divisible by 4 AND not divisible by 100)
if ((year % 400 == 0) || (year % 4 == 0 && year % 100 != 0)) {
return 1;
}
return 0;
}
/**
* @brief Calculates the day of the week for January 1st of the given year.
* * The base date is 01/01/01, which was a Monday (index 1).
* * Day Mapping: 0:Sunday, 1:Monday, 2:Tuesday, 3:Wednesday, 4:Thursday, 5:Friday, 6:Saturday
*/
int main() {
long long year; // Use long long for year input if years far in the future/past are tested
int i;
long long total_days = 0;
int day_index;
// Day names array for output
const char *day_names[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
printf("Enter the year (e.g., 2025): ");
if (scanf("%lld", &year) != 1 || year < 1) {
printf("Invalid year input. Please enter a positive integer year (>= 1).\n");
return 1;
}
// --- Core Logic: Calculate Total Days ---
// We only need to consider the years that have *passed* before the target year.
// So, we count days from the end of year 0 up to the end of year (year - 1).
int years_passed = year - 1;
// 1. Calculate the number of leap days up to the end of year (year - 1)
// Formula based on Gregorian calendar rules for years Y-1:
// (Y-1)/4 - (Y-1)/100 + (Y-1)/400
long long leap_years = years_passed / 4 - years_passed / 100 + years_passed / 400;
// 2. Total days = (Number of years * 365) + (Number of leap years)
// Note: The loop method (below) is more intuitive but the formula is faster.
// We will use the direct formula for efficiency.
total_days = years_passed * 365 + leap_years;
// --- Alternate Loop Method (for conceptual simplicity) ---
/*
for (i = 1; i < year; i++) {
total_days += 365;
if (is_leap(i)) {
total_days += 1; // Add 1 for the leap day
}
}
*/
// --- Determine the Day of the Week ---
// Since 01/01/01 was Monday (index 1), we use the following setup:
// Index 1 corresponds to Monday.
// The calculation gives the number of days *past* the Monday start (01/01/01).
// The modulo operation gives the remainder (0-6).
// 0 days elapsed (Year 1): total_days=0. (0 + 1) % 7 = 1 (Monday). Correct.
// 365 days elapsed (Year 2): total_days=365. (365 + 1) % 7 = 2 (Tuesday). Correct. (365 mod 7 = 1, 1+1 = 2)
day_index = (total_days + 1) % 7;
// Correct the Day Index to match the array (0:Sun, 1:Mon, ..., 6:Sat)
// The +1 adjusts for the Monday starting point (index 1).
printf("\nOn January 1st, %lld, the day was: **%s**.\n", year, day_names[day_index]);
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 (luc018-logic.c):
#include <stdio.h>
/**
* @brief Determines if a given year is a leap year.
* * The rule: A year is a leap year if it is divisible by 4, UNLESS it is
* divisible by 100 but NOT by 400.
* * @param year The year to check.
* @return 1 if it is a leap year, 0 otherwise.
*/
int is_leap(int year) {
// Check if divisible by 400 OR (divisible by 4 AND not divisible by 100)
if ((year % 400 == 0) || (year % 4 == 0 && year % 100 != 0)) {
return 1;
}
return 0;
}
/**
* @brief Calculates the day of the week for January 1st of the given year.
* * The base date is 01/01/01, which was a Monday (index 1).
* * Day Mapping: 0:Sunday, 1:Monday, 2:Tuesday, 3:Wednesday, 4:Thursday, 5:Friday, 6:Saturday
*/
int main() {
long long year; // Use long long for year input if years far in the future/past are tested
int i;
long long total_days = 0;
int day_index;
// Day names array for output
const char *day_names[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
printf("Enter the year (e.g., 2025): ");
if (scanf("%lld", &year) != 1 || year < 1) {
printf("Invalid year input. Please enter a positive integer year (>= 1).\n");
return 1;
}
// --- Core Logic: Calculate Total Days ---
// We only need to consider the years that have *passed* before the target year.
// So, we count days from the end of year 0 up to the end of year (year - 1).
int years_passed = year - 1;
// 1. Calculate the number of leap days up to the end of year (year - 1)
// Formula based on Gregorian calendar rules for years Y-1:
// (Y-1)/4 - (Y-1)/100 + (Y-1)/400
long long leap_years = years_passed / 4 - years_passed / 100 + years_passed / 400;
// 2. Total days =
... (truncated for brevity)