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C Exercises: Convert a given integer to years, months and days

C Basic Declarations and Expressions: Exercise-18 with Solution

Convert days to years, months, days

Write a C program to convert a given integer (in days) to years, months and days, assuming that all months have 30 days and all years have 365 days.

C Programming: Convert a given integer to years, months and days

C Code:

#include <stdio.h>
int main() {
    int ndays, y, m, d; // Declare variables for number of days, years, months, and days
    
    // Prompt user for input number of days and store in 'ndays'
    printf("Input no. of days: ");
    scanf("%d", &ndays);
    
    // Calculate years, months, and remaining days
    y = (int) ndays/365;
    ndays = ndays-(365*y);
    m = (int)ndays/30;
    d = (int)ndays-(m*30);
    
    // Print the result
    printf(" %d Year(s) \n %d Month(s) \n %d Day(s)", y, m, d);
    
    return 0;
}

Sample Output:

Input no. of days: 2535                                                
 6 Year(s)                                                             
 11 Month(s)                                                           
 15 Day(s)

Flowchart:

C Programming Flowchart: Convert a given integer to years, months and days

C Programming Code Editor:

Previous: Write a C program to convert a given integer (in seconds) to hours, minutes and seconds.
Next: Write a C program that accepts 4 integers p, q, r, s from the user where q, r and s are positive and p is even. If q is greater than r and s is greater than p and if the sum of r and s is greater than the sum of p and q print "Correct values", otherwise print "Wrong values".

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