How to print out all combinations? - c

Question

#include <stdio.h>
int main()
{
int day[30],
month[11],
year[2] = { 2000, 2001 };
int combinations[743];
printf("Days:\n");
for (int i = 1; i <= 31; i++)
{
day[i - 1] = i;
printf("%d ", day[i-1]);
}
printf("\n\nMonth:\n");
for (int j = 1; j <= 12; j++)
{
month[j - 1] = j;
printf("%d ", day[j - 1]);
}
printf("\n\nYear:\n%d %d\n\n", year[0], year[1]);
for (int x = 0, y = 0, z = 0, k = 0;
x <= 30, y <= 11, z <= 1, k <= 743;
x++, y++, z++, k++)
{
if (x == 31)
{
x = 0;
}
if (y == 11)
{
y = 0;
}
combinations[k] = day[x],".",month[y],".",year[0];
}
for (int a = 0; a <= 20; a++)
{
printf("Combination: %d \n", combinations[a]);
}
getch();}
I want to make a program that prints out all combinations for birthdays for 2000,2001 people, but I get something weird in the output
It looks like this:
Days:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
Month:
1 2 3 4 5 6 7 8 9 10 11 12
Year:
2000 2001
Combination: 1
Combination: 2
Combination: 3
Combination: 4
Combination: 5
Combination: 6
Combination: 7
Combination: 8
Combination: 9
Combination: 10
Combination: 11
Combination: 12
Combination: 13
Combination: 14
Combination: 15
Combination: 16
Combination: 17
Combination: 18
Combination: 19
Combination: 20
Combination: 21

Answer 1

First of all, quick mistakes there:
In the second for-cycle you are printing the day array, the only reason the outut there is correct is because the numbers are the same and the cycle stop at 12. Anyways, you should fix that, for the sake of good programing.
The last cyle shouldnt be done like that, you are literally asking for bugs. Divide that cycle to 3 different ones. The outside cycle is the year, it will be executed twice (2000, 2001), the inner cycle will iterate 12 times( that will be the month) and the last one will iterate 31 times (that will be the days)
Doing it like this, makes it much easier to read and to do it right, since in your code that cycle will only be executed twice, since Z stops at 2 and you increase it every iteration.
for(year = 0; year < 2 ; year ++)
for(month = 0; month < 12; month ++)
for(day = 0; day < 31; day++)
//combination code goes here
Last things, if you want to save the date you cant make it like an int, unless you sum the days and use it as a reference, if you want something like "11-05-2000" you either creat a struct for that or save it as string. Second you are assuming 31 for every month...
Any further question, as you may have, feel free to ask :)

Answer 2

#include <stdio.h>
int main()
{
//changed days to 31 and months to 12
int day[31],
month[12],
year[2] = { 2000, 2001 };
//made three arrays one for days,one for months,one for year storage, or you can use structure array, structure should have three fields
int combinations1[744];
int combinations2[744];
int combinations3[744];
printf("Days:\n");
for (int i = 1; i <= 31; i++)
{
day[i - 1] = i;
printf("%d ", day[i-1]);
}
printf("\n\nMonth:\n");
for (int j = 1; j <= 12; j++)
{
month[j - 1] = j;
printf("%d ", day[j - 1]);
}
printf("\n\nYear:\n%d %d\n\n", year[0], year[1]);
int x,y,z,k=0;
//this will be code for storing all combination of days, looping through all days of a month of a year, then to next month and finally to next year
for(x = 0; x < 2 ; x ++)
{
for(y = 0; y < 12; y ++)
{
for(z = 0; z < 31; z++)
{
combinations1[k]=z+1;
combinations2[k]=y+1;
combinations3[k]=year[x];
k++;
}
}
}
// printing all combinations
for (int a = 0; a < 744; a++)
{
printf("Combination: %d.%d.%d\n", combinations1[a],combinations2[a],combinations3[a]);
}
//getch();
}
I have explained the code in comments. Please read the comments to understand