Count the number of ways to divide a number in 4 parts
C program to count the number of ways to divide a number in 4 parts or represent n as sum of four positive integers and print them.
For example, num = 6
There are 2 ways
(1, 1, 1, 3)
(1, 1, 2, 2)
Let, num = 14
There are 9 such ways
(1, 1, 1, 7)(1, 1, 2, 6)
(1, 1, 3, 5)(1, 1, 4, 4)
(1, 2, 2, 5)(1, 2, 3, 4)
(1, 3, 3, 3)(2, 2, 2, 4)
(2, 2, 3, 3)
There are 2 ways
(1, 1, 1, 3)
(1, 1, 2, 2)
Let, num = 14
There are 9 such ways
(1, 1, 1, 7)(1, 1, 2, 6)
(1, 1, 3, 5)(1, 1, 4, 4)
(1, 2, 2, 5)(1, 2, 3, 4)
(1, 3, 3, 3)(2, 2, 2, 4)
(2, 2, 3, 3)
Program
#include<stdio.h>
#include<conio.h>
void main()
{
int num, i, j, k, l, count = 0;
printf("Enter the number\t:");
scanf("%d",&num);
printf("\n");
for(i=1;i<num;i++)
{
for(j=i;j<num;j++)
{
for(k=j;k<num;k++)
{
for(l=k;l<num;l++)
{
if(i + j + k + l == num)
{
count++;
printf("(%d, %d, %d, %d)\n",i,j,k,l);
}
}
}
}
}
printf("There are total of %d such combinations",count);
}
Output
Enter the number :12
(1, 1, 1, 9)
(1, 1, 2, 8)
(1, 1, 3, 7)
(1, 1, 4, 6)
(1, 1, 5, 5)
(1, 2, 2, 7)
(1, 2, 3, 6)
(1, 2, 4, 5)
(1, 3, 3, 5)
(1, 3, 4, 4)
(2, 2, 2, 6)
(2, 2, 3, 5)
(2, 2, 4, 4)
(2, 3, 3, 4)
(3, 3, 3, 3)
There are total of 15 such combinations