Express a number as the sum of three or more consecutive numbers in c

[miyakisan]

Program in c, to find the consecutive series of a number, example.
Input:24
Output: 7+8+9=24

Another example
Input:26
Output:5+6+7+8

[Rabbit]

We aren't here to do all the work for you but if you post what you've tried so far, we can help you along.

[AjayGohil]

Hiii,

Refer Below equation:

Sum of first n natural numbers = n * (n + 1)/2

Sum of first (n + k) numbers = (n + k) * (n + k + 1)/2

If N is sum of k consecutive numbers, then
following must be true.

N = [(n+k)(n+k+1) - n(n+1)] / 2

OR

2 * N = [(n+k)(n+k+1) - n(n+1)]


you can try below code:

#include <stdio.h>
void printNumber(int l,int f)
{
    int x;
      for ( x = f; x<= l; x++) 
      {
      printf(" %d +",x);
 
      }
        f++;
}
 
void getNumber(int n)
{
    int last,first;
    for ( last=1; last<n; last++) 
    { 
        for ( first=0; first<last; first++) 
        { 
            if (2*n == (last-first)*(last+first+1)) 
            { 
                printf("%d =",n);
                printNumber(last, first+1); 
                return; 
            } 
        } 
    } 
 
   printf("Not possible");  
}
 
int main()
{
    int n;
    printf("Enter input:");
    scanf("%d",&n);
    getNumber(n);
    return 0;
}

[Rabbit]

There's a slightly more efficient method that uses fewer iterations. Using the formula target = n*x + (n*(n-1)/2), given that x must be an integer, you can test that a solution exists for n consecutive numbers by using the mod operator.

[SioSio]

The answer is not necessarily one set.

void getNumber(int n)
{
    int A;
    int B;
    int C;
    int D;
    int max = n / 3 + 1;
    int Y[1000];
    int Z[1000];
    for (A = 0; A <= max; A++) {
        B = A * (A + 1) / 2;
        Z[A] = B;
        Y[A] = B - n;
    }
    for (C = 0; C <= max; C++) {
        for (D = 0; D <= max; D++) {
            if (Z[C] == Y[D]) {
                printf("from %d to %d\n", C + 1, D);
                break;
            }
        }
    }
}

Or

void getNumber(int n)
{
    int A;
    int B[1000];
    int C[1000];
    int D;
    int max = n / 3 + 1;
    for (A = 0; A <= max; A++)
    {
        B[A] = A * (A + 1) / 2;
        C[A] = B[A] - n;
        for (D = 0; D <= max; D++)
        {
            if (C[A] == B[D]) {
                printf("from %d to %d\n", D+1,A);
                break;
            }
        }
    }
}

[Rabbit]

This will find all solutions in the fewest iterations possible.

void findSolution(int target) {
    int i, solution;
    for(i=3; (i * (i+1))/2 <= target; i++) {
        solution = target - (i * (i-1))/2;
        if(solution % i == 0 && solution > 0){
            solution = solution/i;
            printf("\n%d consecutive numbers starting at %d", i, solution);
        }
    }
}

[jayclayton]

Thanks, i understand it now :)