Generating all distinct partitions of a number

Question

I am trying to write a C code to generate all possible partitions (into 2 or more parts) with distinct elements of a given number. The sum of all the numbers of a g

Answer 1

It is another solution that is based on an iterative algorithm. It is much faster than @imreal's algorithm and marginally faster than @JasonD's algorithm.

Time needed to compute n = 100

$ time ./randy > /dev/null
./randy > /dev/null  0.39s user 0.00s system 99% cpu 0.393 total
$ time ./jasond > /dev/null
./jasond > /dev/null  0.43s user 0.00s system 99% cpu 0.438 total
$ time ./imreal > /dev/null
./imreal > /dev/null  3.28s user 0.13s system 99% cpu 3.435 total

#include 
#include 
#include 

int next_partition(int *a, int* kp) {
    int k = *kp;
    int i, t, b;

    if (k == 1) return 0;
    if (a[k - 1] - a[k - 2] > 2) {
        b = a[k - 2] + 1;
        a[k - 2] = b;
        t = a[k - 1] - 1;
        i = k - 1;
        while (t >= 2*b + 3) {
            b += 1;
            a[i] = b;
            t -= b;
            i += 1;
        }
        a[i] = t;
        k = i + 1;
    } else {
        a[k - 2] = a[k - 2] + a[k - 1];
        a[k - 1] = 0;
        k = k - 1;
    }
    *kp = k;
    return 1;
}

int main(int argc, char* argv[])
{
    int n = 100;
    int m = floor(0.5 * (sqrt(8*n + 1) - 1));
    int i, k;
    int *a;
    a = malloc(m * sizeof(int));
    k = m;
    for (i = 0; i < m - 1; i++) {
        a[i] = i + 1;
    }
    a[m - 1] = n - m*(m-1)/2;

    for (i = 0; i < k; i++) printf("%d ", a[i]);
    printf("\n");

    while (next_partition(a, &k)) {
        for (i = 0; i < k; i++) printf("%d ", a[i]);
        printf("\n");
    }
    free(a);
    return 0;
}