用户输入数字的斐波那契函数

Question

I'm making a program like Fibonacci series, but for numbers up to the value provided by user, example: 0 to 5 (0, 1, 2, 3, 4). Program needs to calculate the sum of numbers, move to right and calculate other five numbers. It's working for Fibonacci series, but not for higher numbers. Does anyone know how to do this?

#include <stdio.h>


int main() {

    int gg; //number which user inputs
    int dg = 0; //first number
    int next = 0;
    int n;
    int i;

    printf("Number of series: ");
    scanf("%d", &gg);

    printf("Positive integer: ");//max number which outputs in printf("Series: \n", gg), etc. 1000;
    scanf("%d", &n);

    printf("Series: \n", gg); //result

    for (i = 0; i < gg; i++) {
        printf("%d, ", i);
    }

    next = gg + dg;

    while(next <= n){
        printf("%d, ", next);
        dg = gg;
        gg = next;
        next = gg + dg;
    }

    return 0;
}

For values 0 to 5 it should output:

0, 1, 2, 3, 4, 10, 20, 39, 76, 149, 294...

But it outputs:

0, 1, 2, 3, 4, 5, 10, 15, 25, 40, 65, 105, 170, 275,

Answer 1

You need save the last nth values, not just the last two values. In your example you need to store the last five values. You can use an array for this:

#include <stdio.h>
#include <stdlib.h>

int main() {
    int series_number;
    int *previous_numbers;
    int marker = 0;
    int max_series_value;
    int next = 0;

    printf("Number of series: ");
    scanf("%d", &series_number);

    printf("Positive integer: ");
    scanf("%d", &max_series_value);

    printf("Series %d: \n", series_number);

    previous_numbers = malloc(sizeof *previous_numbers *series_number);
    for (int i = 0; i < series_number; i++) {
        printf("%d, ", i);
        previous_numbers[i] = i;
        next += i;
    }
    previous_numbers[marker] = next;
    marker = (marker+1) % series_number;

    while(next <= max_series_value){
        printf("%d, ", next);
        next = 0;
        for (int i = 0; i < series_number; i++) {
            next += previous_numbers[(marker+i) % series_number];
        }
        previous_numbers[marker] = next;
        marker = (marker+1) % series_number;
    }
    free(previous_numbers);

    return 0;
}