/*
    file name: fibUsingIterative.c
    Description: Fibonacci numbers count using iterative
    利用迴圈法計算費氏數列

    費氏數列為0,1,1,2,3,5,8,13,21,…
    其中某一項為前二項之和,且第0項為0,第1項為1
*/

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

/* 函數原型宣告 */
long Fibonacci(long);
void flushBuffer(void);

int main()
{
    char ch;
    long n;

    printf("-----Fibonacci numbers Using Iterative-----");
    do
    {
        printf("\nEnter a number(n >= 0): ");
        scanf("%ld", &n);
        flushBuffer();
        /* n 值大於 0 */
        if ( n < 0 )
            printf("Input number must be > 0!\n");
        else
            printf("Fibonacci(%ld) = %ld\n", n, Fibonacci(n));

        printf("Continu(y/n)? ");
        ch = toupper(getchar());
    } while (ch == 'Y');

    return 0;
}

long Fibonacci(long n)
{
    long backitem1;        /*前一項值*/
    long backitem2;        /*前二項值*/
    long thisitem = 0;     /*目前項數值*/
    long i;

    if (n == 0)    /* 費氏數列第0項為0 */
        return (0);
    else if (n == 1)    /* 第二項為1 */
        return (1);
    else {
        backitem2 = 0;
        backitem1 = 1;

        /* 利用迴圈將前二項相加後放入目前項 */
        /* 之後改變前二項的值至到第n項求得 */
        for (i = 2; i <= n; i++) {
            /* F(i) = F(i-1) + F(i-2) */
            thisitem = backitem1 + backitem2;
            /*改變前二項之值*/
            backitem2 = backitem1;
            backitem1 = thisitem;
        }

        return thisitem;
    }
}

void flushBuffer()
{
    while (getchar() != '\n') {
        continue;
    }
}