C 程式設計作業二，選擇邏輯與迴圈應用：解題

這題主要在能解決前兩天的訂貨能被納進來計算，且能更新。所以我們使用兩個分別處理兩天的訂貨量，如下：
arrivedNum = firstOrder;
firstOrder = secondOrder;

如此第一天的的訂貨為到貨量，第二天的訂貨變成第一天的訂貨，而第二天的訂貨則是今天計算後再決定
。

/* C Programming, Project 2 */
#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int orderNum, stockNum, arrivedNum, soldNum;
    int orderMax = 500;
    int orderMin = 300;
    int firstOrder = 0;
    int secondOrder = 0;
    int i;
    
    stockNum = 450;
    for (i=1; i<=10; i++) {
        printf("Enter the sold amount: ");
        scanf("%d", &soldNum);
        arrivedNum = firstOrder;
        firstOrder = secondOrder;
        stockNum = stockNum + arrivedNum - soldNum;
        if (stockNum < 300)
            secondOrder = 500 - stockNum;
        else
            secondOrder = 0;
        printf("Order = %3d, Arrived = %3d, Stock = %3d\n", secondOrder, arrivedNum, stockNum);
    }

    system("pause");
    return 0;
}

作業二題目
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