[英]Restaurant simulation
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#define Empty 1
#define Full 0
float ChanceOfArrival (float CustomersPerMinute)
{
float CustomersArrivalChance;
int i;
/*Customers arriving per second */
CustomersArrivalChance = (CustomersPerMinute / 60) * 100;
printf ("The chance of customers arriving is: %0.3f%%\n", CustomersArrivalChance);
/* each 10 minute interval */
for (i = 0; i <= 18; i++)
{
intervals (CustomersArrivalChance);
}
return CustomersArrivalChance;
}
int intervals (CustomersArrivalChance)
{
int totalCustomers = 0, totalWait = 0, queue = 0, SecondsInterval, waitingCustomers = 0;
int Cash1Salad, Cash1Buger, Cash2Salad, Cash2Burger;
int Cash1 = 1, Cash2 = 1;
int cointoss;
int x, Empty1, Empty2;
int CustomersServed = 0;
float RatePerMinute = 0, AverageWait = 0;
static int intervalNumber;
srand(time(NULL));
/*What could possibly happen every second in a 10 minute interval */
for (SecondsInterval = 0; SecondsInterval <= 600; SecondsInterval++)
{
x = rand() % 101;
if (CustomersArrivalChance >= x)
{
/*Customers Arrive this second */
totalCustomers++;
queue++;
/*Choosing a cash at random */
cointoss = rand()%2;
if (queue > 0)
{
/* Cash 1 is open cash 2 is busy so the customer goes to cash 1 and chooses
at random what they want to eat */
if ((Cash1 == Empty) && (Cash2 != Empty) || (cointoss == 1) )
{
Cash1 = Full;
queue--;
switch ((rand()%2))
{
case 0:
Cash1Salad = rand()% 66 + 55;
totalWait = totalWait + Cash1Salad;
Empty1 = Cash1Salad;
CustomersServed++;
break;
case 1:
Cash1Buger = rand()% 130 + 111;
totalWait = totalWait + Cash1Buger;
Empty1 = Cash2Burger;
CustomersServed++;
break;
}
}
/* Cash 1 is busy cash 2 is open customer goes to cash 2 and chooses what they want */
else if (Cash2 = Empty)
{
Cash2 = Full;
queue--;
switch ((rand()%2))
{
case 0:
Cash2Salad = rand()% 75 + 65;
totalWait = totalWait + Cash2Salad;
Empty2 = Cash2Salad;
CustomersServed++;
break;
case 1:
Cash2Burger = rand()% 140 + 121;
totalWait = totalWait + Cash2Burger;
Empty2 = Cash2Burger;
CustomersServed++;
break;
}
}
/*Both cashes are busy so the customer has to wait until one cash opens */
else
{
totalWait++;
waitingCustomers++;
}
/*Clearing Cash 1 if someone went there */
if (Empty1 > 0)
{
Empty1--;
}
/*empty1 is equal to 0 then cash 1 is empty */
else
{
Cash1 = Empty;
}
/*Clearing cash 2 is someone went there */
if (Empty2 > 0)
{
Empty2--;
}
/*empty2 is equal to 0 then cash 2 is empty */
else
{
Cash2 = Empty;
}
}
}
else
{
/*nothing happens because no customer showed up */
}
}
intervalNumber++;
AverageWait = ((totalWait*1.0)/ (totalCustomers));
printf ("The average waiting time per customer in seconds is %0.2f in the interval %d\n\n", AverageWait, intervalNumber);
printf ("The total customers that arrived in the interval %d is %d\n\n", intervalNumber, totalCustomers);
}
int main (void)
{
float CustomersPerMinute;
printf ("Enter in the number of customers you want to arrive per minute:\n");
scanf ("%f", &CustomersPerMinute);
ChanceOfArrival(CustomersPerMinute);
return 0;
}
嗨,我有这个程序,假设模拟一个只供应沙拉或汉堡的餐厅,只有两个收银员,只有一个排队,客户可以排队等候送达。
我不确定为什么它不起作用但是因为我认为一切都具有逻辑意义,但是当我运行这个程序时,它只打印出平均等待时间。
但问题是每个间隔打印的平均等待时间是相同的。 这不是我想要的。
从逻辑上讲,我认为每个时间间隔的平均等待时间应该不同,因为客户是随机生成的。 然而事实并非如此,因为平均等待时间总是相同的,我该如何解决这个问题呢?
由于客户是随机生成的,因此每个时间间隔的客户总数是相同的,因为客户是随机生成的,我该如何解决这个问题呢?
看起来你正在循环srand。 所以它可能会在给定的秒内给你相同的rand()结果。 堆栈溢出不是一个debuging论坛虽然;)。
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