假设我想证明1d12 (十二边模)服从矩形分布,2d6服从正态分布。
快速和肮脏的方法是统计大约1000个随机生成的数字,将它们放在一个数组中,然后从那里计算平均值和期望值。
但是,如果我想通过使用运行的总计而不是1000个成员数组来节省内存呢?
我可以这样做吗?
for (i =0; i < 1000; i++){
x = Math.Random(1,6);
runningTotal += x;
}
mean = runningTotal / 1000;
SELECT month(dateofappointment), COUNT(*) 'NumberOfAppointments'
FROM appointment
WHERE YEAR(dateofappointment) = '2016'
GROUP BY MONTH(dateofappointment)
这显示了我所有的月份,但是十二月不在那里,因为那一年没有预约。我如何显示十二月是0?
我的apache会议出了点小问题。当我阅读错误日志时,我可以看到以下内容:
[client xxx.xxx.xx.xx] AH01964: Connection to child 1 established (server www.mywebsite.com:443)
[client xxx.xxx.xx.xx] AH01964: Connection to child 6 established (server www.mywebsite.com:443)
[client xxx.xxx.xx.xx] AH01964: Connection to child 10 established (
#!/bin/bash
sum=0
for number in $@
do
echo $number | grep -q "[^a-z]" >> /dev/null
if [ $? != 0 ]
then
echo "Sorry, '$number' is not a number"
else
sum=$((sum + number))
echo "$sum"
fi
done
我的分配要求,如果我输入add2 4 -3 12 9,它将输出
22
但是地雷的产出:
4 1