Private sub command1_click()
Dim n as long
Dim t as integer
Dim a(1 to 9) as integer
n=val(Text1.Text)
List1.clear
Do while n>0
t=n mod 10
If t mod 2=0 then a(t)=a(t)+1
n=n\10
Loop
List1.additem a(2)
End sub
For i = 1 To 8
a(i) = Int(Rnd * 7) + 1
Next i
For i = 1 To 3
For j = 1 To 8 - 2 * i
If a(j) Mod 7 > a(j + 2) Mod 7 Then
t = a(j): a(j) = a(j + 2): a(j + 2) = t
End If
Next j
Next i
For i = 1 To 8
ch(i) = Chr(a(i) + Asc("A") - 1)
Next i
执行该程序段后,ch(1)~ch(8)各元素值不可能的是( )
s = Text1.Text
ans = "": k =4
i = k - 1: j = k + 1: n = Len(s)
Do While k > 0
If Mid(s ,i,1) = Mid(s, j, 1) Then
ans = ans + Mid(s, i,1): k = k - 1
End If
i = (n + i - 2) Mod n + 1
j = j Mod n+1
Loop
Label1.Caption=ans
在文本框Text1中输入“banana”,执行程序后,标签Label1中显示的内容是( )
For i = 2 To 1000
flag = True:s = i
Do While flag And s >1
For j = 2 To Int(Sqr(s))
If s Mod j = 0 Then
Exit For
End If
Next j
Loop
If flag And s<>1 Then
End If
Next i
Label1.caption=str(c)
上述程序段中方框处可选语句为:
①flag = False ②c = c + 1 ③s = s \ 10
则(1)(2)(3)处语句依次可为( )
程序界面如图1所示,在本框Text1中输入COD指标,单击“计算投放量”按钮(Command1)后,程序根据COD指标计算出合适的COD去除剂投放量,并在标签Label3中输出计算结果。
污水处理厂根据COD指标投放COD去除剂的标准如下表所示:
当前COD指标X |
根据COD指标应投放COD去除剂的剂量 |
X<50 |
10 |
X≥50并且X≤500 |
|
X>500 | 3X |
图 1
图2 |
Private Sub Command1_Click( )
Dim x As Single,s As Single
x =Val(Text1. Text)
If x < 50 Then
s=10
ElseIf
Else
s=3*x
End If
Label3. Caption=
End Sub
Private Sub Command1_ Click()
Dims As String, c As String, i As Integer
Dim sum As Single, sum1 As Single, sum2 As Single, k As Integer,j As Integer
s = Text1.Text : sum2=0:k= 1
For i= 1 To Len(s)
c=
If c= "*" Or c= "/" Then
If c="*" Then
sum = Val(Mid(s,i+ 1, 1)) * Val(Mid(s,i- I, 1))
Else
sum = Val(Mid(s, i- 1, 1))/ Val(Mid(s,i+ 1,1))
End If
j=i-2
Do While
c = Mid(s,j, 1)
If c="+" Then sum1 = Val(Mid(s,j-1, 1)) + sum
If c="-" Then sum1 = Val(Mid(s,j- 1, 1)) - sum
sum=sum1
j=j- 1
Loop
If k=1 Then
sum2 = sum2 + sum
Else
c= Mid(s, k, 1)
If c =“+" Then sum2 = sum2 + sum
If c="" Then sum2 = sum2 - sum
End If
k=
End If
sum=0
Next i
Label2.Caption = sum2
End Sub
Private Sub Command1_Click()
Dim i As Integer, j As Integer, t1 As Integer, t2 As Integer
Dim s As String, c As String, st1 As Integer, st2 As Integer, max As Integer
s = Text1.Text: max = 0: st1 = 1
For i = 1 To Len(s) - 1
c = Mid(s, i, 1)
If c = "," Then
t1 = Val(Mid(s, st1, i - st1))
①
st2 = st1
For j = i + 1 To Len(s)
②
If c = "," Then
t2 = Val(Mid(s, st2, j - st2))
st2 = j + 1
End If
Next j
End If
If j>0 Then t2 = Val(Mid(s, st2, j - st2))
If Abs(t1 - t2) > max Then max = Abs(t1 - t2)
Next i
shuchu.Caption = "所有数之间绝对值相差最大为:" + Str(max)
End Sub
① ②
t(1) = 1: t(2) = 1
For i = 3 To 5
t(i) = t(i - 1) + t(i -2)
Next i
Key =13
i=1: j=8: k=6
Do While i <=j
m = i+t(k-1) ‘①
If Key = d ( m ) Then Exit Do ‘Exit Do表示退出循环
If d(m) > Key Then
j= m - 1
k=k-1
Else
i = m + 1
k=k-2
End If
Loop
运行该程序段后,①处的赋值语句共执行( )
s=input("请输入字符串:")
count=0
for I in s:
if i>="0"and i<="9":
count=count+1
print(s[count:count+3])
若输入的字符串为"AB12CCC222GGBD",则程序运行结果为( )
importmath
total=float(input("请输入金额:"))
if total<=500:
total*=0.9
elif total<=1500:
total*=0.85
else
total*=0.75
print(math.floor(total))
运行程序,输入“1024”后回车,程序的输出结果是
def prime(x):#如果x 是素数返回True,否则返回False
flag= True
for i in range(2, x):
if :
flag = False
break
return flag
a = int(input("请输入整数a:"))
b = int(input("请输入整数b:"))
if a > b:
#如果a 大于b,则交换a 和b 的值
c = 0
for j in range(a, b + 1):
if :
print(j)
print(a ,"和",b"之间共找到",c,"个素数!")
For i = 1 To 7
a(i) = Int(Rnd * 9) + 1
Next i
s = 0
For i = 1 To 6
If a(i) < a(i + 1) Then
f(i) = 1
ElseIf a(i) > a(i + 1) Then
f(i) = -1
Else
f(i) = 0
End If
s = s + f(i)
Next i
执行该程序段后,变量s的值为0,则数组a各元素的值可能是( )
L=[8, 10, 9, 14, 13, 4, 9, 13, 10]
k=int(input('请输入 k 值:'))
j=0
for i in range(8):
if L[i]<=k:
L[j]=L[i]
j=j+1
print(L[:j])
import random
a = [0] * 6
for i in range(0,6):
a[i] = random.randint(1,5) * 2 + 1
i = 0
while i < 5:
if a[i] > a[i+1]:
a[i],a[i+1] = a[i+1],a[i]
else:
a[i] += 1
i += 1
print(a)
执行以上程序后,列表变量a可能的是 ( )
DIM A AS INTEGER, B AS INTEGER
DIM C AS INTEGER
A= VAL(INPUTBOX(“请输入第一个数”))
B= VAL(INPUTBOX(“请输入第二个数”))
C= VAL(INPUTBOX(“请输入第三个数”))
IF A>B THEN
T=A
A=B
B=T
IF A>C THEN
T=A
A=C
C=T
IF THEN
PRINT A,B,C
PRINT A,C,B
END IF
FOR I=1 TO 20
IF I MOD 2=0 THEN
S=S+1
ELSEIF I MOD 3 =0 THEN
M=M+1
ELSEIF I MOD 5 =0 THEN
N=N+1
END IF
NEXT I
PRINT S, M, N
这段程序的执行结果是( )
import random
list=[0]*100
count=[0]*11
for i in range(0,100):
num=random.randint(0,20)
list[i]=num
If :
num=10
count[num]+=1
print(‘0~9 及 9 以上依次出现的次数为: ’,count)
程序某次运行的结果如下:
0~9 及 9 以上依次出现的次数为: [3, 5, 4, 6, 5, 5, 2, 5, 9, 4, 52] |
则划线处的代码为:( )
string=input('请输入字符串:')
word=input('请输入查询字符:')
count=0
for i in string:
if i==word:
count+=1
图1 |
图2 |
通过了解当天的是否有风、天气、温度和湿度这4个节点参数即可预测当天是否有人来游乐场。
不同的节点划分顺序可以绘制不同的决策树,为了选出最优的节点划分顺序,需要采用“信息熵”与“信息增益”指标。
信息熵,又称香农熵,被用来度量信息量的大小,信息熵越大表示信息量越大;
信息增益,表示样本经某节点划分后的信息熵变化大小。我们绘制决策树时应当逐次选择信息增益最大的节点作为当前节点。
对于有n个信息的样本D,记第k个信息发生的概率为,信息熵计算公式为E(D)=,
例如游乐场14个样本中“去”(9个)、“不去”(5个),则信息熵==0.940
若样本按“是否有风”节点划分,“是”(6个,其中3个去,3个不去)信息熵==1;
“否”(8个,其中6个去,2个不去)信息熵==0.811;经过此节点划分后的信息增益=原始信息熵按此节点划分后样本信息熵比例和=0.940(0.811)=0.048。
from flask import render_template,request,Flask
import random
app=Flask(__name__) #创建应用实例
@app.route('/') #选择页面路由
def index():
return render_template('')
#加密功能代码略,以下为解密代码:
@app.route('/jiemi1/',methods=["GET","POST"])
def jiemi1():
wb=request.form["wb"] #利用request获取网页文本框内容,返回示例:“1,4,2,3,0”
keyo=request.form["key"] #变量wb存储密文,变量keyo存储密钥
keyn=list(map(int,keyo.split(","))) #将字符串keyo转换为数值列表,示例:[1,4,2,3,0]
result=""
for i in range(len(keyn)):
for j in range(len(keyn)):
if :
break
result+=wb[j]
return render_template("jie.html",WB=wb,KEY=keyo,RESULT=result)
if __name__=="__main__":
def cal(lst): #计算样本lst的信息熵
x,y,z=0,len(lst),0 #x表示该样本信息熵,y表示该样本数量,z表示某信息发生的概率
num={}
for i in lst:
if i not in num:
num[i]+=1
for k in num:
z=num[k]/y #计算该信息发生的概率
x-=z*log(z,2) #根据公式计算信息熵,log(b,a)等价于logab
return x
def check(x,y):
#根据节点x,对样本y进行划分,返回示例:{'否': [1, 1, 0, 0, 1, 1, 1, 1], '是': [1, 1, 0, 1, 0, 0]}
dic={'是否有风': ['否', '否', '否', '否', '否', '否', '否', '否', '是', '是', '是', '是', '是', '是'],
'天气': ['多云', '多云', '晴', '晴', '晴', '雨', '雨', '雨', '多云', '多云', '晴', '晴', '雨', '雨'],
'温度': [28, 27, 29, 22, 21, 21, 20, 24, 18, 22, 26, 24, 18, 21],
'湿度': [78, 75, 85, 90, 68, 96, 80, 80, 65, 90, 88, 63, 70, 80],
'是否前往': [1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0]}
xm=list(dic.keys())
entropy=cal(dic[xm[-1]]) #调用函数计算样本原始信息熵entropy
#计算各节点信息增益
m=0;p=""
col=xm[:-1] #“是否前往”是结果项,不参与计算
for i in col:
size=len(dic[i]);entropy_1=0
zyb= #调用函数对样本按照当前节点进行划分
for j in zyb: #根据划分情况逐个求子样本信息熵并按比例累加
entropy_1+=len(zyb[j])/size*cal(zyb[j])
zy=entropy-entropy_1
print(i,"的信息增益:",zy)
if zy>m: #计算最大信息增益与信息增益最大的节点
m=zy
print("信息增益最大的节点:",p)