# 输入n个数从小到大依次存储到列表nums中,代码略
k = -1
for i in range(n):
if nums[i] < 0:
k = i
else:
break
ans = []
i, j = k, k + 1
while i >= 0 or j < n:
if i < 0:
ans.append()
j += 1
elif j==n:
ans.append(nums[i] * nums[i])
elif :
ans.append(nums[i] * nums[i])
i -= 1
else:
ans.append(nums[j] * nums[j])
j += 1
print(ans)
def isprime(n): #判断n是不是素数
for i in range():
if n%i==0:
else:
return True
def strsum(word): #统计单词的字母值总和
dic={"a":1,"b":2,"c":3,"d":4,"e":5,"f":6,"g":7,"h":8,"i":9,"j":10,
"k":11,"l":12,"m":13,"n":14,"o":15,"p":16,"q":17,"r":18,"s":19,
"t":20,"u":21,"v":22,"w":23,"x":24,"y":25,"z":26}
n=0
for ch in word:
return n
word=input("请输入一个单词:")
s=strsum(word)
if flag:
print("这是一个素单词")
else:
print("这不是一个素单词")
import random
a=[0,0,0,0,0,0]
for i in range(6):
x = random.randint(1, 10)
if i % 2 == 0:
a[i] = 2 * x + 1
elif x % 2 == 0:
a[i] = x // 2
else:
a[i] = x - 1
执行该程序段后,a[0]~a[5]各元素可能的值是( )
a = int(input("请输入第一个数:"))
b = int(input("请输入第二个数:"))
k = 2
Lcd = 1
while a != 1 or b != 1:
if a % k == 0 or b % k == 0:
if
a = a // k
if
b = b // k
else:
print("最小公倍数为:", Lcd)
方框中的代码由以下四部分组成:
①Lcd = Lcd * k ②k = k + 1 ③b % k == 0 ④a % k == 0
则(1)(2)(3)(4)处代码顺序依次为( )
A | B | C | D |
i=0 while i<=10: print(i) i=i+1 | i=10 while i>0: print(i) i=i-1 | for i in range(10): print(i) | for i in range(10,0,-1): print(i) |
图a
为了梳理产品组件的组装顺序,并计算所有组件安装完成所需的最短时间,编写程序模拟组装过程:先同时组装前置总数为0的组件,完成后更新每个组件的前置总数,再重复以上步骤,直至所有组件安装完毕,程序运行结果如下图b所示,请回答下列问题:
图b
Def cal(a, n):
pre=[0]*n
s=[[0 for i in range(n)] for j in range(n)] #创建n×n的二维数组s,元素初始值为0
for i in range(len(a)):
x, y=a[i][0], a[i][1]
s[x][y]=1
pre[y]=
return pre, s
def proc(n, s, pre):
head=tail=0
que=[0]*n
for I in range(n):
if pre[i]==0:
que[tail]=i
tail+=1
while :
x=que[head]
head+=1
for i in range(n):
if s[x][i]==1:
pre[i]-=1
if pre[i]==0:
que[tail]=i
tail+=1
v[i]=max(v[i], )
return v
"""
组装编号0~n-1的单个组件所需时间存入t列表,组件前置关系存入a列表,如第15题图a所需时间t=[2, 5, 2, 4, 3, 5];a=[[0, 2], [2, 3], [1, 3], [3, 5], [3, 4]]
"""
n=len(t)
print(’编号为0~’+str(n-1)+’的组件组装所需单位时间分别为:’, t)
v=t[:]
pre, s=cal(a, n)
v=proc(n, s, pre)
data=[0]*n
result=[i for I in range(n)] #创建列表result=[0,1,2,……,n-1]
for i in range(n):
data[i]=v[i]-t[i] #data[i]表示组件i开始安装时间
for i in range(n-1): #按组件开始安装时间升序排序,开始安装时间相同时按组件序号升序
for j in range(n-1-i):
if data[result[j]]>data[result[j+1]]:
print(‘组件组装顺序:’, result, ‘, 安装完成所需最短时间:’, max(v))
def fun(x, i):
if x<i:
return i
elif x%i==0:
return x
else:
return fun(x-i, i+1)
执行语句k=fun(37, 3)后,k的值为( )
k=1; lmax=1; n=len(d)
for I in range(1, n):
if d[i]>d[i-1]:
k+=1
else:
if k>lmax:
lmax=k
k=1
print(lmax)
该程序段在某些情况下无法得到符合题意的结果,下列4组数据中能测试出这一问题的是( )