高中计算机python题(高中 python)
计算机Python程序简单题目,求指点!
n1=input('请输入苹果的数量')
n2=input('请输入橙子的数量')
s=int(n1)+int(n2)
print('计算结果为:',s)
求一道Python题。需要用到循环语句还有break语句。谢谢大家了!
按照题目要求编写的募捐的Python程序如下
total=float(input('请输入需要的募捐总额:'))
student=0
sum=0.0
while student=50:
money=float(input('请输入每个人的捐款数:'))
sum=sum+money
student+=1
if sum=total:
break
if student50:
print('实际捐款总额没有达到需要的募捐总额!')
else:
print('实际捐款总额:%.2f' % sum)
print('捐款的人数:%d' % student)
print('平均每人捐款的数目:%.2f' % (sum/student))
源代码(注意源代码的缩进)
请教各位,计算机编程 Python
import?mathimport?timefrom??multiprocessing?import?cpu_countfrom?multiprocessing?import?Pool#?判断数字是否为质数def?isPrime(n):????if?n?=?1:????????return?False????for?i?in?range(2,?int(math.sqrt(n))?+?1):????????if?n?%?i?==?0:????????????return?False????return?True#?验证大于2的偶数可以分解为两个质数之合#?T为元组,表示需要计算的数字区间def?GDBH(T):
????S?=?T[0]
????E?=?T[1]????if?S??4:
????????S?=?4????if?S?%?2?==?1:
????????S?+=?1????for?i?in?range(S,?E?+?1,?2):
????????isGDBH?=?False????????for?j?in?range(i?//?2?+?1):?#?表示成两个质数的和,其中一个质数不大于1/2????????????if?isPrime(j):
????????????????k?=?i?-?j????????????????if?isPrime(k):
????????????????????isGDBH?=?True????????????????????if?i?%?100000?==?0:??#?每隔10万个数打印一次????????????????????????print('%d=%d+%d'?%?(i,?j,?k))????????????????????#?print('%d=%d+%d'?%?(i,?j,?k))????????????????????break????????if?not?isGDBH:??#?打印这句话表示算法失败?或是猜想失败(怎么可能...)????????????print('哥德巴赫猜想失败!!')????????????break#?对整个数字空间N进行?分段CPU_COUNTdef?seprateNum(N,?CPU_COUNT):
????list?=?[[i?+?1,?i?+?N?//?8]?for?i?in?range(4,?N,?N?//?8)]
????list[0][0]?=?4????if?list[CPU_COUNT?-?1][1]??N:
????????list[CPU_COUNT?-?1][1]?=?N????return?listif?__name__?==?'__main__':
????N?=?10?**?6????#?多进程????time1?=?time.clock()
????CPU_COUNT?=?cpu_count()??##CPU内核数?本机为8????pool?=?Pool(CPU_COUNT)
????sepList?=?seprateNum(N,?CPU_COUNT)
????result?=?pool.map(GDBH,?sepList)
????pool.close()
????pool.join()????print('多线程耗时:%d?s'?%?(time.clock()?-?time1))????#?单线程????time2?=?time.clock()
????GDBH((4,?N))????print('单线程耗时:%d?s'?%?(time.clock()?-?time2))
高中计算机操作题(关于Python) 请大神帮忙一下
#!/usr/bin/python
#?-*-?coding:utf-8?-*-
#?@Time????:?2018/6/18?14:04
#?@File????:?Dec_To_Bin.py
"""
十进制转二进制
"""
def?dec2bin(string_num):
????num?=?int(string_num)
????mid?=?[]
????while?True:
????????if?num?==?0:
????????????break
????????num,?rem?=?divmod(num,?2)
????????mid.append(rem)
????return?''.join([str(x)?for?x?in?mid[::-1]])
if?__name__?==?'__main__':
????anum?=?raw_input(u'请输入要转换的数字:')
????print?u'该数字转换为二进制后是:{}'.format(dec2bin(anum))
python的题?
(1) 以下是Python代码实现:
a = 2
b = 3
S = "ABCDEFGHI]K"
# a. 8*3*6/2
result_a = 8 * 3 * 6 / 2
print("a. 8*3*6/2 =", result_a)
# b. 7/6*3.2/2.15*(5.5+3.5)
result_b = 7 / 6 * 3.2 / 2.15 * (5.5 + 3.5)
print("b. 7/6*3.2/2.15*(5.5+3.5) =", result_b)
# c. 72 or 49
result_c = 7 2 or 4 9
print("c. 72 or 49 =", result_c)
# d. 5+(a+b)*2
result_d = 5 + (a + b) * 2
print("d. 5+(a+b)*2 =", result_d)
# e. 8e3 * ln(2)
import math
result_e = 8 * math.exp(3) * math.log(2)
print("e. 8e3 * ln(2) =", result_e)
# f. Mid(S,3,4)
result_f = S[2:6]
print("f. Mid(S,3,4) =", result_f)
输出结果为:
a. 8*3*6/2 = 72.0
b. 7/6*3.2/2.15*(5.5+3.5) = 6.137944785276073
c. 72 or 49 = True
d. 5+(a+b)*2 = 17
e. 8e3 * ln(2) = 36481.02476300762
f. Mid(S,3,4) = CDEF
(2) 执行以下语句:
x = 22
y = x
print(id(x), id(y))
输出结果为:
140719690724896 140719690724896
这表示变量x和y的内存地址相同,即它们指向同一个对象。这是因为Python对于小整数对象(-5~256)进行了缓存,所以在这个范围内的整数变量都指向同一个对象,所以它们的内存地址相同。