day 04 作业 循环和分之
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day 04 作业 循环和分之
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# 讀程序,總結(jié)程序的功能:
"""
1.求2 的20次方
numbers=1
for i in range(0,20):
numbers*=2
print(numbers)
2.求100以內(nèi)能整除3 或 7的數(shù),并21不能整除的數(shù)
summation = 0
num = 1
while num<=100:
if (num%3==0 or num%7==0) and num%21!=0:
summation += 1
num +=1
print(summation)
"""
# 1. 求1到100之間所有數(shù)的和、平均值.
# #for
"""
num = 0
for x in range(1, 101, 1):
num += x
print(num)
print(num/x)
#while
b = 0
a = 0
while a <= 99:
a += 1
b += a
print(b)
print(b/100)
"""
#2. 計(jì)算1-100之間能3整除的數(shù)的和
"""
#for
a = 0
for x in range(1, 101, 1):
if x % 3 == 0:
a += x
print(a)
#while
b = 0
a = 1
while a <= 100:
if a % 3 == 0:
b += a
a += 1
print(b)
"""
#3. 計(jì)算1-100之間不不能被7整除的數(shù)的和
"""
#for
a = 0
for x in range(1, 101, 1):
if x % 7 != 0:
a += x
print(a)
#while
b = 0
a = 1
while a <= 100:
if a % 7 != 0:
b += a
a += 1
print(b)
"""
# 稍微困難
# 1. 求斐波那契數(shù)列列中第n個(gè)數(shù)的值:1,1,2,3,5,8,13,21,34....
"""
d = 0
c = 0
b = 1
_ = 1
while _ <= 3:
d = c + b
print(d)
b = d + c
print(b)
c = b + d
print(c)
_ += 1
"""
# 2. 判斷101-200之間有多少個(gè)素?cái)?shù),并輸出所有素?cái)?shù)。判斷素?cái)?shù)的方法:用一個(gè)數(shù)分別除2到sqrt(這個(gè)
#數(shù)),如果能被整除,則表明此數(shù)不不是素?cái)?shù),反之是素?cái)?shù)
""" a = 0 for fan_wei in range(101, 200, 1):
j = 0
for num in range(2, fan_wei,1):
if fan_wei % num == 0:
j = 1
break
if j == 0:
print(fan_wei)
a += 1
print(a)
"""
# 3. 打印出所有的水仙花數(shù),所謂水仙花數(shù)是指一個(gè)三位數(shù),其各位數(shù)字立方和等于該數(shù)本身。例例如:153是
# 一個(gè)水仙花數(shù),因?yàn)?53 = 1^3 + 5^3 + 3^3
"""
for f in range(100, 1000):
a = f % 10
b = f // 100
c = f // 10 % 10
if f == a**3 + b**3 + c**3:
print(f)
"""
# 4. 有一分?jǐn)?shù)序列:2/1,3/2,5/3,8/5,13/8,21/13...求出這個(gè)數(shù)列的第20個(gè)分?jǐn)?shù)
# 分子:上一個(gè)分?jǐn)?shù)的分子加分母 分母: 上一個(gè)分?jǐn)?shù)的分子 fz = 2 fm = 1 fz+fm / fz
"""
time = 0
fz = 2
fm = 1
for a in range(0, 19):
fz = fz + fm
fm = fz - fm
print(fz,"/",fm)
time += 1
print(time)
"""
# 5. 給一個(gè)正整數(shù),要求:1、求它是幾位數(shù) 2.逆序打印出各位數(shù)字
"""time = 0
b = 55414154351
while b > 0:
n = b % 10
print(n)
b //= 10
time += 1
print(time,"個(gè)數(shù)")
"""
"""
1.求2 的20次方
numbers=1
for i in range(0,20):
numbers*=2
print(numbers)
2.求100以內(nèi)能整除3 或 7的數(shù),并21不能整除的數(shù)
summation = 0
num = 1
while num<=100:
if (num%3==0 or num%7==0) and num%21!=0:
summation += 1
num +=1
print(summation)
"""
# 1. 求1到100之間所有數(shù)的和、平均值.
# #for
"""
num = 0
for x in range(1, 101, 1):
num += x
print(num)
print(num/x)
#while
b = 0
a = 0
while a <= 99:
a += 1
b += a
print(b)
print(b/100)
"""
#2. 計(jì)算1-100之間能3整除的數(shù)的和
"""
#for
a = 0
for x in range(1, 101, 1):
if x % 3 == 0:
a += x
print(a)
#while
b = 0
a = 1
while a <= 100:
if a % 3 == 0:
b += a
a += 1
print(b)
"""
#3. 計(jì)算1-100之間不不能被7整除的數(shù)的和
"""
#for
a = 0
for x in range(1, 101, 1):
if x % 7 != 0:
a += x
print(a)
#while
b = 0
a = 1
while a <= 100:
if a % 7 != 0:
b += a
a += 1
print(b)
"""
# 稍微困難
# 1. 求斐波那契數(shù)列列中第n個(gè)數(shù)的值:1,1,2,3,5,8,13,21,34....
"""
d = 0
c = 0
b = 1
_ = 1
while _ <= 3:
d = c + b
print(d)
b = d + c
print(b)
c = b + d
print(c)
_ += 1
"""
# 2. 判斷101-200之間有多少個(gè)素?cái)?shù),并輸出所有素?cái)?shù)。判斷素?cái)?shù)的方法:用一個(gè)數(shù)分別除2到sqrt(這個(gè)
#數(shù)),如果能被整除,則表明此數(shù)不不是素?cái)?shù),反之是素?cái)?shù)
""" a = 0 for fan_wei in range(101, 200, 1):
j = 0
for num in range(2, fan_wei,1):
if fan_wei % num == 0:
j = 1
break
if j == 0:
print(fan_wei)
a += 1
print(a)
"""
# 3. 打印出所有的水仙花數(shù),所謂水仙花數(shù)是指一個(gè)三位數(shù),其各位數(shù)字立方和等于該數(shù)本身。例例如:153是
# 一個(gè)水仙花數(shù),因?yàn)?53 = 1^3 + 5^3 + 3^3
"""
for f in range(100, 1000):
a = f % 10
b = f // 100
c = f // 10 % 10
if f == a**3 + b**3 + c**3:
print(f)
"""
# 4. 有一分?jǐn)?shù)序列:2/1,3/2,5/3,8/5,13/8,21/13...求出這個(gè)數(shù)列的第20個(gè)分?jǐn)?shù)
# 分子:上一個(gè)分?jǐn)?shù)的分子加分母 分母: 上一個(gè)分?jǐn)?shù)的分子 fz = 2 fm = 1 fz+fm / fz
"""
time = 0
fz = 2
fm = 1
for a in range(0, 19):
fz = fz + fm
fm = fz - fm
print(fz,"/",fm)
time += 1
print(time)
"""
# 5. 給一個(gè)正整數(shù),要求:1、求它是幾位數(shù) 2.逆序打印出各位數(shù)字
"""time = 0
b = 55414154351
while b > 0:
n = b % 10
print(n)
b //= 10
time += 1
print(time,"個(gè)數(shù)")
"""
轉(zhuǎn)載于:https://www.cnblogs.com/liyaobin/p/9884233.html
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