1.輸出2的20次方
2.輸出0到100之間能被3或7整除疗疟，但不能被3或7同時整除的數(shù)字
1.求1到100之間所有數(shù)的和如失、平均值
num = 1
sum1 = 0
for i in range(1,101):
sum1+=num
num+=1
print("所有數(shù)的和",sum1)
print("平均值",sum1/100)
sum_1 = 0
num = 1
while num<=100:
sum_1+=num
num+=1
print("所有數(shù)的和",sum_1)
print("平均值",sum_1/100)
2.計算1-100之間能3整除的數(shù)的和
num = 1
sum2 = 0
for i in range(1,101):
if num % 3 == 0:
sum2 += num
num +=1
print("能3整除的數(shù)的和為",sum2)
num = 1
sum_2 = 0
while num<101:
if num % 3 == 0:
sum_2 +=num
num+=1
print("能3整除的數(shù)的和為",sum2)
3. 計算1-100之間不能被7整除的數(shù)的和
num = 1
sum3 = 0
for i in range(1,101):
if num % 7 != 0:
sum3 += num
num+=1
print("不能被7整除的數(shù)的和",sum3)
num = 1
sum_3 = 0
while num<101:
if num % 7 != 0:
sum_3 += num
num+=1
print("不能被7整除的數(shù)的和",sum3)
求斐波那契數(shù)列中第n個數(shù)的值：1，1限嫌，2靴庆，3，5怒医，8炉抒，13，21稚叹，34....
a1 = 1
a2 = 1
for i in range(8):
t=a1
a1 = a2
a2 = t+a2
print(a1)
判斷101-200之間有多少個素數(shù)焰薄，并輸出所有素數(shù)拿诸。
num = 2
count = 0
for i in range(101,200):
for j in range(2,i):
if i % j == 0:
break
else:
count+=1
print("素數(shù)",i)
print(count)
打印出所有的水仙花數(shù),所謂水仙花數(shù)是指一個三位數(shù)，其各位數(shù)字立方和等于該數(shù)本身塞茅。例如：153是1個水仙花數(shù),因為153 = 1^3 + 5^3 + 3^3
for num in range(100,1000):
if (num%10)**3+(num//10%10)**3+(num//100)**3==num:
print("水仙花",num)
有分?jǐn)?shù)序列：2/1,3/2,5/3,8/5,13/8,21/13...求出這個數(shù)列的第20個分?jǐn)?shù) 分子：上一個分?jǐn)?shù)的分子加分母 分母: 上一個分?jǐn)?shù)的分子fz = 2 fm = 1 fz+fm /fz
x = 2
y = 1
for i in range(1,20):
t = x
print(t)
x = x+y
print(x)
y= t
print(x,"/",y)
給一個正整數(shù)亩码，要求：1、求它是幾位數(shù) 2.逆序打印出各位數(shù)字
num = int(input('輸入一個數(shù)'))
count = 0
while num>0:
n = num%10
print(n)
count+=1
print("位數(shù)",count) num =int(num/10)
