Day4-作业

2019-07-26  本文已影响0人  略略略_29fd

1.读程序,总结程序的功能:
(1).

numbers=1
for i in range(0,20):
 numbers*=2
print(numbers)

答:求2^20的值

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)

答:求100以内能被3,7整除不能被21整除的数的个数。
2.编程实现(for和while各写⼀遍):
(1). 求1到100之间所有数的和、平均值

sum1 = 0
for num in range(1, 101):
    sum1 += num
ave = sum1 / 100
print(sum1, ave)

sum1 = 0
num = 1
while num in range(1, 101):
    sum1 += num
    num += 1
ave = sum1 / 100
print(sum1, ave)

(2). 计算1-100之间能3整除的数的和

sum1 = 0
for num in range(1, 101):
    if num % 3 == 0:
        sum1 += num
print(sum1)

sum1 = 0
num = 1
while num in range(1, 101):
    if num % 3 == 0:
        sum1 += num
    num += 1
print(sum1)

(3). 计算1-100之间不能被7整除的数的和

sum1 = 0
for num in range(1, 101):
    if num % 7 != 0:
        sum1 += num
print(sum1)

sum1 = 0
num = 1
while num in range(1, 101):
    if num % 7 != 0:
        sum1 += num
    num += 1
print(sum1)
  1. 求斐波那契数列中第n个数的值:1,1,2,3,5,8,13,21,34....
a = 1
b = 1
sum1 = 0
n = int(input('请输入一个数'))
for sum1 in range(0, n-2):
    x = a +b
    a = b
    b = x
    sum1 += 1
print(b)

  1. 判断101-200之间有多少个素数,并输出所有素数。判断素数的⽅法:⽤⼀个数分别除2到sqrt(这个
    数),如果能被整除,则表明此数不是素数,反之是素数
import math
for i in range(101, 201):
    # for j in range(2, i):
    for j in range(2, int(math.sqrt(i))):
        if i % j == 0:
            print(i, '不是素数')
            break
    else:
        print(i, '是素数')


# 方法2
import math
for i in range(101, 201):
    flag = True     #假设当前数是素数
    for j in range(2, int(math.sqrt(i))):
         if i % j == 0:
             flag = False # 如果在2-i之间遇到一个能整除的假设不成立
             break
    if flag:
        print(i, '是素数')
  1. 打印出所有的⽔仙花数,所谓⽔仙花数是指⼀个三位数,其各位数字⽴⽅和等于该数本身。例如:153是
    ⼀个⽔仙花数,因为153 = 1^3 + 5^3 + 3^3
for n in range(100,1000):
    if n == (n // 100) ** 3 + (n % 100 // 10)**3 \
            + (n % 100 % 10) ** 3:
        print(n)
  1. 有⼀分数序列:2/1,3/2,5/3,8/5,13/8,21/13...求出这个数列的第20个分数
    分⼦:上⼀个分数的分⼦加分⺟ 分⺟: 上⼀个分数的分⼦ fz = 2 fm = 1 fz+fm / fz
fz = 1
fm = 1
for n in range(20):
    fz, fm = fz + fm, fz
    # sum1 = fz + fm
    # fm = fz
    # fz = sum1
print(fz, '/', fm)
  1. 给⼀个正整数,要求:1、求它是⼏位数 2.逆序打印出各位数字
m = int(input("请输入:"))
string = ""
count = 0
while True:
        string += str(m % 10) + " "
        m = int(m / 10)
        count += 1
        if m == 0:
            break
print("m是" + str(count) + "位数")
print(string)
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