**-1

2018-11-09  本文已影响0人  憧憬001

读程序,总结程序的功能:

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

求2的20次方的结果

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)

记录1~100中能被3或7整除但不能被21整除的数的个数

编程实现(for和while各写⼀一遍):
# for 循环
sum1 = 0  # 存储和
avg1 = 0    # 存储平均值
num1 = 0   # 记录次数
for i in range(101):
    sum1 += i
    num1 += 1
print("1到100之间所有数的和: %d" % sum1)
print("1到100之间所有数的平均值:%.2f" % (sum1 / num1))

# while 循环
sum2 = 0
i = 1
while i <= 100:
    sum2 += i
    i += 1
print("1到100之间所有数的和: %d" % sum2)
print("1到100之间所有数的平均值: %.2f" % (sum2/i))

>>>>
1到100之间所有数的和: 5050
1到100之间所有数的平均值:50.00
1到100之间所有数的和: 5050
1到100之间所有数的平均值: 50.00
# for循环
sum1 = 0   # 存储和
for i in range(101):
    if i % 3 == 0:
        sum1 += i
print("1-100之间能3整除的数的和: %d" % sum1)

# while 循环
sum2 = 0
i = 1
while i <= 100:
    if i % 3 == 0:
        sum2 += i
    i += 1
print("1-100之间能3整除的数的和: %d" % sum2)

>>>>
1-100之间能3整除的数的和: 1683
1-100之间能3整除的数的和: 1683

# for循环
sum1 = 0   # 存储和
for i in range(101):
    if i % 7 != 0:
        sum1 += i
print("1-100之间不不能被7整除的数的和: %d" % sum1)

# while 循环
sum2 = 0
i = 1
while i <= 100:
    if i % 7 != 0:
        sum2 += i
    i += 1
print("1-100之间不不能被7整除的数的和: %d" % sum2)

>>>>
1-100之间不不能被7整除的数的和: 4315
1-100之间不不能被7整除的数的和: 4315
4.求斐波那契数列列中第n个数的值:1,1,2,3,5,8,13,21,34....
num = int(input("请输入你需要查阅斐波那契数列列中第n个数的值"))
# n1,n2记录最开始的两个数
n1 = 1
n2 = 1
result = 0
if num <= 0:
    print("请输入正整数")
elif num == 1:
    print("斐波那契数列列中第1个数的值是1")
elif num == 2:
    print("斐波那契数列列中第2个数的值是1")
else:
    for _ in range(2, num):
        result = n1 + n2
        n1 = n2
        n2 = result
    print("斐波那契数列列中第%d个数的值是:%d" % (num, result))

# 简写
n1 = 1
n2 = 1
result = 1
num = int(input("请输入n的值"))
for _ in range(3, num + 1):
    result = n1 + n2
    n1, n2 = n2, result

print("斐波那契数列中第%d个数的值是:%d" % (num, result))
5.判断101-200之间有多少个素数,
for i in range(101, 201):
    for j in range(2, i):
        if i % j == 0:
            break
    else:
        print(i)
6. 打印出所有的水仙花数,所谓⽔仙花数是指一个三位数,
for i in range(100, 1000):
    gewei = i % 10
    shiwei = i // 10  % 10
    baiwei = i // 100 % 10
    if gewei**3 + shiwei**3 + baiwei**3 == i:
        print(i)
# 方法二
for i in range(100, 1001):
    if int(str(i)[0])**3 + int(str(i)[1])**3 + int(str(i)[2])**3 == i:
        print(i)
>>>>
153
370
371
407
7.有一分数序列:2/1,3/2,5/3,8/5,13/8,21/13...
fz = temp = 2
fm = 1
for _ in range(1, 20):

    fz += fm
    fm = temp
    temp = fz
print("%d/%d" % (fz, fm))

>>>>
17711/10946
8.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
num = input("请输入一个正整数")
len1 = len(num)
print("%s是%d位数" % (num, len1))
print(num[::-1])
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