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周末作业-循环练习题(2)
2022-08-05 05:13:00 【非鱼丶丶】
- 判断101-200之间有多少个素数,并输出所有素数。
count = 0
for num in range(100, 201):
for x in range(2, int(num ** 0.5 +1)):
if num % x == 0:
break
else:
count += 1
print(num)
print(count)
- 求整数1~100的累加值,但要求跳过所有个位为3的数。
sum1 = 0
for x in range(1 , 100):
if x % 10 != 3:
sum1 += x
print(sum1)
- 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
a = 2
b = 1
for x in range(1, 20):
a, b = a + b, a
print(a, '/', b, sep='')
- 写程序计算 n 的阶乘 n! 的结果
qua1 = 1
n = int(input('求n的阶乘'))
for x in range(1, n + 1):
qua1 *= x
print(qua1)
- 求1+2!+3!+…+20!的和
qua1 = 1
sum1 = 0
for n in range(1, 20):
qua1 *= n
sum1 += qua1
print(sum1)
# 方法2
sum1 = 0
for n in range(1, 20):
for x in range(1, n):
n *= x
sum1 += n
print(sum1)
写程序求表达式 a + aa + aaa + aaaa+ … 的结果,其中a是1~9的数字,求和的项数用n来控制。(a和n可以用变量来表示)
例如:a为3, n为5的时候: 3 + 33 + 333 + 3333 + 33333
n = int(input('n'))
a = int(input('a'))
i = 0
for x in range(1, n+1):
i = a * 10**(x - 1) + i
if x == n:
print(i)
else:
print(i, '+', sep='', end='')
# 方法2
n = int(input("n"))
a = input("a")
for x in range(1, n+1):
res = a * x
if x == n:
print(res)
else:
print(res, "+", sep="", end="")
控制台输出三角形
a.根据n的值的不同,输出相应的形状 n = 5时 n = 4 ***** **** **** *** *** ** ** * * b.根据n的值的不同,输出相应的形状(n为奇数) n = 5 n = 7 * * *** *** ***** ***** ******* c. 根据n的值的不同,输出相应的形状 n = 4 1 121 12321 1234321 n = 5 1 121 12321 1234321 123454321小明单位发了100元的购物卡,小明到超市买三类洗化用品,洗发水(15元),香皂(2元),牙刷(5元)。要把100元正好花掉,可有哪些购买结合?
for shampoo in range(1, 7):
for soap in range(1, 51):
for brush in range(1, 21):
if shampoo * 15 + soap * 2 + brush * 5 == 100:
print('洗发水', shampoo, '香皂', soap, '牙刷', brush)
- 一张纸的厚度大约是0.08mm,对折多少次之后能达到珠穆朗玛峰的高度(8848.13米)?
x = 0.00008
n = 0
while x <= 8848.13:
x = x * 2
n += 1
print(n)
- 古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少?
- 将一个正整数分解质因数。例如:输入90,打印出90=2x3x3x5。
num = int(input("整数"))
x = 2
print(num, '=', sep='', end='')
while num != 1:
if num % x == 0:
num = num / x
if num == 1:
print(x, sep='', end='')
else:
print(x, '*', sep='', end='')
else:
x += 1
- 某个公司采用公用电话传递数据,数据是四位的整数,在传递过程中是加密的,加密规则如下:每位数字都加上5,然后用和除以10的余数代替该数字,再将第一位和第四位交换,第二位和第三位交换。求输入的四位整数加密后的值
num =int(input("整数"))
a = num // 1000
b = num // 100 % 10
c = num % 100 // 10
d = num % 10
a = (a + 5) % 10
b = (b + 5) % 10
c = (c + 5) % 10
d = (d + 5) % 10
print(d, c, b, a, sep='')
# fang 2
num =input("整数")
total = ""
for item in str(num):
res = str((int(item)+5) % 10)
total = res + total
print(int(total))
- 本金10000元存入银行,年利率是千分之三。每过1年,将本金和利息相加作为新的本金。计算5年后,获得的本金是多少。
x = 10000
for i in range(1, 6):
x = x + x * 0.003
print(x)
- 输入一个整数,计算它各位上数字的和。(注意:输入的整数可以是任意位)
sum1 = 0
x = input('整数')
n = len(x)
for i in range(0, n):
sum1 += int(x[i])
print(sum1)
- 求两个数的最大公约数和最小公倍数。(提示:公约数一定小于等于两数中的小的那个数,且能同时被两个数整除;公倍数一定大于等于两数中的大数,且是大数的倍数又能被两数中的小数整除)
a = int(input("a"))
b = int(input("b"))
c = b if a >= b else a
# if a >= b:
# c = b
# else:
# c = a
for x in range(c, 0, -1):
if a % x == 0 and b % x == 0:
print(x)
res1 = a / x
res2 = b / x
res = res1 * res2 * x
print(res)
break
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