前言

这些代码都是校本教材上面的习题，有的代码我可能偷懒就直接用轮子了，不保证和老师要求的一模一样。

本文将持续更新

第一节

T1

依次打出结果即可

a = int(input())
print(a + 8, int((a + 8) / 5), int((a + 8) / 5 - 10), int(((a + 8) / 5 - 10) * 10))

main = do
    a :: Int <- fmap read getLine
    putStrLn (show (a + 8) ++ " " ++ show ((a + 8) `div` 5) ++ " " ++ show ((a + 8) `div` 5 - 10) ++ " " ++ show (((a + 8) `div` 5 - 10) * 10))

T2

依次打出结果即可，Python 当中记得将分隔符 sep 改为逗号。

a = int(input())
b = int(input())
print(a + b, a - b, a * b, a / b, sep=",")

main = do
    a <- fmap read getLine
    b <- fmap read getLine
    putStrLn (show (a + b) ++ "," ++ show (a - b) ++ "," ++ show (a * b) ++ "," ++ show (a / b))

T3

按要求格式打出即可。

a = int(input())
b = int(input())
print(str(a) + "+" + str(b) + "=" + str(a + b))

main = do
    a <- fmap read getLine
    b <- fmap read getLine
    putStrLn (show a ++ "+" ++ show b ++ "=" ++ show (a + b))

T4

按要求分行打出。

a = int(input())
b = int(input())
c = int(input())
print(a * b, a * c, b * c)
print(a * b * c)

main = do
    a <- fmap read getLine
    b <- fmap read getLine
    c <- fmap read getLine
    putStrLn (show (a * b) ++ " " ++ show (a * c) ++ " " ++ show (b * c))
    print (a * b * c)

第二节

T1

根据公式计算即可。

r = float(input())
print(2 * 3.14 * r)
print(3.14 * r * r)

main = do
    r <- fmap read getLine
    print (3.14 * r * 2)
    print (3.14 * r * r)

T2

根据公式计算即可。

r = float(input())
print(4 * 3.14 * r)
print(4 * r * r * r * 3.14 / 3)

main = do
    r <- fmap read getLine
    print (4 * 3.14 * r)
    print (4 * r * r * r * 3.14 / 3)

T3

首先如果我们令原数为 a。a 的个位是(a%10)，将其乘上 100；a 的十位是(a//10)%10，将其乘上 10；a 的百位是(a//100)，将其乘上 1：这三个值相加得到的结果，就是 a 百位和个位调换的结果（可以自己模拟一下），然后加上 a 输出就得到答案了。

a = int(input())
print(a + (a % 10) * 100 + (a // 10) % 10 * 10 + a // 100)

main = do
    a <- fmap read getLine
    print (a + (a `mod` 10) * 100 + (a `div` 10) `mod` 10 * 10 + a `div` 100)

T4

以一个正常身份证号码为例，应该是 xxxxxxyyyymmddxxxx 的格式（x 代表其他字符，yyyy 代表年份，mm 代表月份，dd 代表日期），如果我们将这个东西除以 10000 取整，则得到 xxxxxxyyyymmdd，如果再将这个东西模 100000000，即可得到 yyyymmdd，即为所求。

id = int(input())
print(id // 10000 % 100000000)

main = do
    id <- fmap read getLine
    print ((id `div` 10000) `mod` 100000000)

第三节

T1

根据题意判断输出即可，此处我用了三目表达式，格式是：(statement-true) if (condition) else (statement-false)。

a = int(input())
b = int(input())
print(1 if a > b else 2 if a < b else 3)

main = do
    a :: Int <- fmap read getLine
    b :: Int <- fmap read getLine
    print
        ( if a > b
            then 1
            else
                if a < b
                    then 2
                    else 3
        )

T2

正常的做法应该是打擂法，但是不够快，直接调max()函数就得了。

a = int(input())
b = int(input())
c = int(input())
print(max(a, b, c))

main = do
    [a :: Int, b :: Int, c :: Int] <- traverse (fmap read) [getLine, getLine, getLine]
    print (maximum [a, b, c])

T3

分层输出。

score = int(input())
if score >= 90:
    print("A")
elif score >= 80:
    print("B")
elif score >= 70:
    print("C")
elif score >= 60:
    print("D")
else:
    print("E")

-- along with this program.  If not, see <https://www.gnu.org/licenses/>.
getGrade :: Int -> Char
getGrade score
    | score >= 90 = 'A'
    | score >= 80 = 'B'
    | score >= 70 = 'C'
    | score >= 60 = 'D'
    | otherwise = 'E'

main = do
    score <- fmap read getLine
    putChar $ getGrade score
    putChar '\n'

T4

如果是 Haskell 就直接 sort 后打印就得了，Python 的做法和教材上写的一样。

a = int(input())
b = int(input())
c = int(input())
print(min(a, b, c), a + b + c - min(a, b, c) - max(a, b, c), max(a, b, c))

import Data.List

main = do
    [a :: Int, b :: Int, c :: Int] <- traverse (fmap read) [getLine, getLine, getLine]
    putStrLn (foldl (\cur x -> cur ++ show x ++ " ") [] (sort [a, b, c]))

第四节

T1

和前面一样，我这里用了三目，分开判断输出即可。

a = int(input())
print("YES" if a <= 999 and a >= 100 and a % 2 == 1 else "NO")

main = do
    a <- fmap read getLine
    if a <= 999 && a >= 100 && a `mod` 2 == 1
        then putStrLn "YES"
        else putStrLn "NO"

T2

根据题意打印即可。

lucky = int(input())
good = int(input())
if lucky >= 10 or good >= 20:
    print(1)
else:
    print(2)

main = do
    lucky <- fmap read getLine
    good <- fmap read getLine
    if lucky >= 10 || good >= 20
        then putStrLn "1"
        else putStrLn "0"

T3

我使用了max和min先分别搞出最大最小值，然后综合减去最大最小就是中间值，判断输出即可。

a = int(input())
b = int(input())
c = int(input())
mi = min(a, b, c)
ma = max(a, b, c)
mid = a + b + c - mi - ma
print("YES" if mi + mid > ma else "NO")

import Data.List

main = do
    [a, b, c] <- traverse (fmap read) [getLine, getLine, getLine]
    let chk = sum $ take 2 (sort [a, b, c])
    if c < chk
        then putStrLn "YES"
        else putStrLn "NO"

T4

按照闰年立法判断即可。

year = int(input())
print("YES" if (year % 400 == 0) or (year % 100 != 0 and year % 4 == 0) else "NO")

main = do
    year <- fmap read getLine
    if (year `mod` 400 == 0) || (year `mod` 100 /= 0 && year `mod` 4 == 0)
        then putStrLn "YES"
        else putStrLn "NO"

T5

算出最大最小，减去然后除以二输出即可。

a = int(input())
b = int(input())
c = int(input())
d = int(input())
print((a + b + c + d - min(a, b, c, d) - max(a, b, c, d)) / 2)

main = do
    [a, b, c, d] <- traverse (fmap read) [getLine, getLine, getLine, getLine]
    print ((a + b + c + d - maximum [a, b, c, d] - minimum [a, b, c, d]) / 2)

第五课

T1

先利用三目判断一下奇偶性，如果是奇数设为 n-1，如果是偶数则保留，然后以-2 为步进向前循环。

n = int(input())
for x in range(n - 1 if n % 2 == 1 else n, 1, -2):
    print(x)

main = do
    n <- fmap read getLine
    let realN = n `div` 2
    putStrLn $ foldl (\cur x -> cur ++ show x ++ " ") "" [x * 2 | x <- [realN, realN - 1 .. 1]]

T2

for 输出[a,b)之间所有数，以逗号隔开，然后对 b 进行特殊处理（防止末尾多一个逗号）

a = int(input())
b = int(input())
for x in range(a, b):
    print(x, end=",")
print(b)

main = do
    [a, b] <- traverse (fmap read) [getLine, getLine]
    putStr $ foldl (\cur x -> cur ++ show x ++ ",") "" [a, a + 1 .. b - 1]
    print b

T3

可以依次以对应的值为步进进行 for，但是我这里不太一样，我是枚举 size 的系数，然后对应相乘后相加。

begin = int(input())
size = int(input())
for x in range(1, 10):
    print(begin + size * x, end=",")
print(begin + size * 10)

main = do
    [begin, dep] <- traverse (fmap read) [getLine, getLine]
    putStr $ foldl (\cur x -> cur ++ show x ++ ",") "" [begin, begin + dep .. begin + 9 * dep]
    print (begin + 10 * dep)

T4

暴力办法，对 x 的因子进行一个个判断。

当然，这不是最好的方法，只是在学考（甚至选考）当中最简洁明了的办法。理论上由于因子成对出现，所以我只要循环到$\sqrt{n}$即可，但是为了保证顺序输出，这样子就需要存储结构暂时未学，我也就不提了。

n = int(input())
for x in range(2, n):
    if n % x == 0:
        print(x)

main = do
    n <- fmap read getLine
    putStrLn $ foldl (\cur x-> cur++show x++" ") "" [x|x<-[2..n-1],n`mod`x==0]

T5

判定素数首先也只要判断到$\sqrt{n}$就行了，然后这个方法仍然不够快（当然绝对准确），更加快的方法应该是使用 Miller-Rabin 素性检验（可能会有例外，但是根据生日悖论意外应该是很少很少的）。还有类似 Pollard-Rho 加上各种各样的消环等等，不过那些都是 OI 的内容了我也就提一下下。

n = int(input())
for x in range(2, n):
    if n % x == 0:
        print("NO")
        exit()
print("YES")

main = do
    n <- fmap read getLine
    if null  [x|x<-[2..n-1],n`mod`x==0]
        then putStrLn "YES"
    else putStrLn "NO"

第六课

T1

while 死除就行

n = int(input())
cnt = 0
while n > 0:
    n = n // 10
    cnt = cnt + 1
print(cnt)

log10 :: Int -> Int
log10 0 = 0
log10 a = log10 (a `div` 10) + 1

main = do
    a <- fmap read getLine
    print (log10 a)

T2

n = int(input())
x = 0
while n % (2 ** (x + 1)) == 0:
    x = x + 1
print(x)

func :: Int -> Int -> Int
func n a =
    if n `mod` (2 ^ (a + 1)) == 0
        then func n (a + 1)
        else a

main = do
    a <- fmap read getLine
    print (func a 1)

T3

n = int(input())
sum = 0
while n > 0:
    sum = sum + n % 10
    n = n // 10
print(sum)

numSum :: Int -> Int
numSum 0 = 0
numSum n = numSum (n `div` 10) + (n `mod` 10)

main = do
    a <- fmap read getLine
    print $ numSum a

T4

n = int(input())
rev = 0
while n > 0:
    rev = rev * 10
    rev = rev + n % 10
    n = n // 10
print(rev)

main = do
    a <- getLine
    putStrLn $ reverse a

暑假作业

T1

n = int(input())
for x in range(2, n):
    if n % x == 0:
        print(x)

T2

# 由于题设要求，我这段代码用了搜索法，但是这不是个好办法，正确的方法应该用辗转相除
a = int(input())
b = int(input())
m = min(a, b)
ans = 0
for x in range(1, m):
    if (a % x == 0) & (b % x == 0):
        ans = max(ans, x)
print(ans, end="\n")

T3

# Note: This is a O(n) program to judge a prime(Obviously correct), but
# Miller-Rabin Algorithm(May not correct) will be a better choice
n = int(input())
for x in range(2, n):
    if n % x == 0:
        print("NO")
        exit()
print("YES")

T4

n = int(input())
sum = 0
for x in range(1, n):
    if n % x == 0:
        sum += x
if sum == n:
    print("YES")
else:
    print("NO")

T5

n = int(input())
for x in range(100, n + 1):
    if (
        (x // 100) * (x // 100) * (x // 100)
        + (x // 10 % 10) * (x // 10 % 10) * (x // 10 % 10)
        + (x % 10) * (x % 10) * (x % 10)
    ) == x:
        print(x)

T6

a = int(input())
b = int(input())
cnt = 0
for x in range(a, b + 1):
    if x // 100 == 1:
        cnt += 1
    if x // 10 % 10 == 1:
        cnt += 1
    if x % 10 == 1:
        cnt += 1
print(cnt)

T7

n = int(input())
flag = True
ans = 0
for x in range(1, n + 1):
    if flag:
        ans += x
        flag = False
    else:
        ans -= x
        flag = True
print(ans)

T8

# This algorithm is naive. Use Euler's Sieve or Eratosenes' Sieve will be a better choice
n = int(input())
for x in range(2, n):
    if n % x == 0:
        flag = True
        for y in range(2, x):
            if x % y == 0:
                flag = False
                break
        if flag:
            print(x, end=" ")
print()

T9

l1 = 1
l2 = 1
n = int(input())
for x in range(1, n):
    tmp = l1 + l2
    l1 = l2
    l2 = tmp
print(l1)

T10

def sol(lis, res: int, cnt: int):
    if cnt > 5:
        return
    sum = 0
    for x in lis:
        sum += x
    if sum > res:
        return
    if cnt == 5 and sum == res:
        print(lis)
        return
    sol(lis + [10], res, cnt + 1)
    sol(lis + [5], res, cnt + 1)
    sol(lis + [2], res, cnt + 1)
    sol(lis + [1], res, cnt + 1)


sol([], 7, 0)

T11

def sol(res: int, lis):
    if res == 0:
        print(lis)
        return
    if res < 5:
        return
    sol(res - 5, lis + [5])
    sol(res - 12, lis + [12])


sol(120, [])

T12

n: int = int(input())
k: int = int(input())
ans: int = 0
while n > k:
    ans += n
    n = n % k + n // k
print(ans)

T13

a = int(input())
Nega = False
if a < 0:
    Nega = True
    a = -a
while a % 10 == 0:
    a //= 10
lis = []
while a > 0:
    lis.append(a % 10)
    a //= 10
if Nega:
    print("-", end="")
for x in lis:
    print(x, end="")
print()

T14

def gcd(a: int, b: int) -> int:
    if a == 0:
        return b
    elif b == 0:
        return a
    else:
        return gcd(b, a % b)


a = int(input())
b = int(input())
print(gcd(a, b))

T15

n = int(input())
for x in range(1, n + 1):
    for y in range(n - x):
        print(end=" ")
    for y in range(1, 2 * x):
        print(y, end="")
    print()

T16

n = int(input())
for x in range(1, n + 1):
    for y in range(n - x):
        print(end=" ")
    for y in range(2 * x - 1):
        print("*", end="")
    print()
for x in range(n - 1, 0, -1):
    for y in range(n - x):
        print(end=" ")
    for y in range(2 * x - 1):
        print("*", end="")
    print()

LICENSE

本文出现的所有代码均以 GPLv3 开源。

Copyright (C) 2023  Andy Shen
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program.  If not, see <https://www.gnu.org/licenses/>.