从整数中删除数字

我遇到了一个家庭作业问题:

输入一个大于或等于0的整数。将其乘以5,再将6加到乘积上,再乘以4,再乘以9,再乘以5。然后,删除最终数字的最后两位数字并减去1。输出答案。

这是我的代码:

1  main = do
2      putStrLn "enter a non-negative integer: "
3      input <- getLine
4      let i = (read input :: Int)
5      print ((((i * 5) + 6) * 4 + 9) * 5)/100-1

这里,我试图通过除以100来截断最后两个数字,因为这在其他语言中也适用。 然而,它给了我这个错误,我不确定这意味着什么:

$ runhaskell Computations.hs

Computations.hs:5:5: error:
    • No instance for (Fractional (IO ())) arising from a use of ‘/’
    • In the first argument of ‘(-)’, namely
        ‘print ((((x * 5) + 6) * 4 + 9) * 5) / 100’
      In a stmt of a 'do' block:
        print ((((x * 5) + 6) * 4 + 9) * 5) / 100 - 1
      In the expression:
        do { putStrLn "enter a non-negative integer: ";
             input1 <- getLine;
             let x = (read input1 :: Int);
             print ((((x * 5) + 6) * 4 + 9) * 5) / 100 - 1 }

Computations.hs:5:5: error:
    • No instance for (Num (IO ())) arising from a use of ‘-’
    • In a stmt of a 'do' block:
        print ((((x * 5) + 6) * 4 + 9) * 5) / 100 - 1
      In the expression:
        do { putStrLn "enter a non-negative integer: ";
             input1 <- getLine;
             let x = (read input1 :: Int);
             print ((((x * 5) + 6) * 4 + 9) * 5) / 100 - 1 }
      In an equation for ‘main’:
          main
            = do { putStrLn "enter a non-negative integer: ";
                   input1 <- getLine;
                   let x = ...;
                   .... }

那么,有人能解释一下错误的含义吗?

有没有其他解决方案比除数法更好?

如何删除任意位置的数字?例如,从整数中删除第一、第二和第六个数字?

ps:如果操作正确,输入和输出应该匹配。

更新:我已将“/”更改为“div”,它给了我以下信息:

Computations.hs:5:12: error:
    • Couldn't match expected type ‘(Integer -> Integer -> Integer)
                                    -> Integer -> Integer’
                  with actual type ‘Int’
    • The function ‘(((i * 5) + 6) * 4 + 9) * 5’
      is applied to two arguments,
      but its type ‘Int’ has none
      In the first argument of ‘(-)’, namely
        ‘((((i * 5) + 6) * 4 + 9) * 5) div 100’
      In the first argument of ‘print’, namely
        ‘(((((i * 5) + 6) * 4 + 9) * 5) div 100 - 1)’

这一部分很奇怪:“函数‘(((i*5)+6)*4+9)*5‘应用于两个参数”,为什么haskell将其解释为函数?