How to shorten the python code?

Question

I'm writing a program that takes a number from input (call it x), loops from 0 to (x - 1) (let's call our iterator y) and prints \$y^x+x\$ for each y.

After the loop, it then prints "SUM: N" (without a newline) where the N is the sum of the all of the previously printed values.

How can I make the code I already have shorter?

Code (Python 3, 98 bytes):

E=print
A=int(input())
B=[]
for D in range(A):C=D**A+A;B.append(C);E(C)
E(end='SUM: '+str(sum(B)))

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Answer 1

Just a different approach to reduce your code to 86 bytes. You don't need to rename the print function. The only thing you need is to first compute all the values and only then print them:

A=int(input())
B=[D**A+A for D in range(A)]
print(*B,f'SUM: {sum(B)}',sep='\n',end='')

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Answer 2

My attempt, 86 79 bytes

E=print
A=int(input())
E(end='SUM: %d'%sum(E(C:=D**A+A)or C for D in range(A)))

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Thanks to dingledooper for shaving off 7 bytes by removing the need for B.

We embed the assignment in a list comprehension for -9 bytes. print returns a falsy value, so E(C:=D**A+A)or C returns C which is collected into the generator E(C:=D**A+A)or C for D in range(A) that is passed to sum.

We also format the output with a %d to save bytes off str.

Answer 3

Doing it in a list comprehension is a waste of time, it seems, mainly since you have to have a separate print statement for the final sum (due to the no-newline limit).

97 bytes:

E=print
f=lambda x:[E(g:=i**x+x)or g for i in range(x)]
E(end='SUM: '+str(sum(f(int(input())))))

Instead, there's a couple changes I noted.

E=print
A=int(input())
B=[]
for D in range(A):C=D**A+A;B.append(C);E(C)
E(end='SUM: '+str(sum(B)))

First, you can do B+=[C] instead of B.append(C). That loses 5 bytes for 93. But we can also just use B as an integer and add - that saves us from having to do a sum later, so we can shave off a total of 11 for a total of 85.

E=print
A=int(input())
B=0
for D in range(A):C=D**A+A;B+=C;E(C)
E(end='SUM: '+str(B))

We can then do an inline variable assignment when printing D**A+A to save bytes as well, from C=D**A+A;B+=C;E(C) to E(G:=D**A+A);B+=G. This saves one byte, for 84 bytes.

E=print
A=int(input())
B=0
for D in range(A):E(G:=D**A+A);B+=G
E(end='SUM: '+str(B))

Lastly, calling str is long and unwieldy. There's no backticks to call repr like in Python 2, but we can just use autocasting print arguments to do it anyway. From E(end='SUM: '+str(B)) to just E('SUM: ',B,end=''). We add two single quotes and a comma, but lose out on 5 bytes from the str(), for a total of -2 and a final byte count of 83.

E=print
B=0
A=int(input())
for D in range(A):E(G:=D**A+A);B+=G
E('SUM: ',B,end='')

Answer 4

Instead of adding all the terms to a list and calling sum, it's a lot shorter to keep a running total. This gets the solution down to 85 bytes:

E=print
A=int(input())
S=0
for D in range(A):C=D**A+A;S+=C;E(C)
E(end='SUM: '+str(S))

From here, two more optimizations can be made. The first is to make use of f-strings, saving three bytes. The second is to convert the for loop to a while loop, which saves another byte. With this we get an 81-byte solution:

E=print
A=int(input())
S=D=0
while D<A:C=D**A+A;S+=C;E(C);D+=1
E(end=f'SUM: {S}')

For Python >=3.8, you can also utilize the walrus operator to save 1 more byte.

E=print
A=int(input())
S=D=0
while D<A:E(C:=D**A+A);S+=C;D+=1
E(end=f'SUM: {S}')