5 Clever Python Ways to Compute 1‑2+3‑4+…+99

Introduction

In a Python community a fan named dcpeng asked how to compute the alternating sum 1‑2+3‑4+…+99 using a loop that accumulates the result in a single variable.

The problem can be solved with simple arithmetic, but the focus is on implementing it correctly in Python.

Solutions

Method 1: dcpeng’s answer

odd = 0
even = 0
for i in range(100):
    if i % 2 == 1:
        odd += i
    else:
        even += i
print(odd - even)

This works but uses two variables, which deviates from the requirement of a single accumulator.

Method 2: dcpeng’s refined answer

count = 1
sum = 0
while count <= 99:
    if count % 2 == 1:
        sum += count
    else:
        sum -= count
    count += 1
print(sum)

This version follows the single‑variable rule and produces the correct result.

Method 3: Eternal from Budapest

Uses the range() function in a straightforward loop.

s = 0
for i in range(1, 100):
    if i % 2 == 0:
        s -= i
    else:
        s += i
print(s)

Method 4: Moon God

A compact two‑line solution leveraging itertools.accumulate.

from itertools import accumulate
list(accumulate((i if i % 2 else -i for i in range(1, 100))))

It can be simplified further with sum:

from itertools import accumulate
print(sum(accumulate((i if i % 2 else -i for i in range(1, 100)))))

Method 5: Yuliang teacher

A one‑liner that directly uses sum with a generator expression.

print(sum(i if i % 2 else -i for i in range(1, 100)))

Conclusion

The article collected five Python implementations for the alternating sum problem, ranging from explicit loops with separate variables to elegant one‑liners using itertools.accumulate and sum. Each solution demonstrates different Pythonic techniques and helps learners deepen their understanding of loops, conditionals, and functional tools.