5 Python Techniques to Compute the Alternating Sum 1‑2+3‑4+…+99

Preface

The author, a Python enthusiast, shares a question asked in a Python community about computing the alternating sum 1‑2+3‑4+…+99 using a loop that accumulates the result in a single variable.

Solution Process

Method 1: dcpeng's answer

Code:

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 accumulating variable.

Method 2: dcpeng's answer

Code:

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

This implementation follows the requirement by using only one accumulator.

Method 3: Eternal of Budapest's answer

Uses the range() function:

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

Method 4: Moon God

A two‑line solution with itertools.accumulate:

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

It can be simplified by applying sum directly:

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

Method 5: Teacher Yuliang

A concise one‑liner based on method 4:

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

Conclusion

These five solutions demonstrate different Python approaches to compute the alternating sum 1‑2+3‑4+…+99, ranging from explicit loops with separate variables to concise one‑liners using itertools and generator expressions, all satisfying the requirement of accumulating the result in a single variable.