Editorial for Range Additions

Approach

Applying each query directly to every element in $[l, r]$ can take $O(nq)$, which is too slow.

Instead, use a difference array:

• To add $x$ to every element in $[l, r]$, add $x$ at position $l$.
• Subtract $x$ at position $r + 1$ if that position exists.
• After processing all queries, take a prefix sum over the difference array to recover how much should be added to each index.

Then add that running total to the original array.

This processes all updates in $O(q)$ and the final reconstruction in $O(n)$, so the total time complexity is $O(n + q)$.

Solution (Python)

n, q = map(int, input().split())
a = list(map(int, input().split()))
diff = [0] * (n + 1)

for _ in range(q):
    l, r, x = map(int, input().split())
    l -= 1
    diff[l] += x
    if r < n:
        diff[r] -= x

cur = 0
ans = []
for i in range(n):
    cur += diff[i]
    ans.append(str(a[i] + cur))

print(" ".join(ans))