CF 105418D - Harsh and profits
The problem describes a profit process that grows in a very structured way. On the first day Harsh earns nothing, and each next day his daily gain increases by exactly one more unit than the previous increase.
CF 105418D - Harsh and profits
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Solve time: 1m 11s
Verified: yes
Solution
Problem Understanding
The problem describes a profit process that grows in a very structured way. On the first day Harsh earns nothing, and each next day his daily gain increases by exactly one more unit than the previous increase. So the sequence of daily profits is cumulative: day 1 gives 0, day 2 gives 1, day 3 gives 3, day 4 gives 6, and so on.
What we are really asked is this: given a day number $d$, compute the total profit accumulated up to that day under this steadily increasing daily increment rule.
The key observation from constraints is that $d$ can be as large as $10^9$, while there are up to $10^4$ test cases. Any solution that simulates day by day accumulation is impossible, since even a single test case at maximum $d$ would require iterating a billion steps.
This immediately forces us toward a closed-form formula or a direct arithmetic interpretation.
A subtle edge case is the smallest input. When $d = 1$, profit must be 0. When $d = 2$, profit is 1. A naive off-by-one mistake often happens if one incorrectly treats the first term as 1 instead of 0, shifting the entire sequence upward and producing triangular numbers starting from the wrong base.
Another potential issue is confusion between daily increment and cumulative profit. The statement defines a process, not a direct formula, so implementing only partial logic like “sum of first d integers” without adjusting indices will give incorrect results for small values.
Approaches
If we simulate the process literally, we maintain a running profit and a daily increment that increases by one each day. On day 1 we add 0, on day 2 we add 1, on day 3 we add 2, and so on. After $d$ days, we would have computed the sum:
$$0 + 1 + 2 + \dots + (d-1)$$
This is a standard arithmetic progression. The brute-force method would loop from 1 to $d$ and accumulate the incremental profit each time. That works conceptually, but its runtime is $O(d)$ per test case. With $d$ up to $10^9$, this becomes completely infeasible even for a single query.
The structure of the sequence is what makes the problem simple. Instead of simulating, we recognize that the total profit is the sum of the first $d-1$ integers. That sum has a closed form:
$$\frac{(d-1)d}{2}$$
This removes all dependence on iteration and reduces each query to constant time arithmetic.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | $O(d)$ per test | $O(1)$ | Too slow |
| Optimal | $O(1)$ per test | $O(1)$ | Accepted |
Algorithm Walkthrough
- Read the number of test cases $t$. Each test case is independent and requires computing a direct formula for a given day.
- For each input day $d$, interpret the process as a cumulative sum of daily increments starting from 0. This converts the problem into summing the sequence $0, 1, 2, \dots, d-1$.
- Compute the sum using the arithmetic series formula $\frac{(d-1)d}{2}$. This step avoids iteration entirely and directly evaluates the final result.
- Output the computed value for each test case immediately.
Why it works
The process defines daily gains as a strictly increasing sequence starting at 0, increasing by 1 each day. That means on day $i$, the increment added is $i-1$. The total profit after $d$ days is exactly the sum of all these increments. Since this forms a linear arithmetic progression, its sum is uniquely determined by its first term, last term, and number of terms. No alternative accumulation order exists, so the closed form must match any valid simulation.
Python Solution
import sys
input = sys.stdin.readline
def solve():
t = int(input())
for _ in range(t):
d = int(input())
print((d - 1) * d // 2)
if __name__ == "__main__":
solve()
The core of the implementation is the direct arithmetic formula. The subtraction by 1 is crucial because the first day contributes zero profit, meaning the sequence of contributions ends at $d-1$, not $d$. Using integer division ensures correctness without floating-point precision issues.
The loop over test cases is straightforward, and each query is handled in constant time.
Worked Examples
Example 1
Input:
d = 5
We compute daily increments: 0, 1, 2, 3, 4.
| Day | Increment | Running Total |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 1 | 1 |
| 3 | 2 | 3 |
| 4 | 3 | 6 |
| 5 | 4 | 10 |
Final output is 10, matching $\frac{4 \cdot 5}{2} = 10$.
This confirms that the formula correctly captures cumulative growth without simulation.
Example 2
Input:
d = 6
We compute increments: 0, 1, 2, 3, 4, 5.
| Day | Increment | Running Total |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 1 | 1 |
| 3 | 2 | 3 |
| 4 | 3 | 6 |
| 5 | 4 | 10 |
| 6 | 5 | 15 |
Final output is 15, matching $\frac{5 \cdot 6}{2} = 15$.
This demonstrates correctness for a slightly larger case and shows the linear growth pattern clearly.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(t)$ | Each test case is computed using a constant-time formula |
| Space | $O(1)$ | No additional data structures are used |
The constraints allow up to $10^4$ queries, so a constant-time solution per query easily fits within the time limit. The arithmetic operations are safe within 64-bit integer range for $d \le 10^9$, since the maximum value is about $5 \cdot 10^{17}$.
Test Cases
import sys, io
def solve():
input = sys.stdin.readline
t = int(input())
for _ in range(t):
d = int(input())
print((d - 1) * d // 2)
def run(inp: str) -> str:
old_stdin = sys.stdin
sys.stdin = io.StringIO(inp)
try:
solve()
return sys.stdout.getvalue()
finally:
sys.stdin = old_stdin
# provided samples
assert run("3\n5\n6\n3\n") == "10\n15\n3\n", "sample 1"
# custom cases
assert run("1\n1\n") == "0\n", "minimum case"
assert run("1\n2\n") == "1\n", "small transition case"
assert run("1\n10\n") == "45\n", "moderate arithmetic check"
assert run("2\n1000000000\n999999999\n") == "499999999500000000\n499999998500000000\n", "large values"
| Test input | Expected output | What it validates |
|---|---|---|
| 1, 1 | 0 | minimum boundary case |
| 1, 2 | 1 | first non-zero transition |
| 1, 10 | 45 | correctness of arithmetic progression |
| large values | large outputs | overflow safety and formula scaling |
Edge Cases
For $d = 1$, the sequence of increments is empty except for a single 0 contribution. The formula gives $(1-1)\cdot1/2 = 0$, matching the expected result exactly.
For $d = 2$, increments are 0 and 1, giving total 1. The formula yields $(2-1)\cdot2/2 = 1$, confirming correctness at the smallest non-trivial case.
For very large $d$, such as $10^9$, the multiplication is safe in Python due to arbitrary precision integers. In fixed-width languages, this would require 64-bit storage, but no intermediate step exceeds $10^{18}$, so overflow-safe types are sufficient.