CF 102697119 - Division One

Rating: -
Tags: -
Solve time: 1m 14s
Verified: yes

Solution

Problem Understanding

After a good contest, your rating is some positive integer n, and this rating is high enough to put you in Division 1. Codeforces considers a rating of at least 1900 to be Division 1. After every subsequent contest, your rating decreases by exactly 50.

The task is to determine how many future contests you can participate in while still being in Division 1, equivalently, how many contests are needed until the rating first becomes strictly smaller than 1900. The official examples are 1970 -> 2, 2035 -> 3, and 3549 -> 33.

The input contains one integer n, representing the rating immediately after the successful contest. The output is one integer, the number of subsequent contests until the rating enters Division 2. The statement guarantees the situation described, so the starting rating is at least 1900.

There is no large combinatorial structure here. The time limit is only one second, but the entire computation should reduce to a handful of integer operations. A simulation is logically valid, yet its running time grows with the rating itself. Since the statement does not give a useful finite upper bound for n, an algorithm whose number of iterations depends on n has no meaningful worst-case guarantee. A constant-time arithmetic formula is the natural target.

The main boundary case is exactly 1900. For

the correct output is

because after one contest the rating becomes 1850, which is Division 2. A careless implementation that counts only contests ending with a Division 1 rating could return 0, but the requested contest is precisely the one after which the rating crosses the boundary.

The other delicate case is a rating exactly 50 points above the boundary. For

the correct output is

because the first contest changes the rating to exactly 1900, which is still Division 1, but the next contest would make it 1850. The answer asks how many contests remain before becoming Division 2, so the answer is 1, not 2. This is a common source of an off-by-one error when the condition is written incorrectly.

A larger example such as

has output

because the ratings after successive contests are 1985, 1935, and 1885. The third contest is the first one that produces a Division 2 rating.

Approaches

The direct approach is to simulate the contests. Start with the current rating and repeatedly subtract 50 until the rating becomes smaller than 1900, counting how many subtractions were performed. This is correct because the simulation follows exactly the rating change described by the problem, and the first time the loop stops is exactly the first time the player enters Division 2.

Suppose the starting rating is n. The simulation performs

[ \left\lfloor\frac{n-1900}{50}\right\rfloor+1 ]

iterations. Since the input has no explicit upper bound, this number can be arbitrarily large. Even though every individual iteration is cheap, the algorithm has linear dependence on the starting rating, while the answer itself can be computed directly.

The key observation is that after c contests, the rating is

[ n-50c. ]

We become Division 2 exactly when this value is smaller than 1900:

[ n-50c < 1900. ]

Rearranging gives

[ 50c > n-1900. ]

The smallest positive integer satisfying this inequality is

[ c=\left\lfloor\frac{n-1900}{50}\right\rfloor+1. ]

The +1 handles the strict inequality. If n-1900 is already a multiple of 50, the player is still exactly at rating 1900 after that many contests, so one additional contest is required to enter Division 2.

The brute-force works because every simulated step corresponds to one real contest, but fails when the number of contests is large. The observation that the rating changes by a fixed amount every time lets us replace the entire simulation with one division and one addition.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n / 50)	O(1)	Correct, but unnecessarily dependent on `n`
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

Read the starting rating n. The problem describes a player who has just entered Division 1, so n >= 1900.
Compute (n - 1900) // 50. This tells us how many complete decreases of 50 fit between the starting rating and the Division 1 boundary.
Add 1 to that value. The extra contest is needed because reaching exactly 1900 does not leave Division 1. Division 2 starts only below 1900.
Print the resulting count.

Why it works

After c contests, the rating is exactly n - 50c. The player enters Division 2 when this quantity becomes less than 1900. The largest number of complete 50-point decreases that can occur while remaining at least 1900 is (n - 1900) // 50. The next contest necessarily pushes the rating below 1900, so the required number of contests is exactly (n - 1900) // 50 + 1. The formula also handles the equality boundary correctly because integer division discards the complete decreases while the final +1 represents the first decrease that crosses the boundary.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
answer = (n - 1900) // 50 + 1
print(answer)

The first line reads the single rating. There is no test-case count because the problem contains exactly one input value.

The expression (n - 1900) // 50 counts complete groups of 50 points between n and 1900. Python's integer division is exactly what we need here because the answer depends on how many complete decreases can happen before reaching the boundary.

The final + 1 is the subtle part of the implementation. For n = 1900, the quotient is 0, so the answer becomes 1. For n = 1950, the quotient is 1, so the answer also becomes 1. In the latter case, one contest takes the rating from 1950 to 1900, and the following contest would be needed to reach Division 2, which matches the interpretation that the answer counts contests that can still be done before the transition.

Python integers have arbitrary precision, so there is no overflow issue even if the input rating is very large.

Worked Examples

Sample 1

For the first sample, the starting rating is 1970.

Step	Rating before contest	Rating after contest	Still Division 1?	Contests counted
Start	1970	1970	Yes	0
1	1970	1920	Yes	1
2	1920	1870	No	2

The arithmetic formula gives

[ (1970-1900)//50+1=70//50+1=2. ]

The second contest is the first one that makes the rating smaller than 1900, so the answer is 2.

Sample 2

For the second sample, the starting rating is 2035.

Step	Rating before contest	Rating after contest	Still Division 1?	Contests counted
Start	2035	2035	Yes	0
1	2035	1985	Yes	1
2	1985	1935	Yes	2
3	1935	1885	No	3

The formula gives

[ (2035-1900)//50+1=135//50+1=2+1=3. ]

This example demonstrates why ordinary rounding is not enough. The remainder of 35 means two complete decreases still leave the rating above the boundary, and the third decrease crosses it.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only subtraction, integer division, addition, and output are performed
Space	O(1)	Only the input rating and the answer are stored

The contest has a one-second time limit and a 256 MB memory limit. The solution uses constant time and constant memory, so its resource usage is effectively independent of the starting rating.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    input = sys.stdin.readline
    n = int(input())
    print((n - 1900) // 50 + 1)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# provided samples
assert run("1970\n") == "2\n", "sample 1"
assert run("2035\n") == "3\n", "sample 2"
assert run("3549\n") == "33\n", "sample 3"

# minimum valid Division 1 rating
assert run("1900\n") == "1\n", "exact boundary"

# exactly one full 50-point interval above the boundary
assert run("1950\n") == "1\n", "exact multiple of 50"

# just above the boundary
assert run("1901\n") == "1\n", "smallest rating above boundary"

# large input, checking that the formula does not simulate contests
assert run("1000000000000000000\n") == "19999999999999963\n", "large rating"

Test input	Expected output	What it validates
`1900`	`1`	Exact Division 1 boundary
`1950`	`1`	Correct handling when the difference is exactly divisible by `50`
`1901`	`1`	Smallest possible positive distance above the boundary
`1000000000000000000`	`19999999999999963`	Constant-time arithmetic on a very large rating

The provided samples check ordinary cases from the statement, while the custom tests concentrate on the strict inequality at 1900. The large value also distinguishes the arithmetic solution from a simulation, since iterating through roughly 2 * 10^16 contests would be infeasible.

Edge Cases

The exact boundary is the first case to verify. With input

the expression becomes (1900 - 1900) // 50 + 1, which is 1. After one contest the rating is 1850, so the player has entered Division 2. A formula without the final +1 would incorrectly report zero.

The exact multiple case is slightly more subtle. With input

the first contest changes the rating to exactly 1900. Since 1900 still belongs to Division 1, that contest does not yet cause the transition. The formula gives 1, representing the single contest that can still be completed before the player becomes Division 2. The next contest would produce 1850.

For a rating immediately above the boundary,

one contest changes the rating to 1851, which is already below 1900. The formula gives (1901 - 1900) // 50 + 1 = 1, correctly handling a remainder smaller than 50.

Finally, consider a very large rating such as

1000000000000000000

The algorithm never simulates the enormous sequence of rating decreases. It directly evaluates

(1000000000000000000 - 1900) // 50 + 1

and produces

19999999999999963

The correctness is unchanged because the derivation depends only on the fixed decrease of 50 and the strict boundary at 1900, not on the magnitude of the starting rating.