CF 102697032 - Prison Break

The lock has (n) independent dials. For dial (i), the allowed values form every integer from its lower limit (Li) through its upper limit (Ri), inclusive. A complete lock combination chooses exactly one allowed value for every dial.

codeforcescompetitive-programming

CF 102697032 - Prison Break

Rating: -
Tags: -
Solve time: 1m 3s
Verified: yes

Solution

Problem Understanding

The lock has (n) independent dials. For dial (i), the allowed values form every integer from its lower limit (L_i) through its upper limit (R_i), inclusive. A complete lock combination chooses exactly one allowed value for every dial.

The task is to count how many different complete combinations are possible. Since the choices of one dial do not restrict the choices of another, the answer is the product of the number of available values on every dial. The archived statement gives the input format and examples, but does not publish explicit numerical bounds for (n) or for the dial limits. It does specify a one second time limit and a 256 MB memory limit.

For a single dial with limits (L_i) and (R_i), the number of possible settings is

[ R_i-L_i+1. ]

The (+1) is necessary because both endpoints are allowed. For example, the range (3) through (7) contains (3,4,5,6,7), which is five values.

The key constraint consequence is that we should not construct the combinations themselves. If every dial has only (k) possible values, there are (k^n) complete combinations. Even with (k=2), this is already (2^n), so enumerating combinations becomes infeasible as (n) grows. The optimal solution only needs to read each dial once, giving linear time in the number of dials.

There are several small cases where an implementation can silently make an off-by-one mistake. With input 1 followed by 3 3, the correct output is 1, because the only possible setting is (3). A formula such as (R-L) would incorrectly produce zero.

With input 1 followed by 1 2, the correct output is 2, because the dial can be set to either (1) or (2). Again, forgetting the inclusive upper endpoint would produce (1).

A second common mistake is initializing the product to zero. For input 1 followed by 5 5, multiplying the first count by zero would keep the answer at zero forever. The product must start at (1), which is the multiplicative identity.

Approaches

The direct brute-force approach is to generate every possible combination of dial settings and count it. This is correct because every valid combination is generated exactly once, and every generated combination corresponds to one choice from every allowed range.

The problem is the number of combinations. If dial (i) has (c_i=R_i-L_i+1) possible values, the total number of combinations is

[ C=\prod_{i=1}^{n}c_i. ]

Generating all of them already requires (\Theta(C)) combinations, and if the implementation explicitly constructs an (n)-value tuple for every combination, it performs (\Theta(nC)) work. For example, if every dial has two choices, that becomes (\Theta(n2^n)), which quickly becomes impossible.

The observation that removes this exponential work is that the dials are independent. Choosing a value for one dial never changes the available values of another dial. When there are (c_1) choices for the first dial, (c_2) choices for the second, and so on, every first-dial choice can be paired with every second-dial choice, giving (c_1c_2) possibilities. Repeating the same reasoning for the remaining dials gives the full product.

So instead of constructing any combination, we only count the choices contributed by each dial and multiply them into the answer. This reduces the problem from enumerating an exponential number of objects to one arithmetic operation per input dial.

Approach Time Complexity Space Complexity Verdict
Brute Force (\Theta(nC)), where (C=\prod_i(R_i-L_i+1)) (\Theta(n)) per stored combination Too slow
Optimal (O(n)) (O(1)) apart from input storage Accepted

Algorithm Walkthrough

  1. Initialize answer to (1). The result is a product, so (1) is the correct neutral starting value.
  2. Read the lower and upper limits (L_i) and (R_i) for the current dial.
  3. Compute the number of legal settings for this dial as (R_i-L_i+1). Both endpoints belong to the range, so the difference alone is one too small.
  4. Multiply answer by this number of settings. Every existing partial combination can be paired with every setting of the new dial, so multiplication counts all newly formed combinations exactly once.
  5. Repeat the previous three steps for all (n) dials.
  6. Print answer after all dials have been processed. At this point it equals the product of the number of choices for every dial, which is exactly the number of complete lock combinations.

Why it works

After processing the first (k) dials, maintain the invariant that answer equals the number of possible settings for those (k) dials. Initially, no dials have been processed and there is exactly one empty partial choice, so answer = 1. When processing dial (k+1), suppose it has (c) legal values. Every one of the existing answer partial combinations can be extended using each of those (c) values, producing exactly answer * c new combinations. No combination is duplicated because its value on the new dial uniquely identifies which extension was chosen. Thus the invariant remains true after every dial, and after all (n) dials the product is exactly the required answer.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    answer = 1

    for _ in range(n):
        lower, upper = map(int, input().split())
        answer *= upper - lower + 1

    print(answer)

if __name__ == "__main__":
    solve()

The first line reads the number of dials. The variable answer starts at (1) because the solution is a product, not a sum.

For every dial, upper - lower + 1 calculates its exact number of possible settings. The multiplication immediately incorporates that dial into the total number of combinations, so there is no need to store previous ranges or generate any actual lock combinations.

Python integers automatically grow as needed, so the implementation does not have the fixed-width integer overflow issue that languages using 32-bit or 64-bit integer types can encounter. This is particularly useful here because the product can become much larger than an individual dial limit.

The inclusive range is the main boundary detail. Writing upper - lower would fail whenever a range contains exactly one value, and would undercount every other range by one.

The loop processes exactly (n) input lines. There is no nested loop over possible dial values, which is the reason the implementation remains fast even when the number of combinations is enormous.

Worked Examples

For the first sample, the four dials have (9), (7), (6), and (4) possible settings respectively.

Dial Lower Upper Choices Answer
1 2 10 9 9
2 3 9 7 63
3 11 16 6 378
4 5 8 4 1512

The final value is (9\cdot7\cdot6\cdot4=1512), matching the sample output.

The trace demonstrates the multiplication invariant directly. After each row, answer is exactly the number of possible partial combinations using the dials processed so far.

For the second sample, the ranges contain (2), (3), and (2) values.

Dial Lower Upper Choices Answer
1 1 2 2 2
2 1 3 3 6
3 11 12 2 12

The result is (2\cdot3\cdot2=12), again matching the sample.

This example also shows why the actual numerical values of the limits do not matter beyond their distance. The range 11 12 contributes two choices just like 1 2.

Complexity Analysis

Measure Complexity Explanation
Time (O(n)) Each dial is read and processed exactly once.
Space (O(1)) Only the running product and the current pair of limits are stored.

The algorithm does not depend on the number of complete combinations. That number can be exponential in (n), while the program still performs only one constant amount of arithmetic per dial. The stated time limit is therefore easily compatible with the intended linear solution.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline

    n = int(input())
    answer = 1

    for _ in range(n):
        lower, upper = map(int, input().split())
        answer *= upper - lower + 1

    print(answer)

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

    try:
        sys.stdin = io.StringIO(inp)
        sys.stdout = io.StringIO()

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

# Provided sample 1
assert run("""4
2 10
3 9
11 16
5 8
""") == "1512\n", "sample 1"

# Provided sample 2
assert run("""3
1 2
1 3
11 12
""") == "12\n", "sample 2"

# Minimum-size input and a single possible value
assert run("""1
5 5
""") == "1\n", "single value"

# All dials have the same range
assert run("""4
1 3
1 3
1 3
1 3
""") == "81\n", "all equal ranges"

# Boundary case where every range has exactly two values
assert run("""5
0 1
10 11
100 101
1000 1001
10000 10001
""") == "32\n", "two choices per dial"

# Large number of dials, each contributing exactly one choice
assert run("100000\n" + "7 7\n" * 100000) == "1\n", "large n"

# Large product, checks that arithmetic is not restricted to 32-bit integers
assert run("""10
1 100
1 100
1 100
1 100
1 100
1 100
1 100
1 100
1 100
1 100
""") == "100000000000000000000\n", "large answer"
Test input Expected output What it validates
1 / 5 5 1 Minimum-size input and inclusive single-value range
Four ranges 1 3 81 All-equal ranges and repeated multiplication
Five two-value ranges 32 Boundary handling of inclusive endpoints
100000 identical single-value ranges 1 Linear processing with a large number of dials
Ten ranges 1 100 100000000000000000000 Large product and integer arithmetic

Edge Cases

A dial whose lower and upper limits are equal contributes exactly one choice. For the input

1
5 5

the algorithm computes (5-5+1=1), multiplies the initial answer (1) by it, and prints 1. A formula without the +1 would incorrectly claim that there are zero settings.

A range containing exactly two values exposes the same inclusive-boundary issue. For

1
1 2

the algorithm computes (2-1+1=2), corresponding to the settings (1) and (2). The output is 2.

If every dial has only one possible setting, the entire lock has only one combination regardless of how many dials exist. For

3
4 4
8 8
20 20

each dial contributes (1), so the running product remains (1) and the output is 1. This also confirms why the initial product must be (1).

A large number of dials does not change the algorithmic structure. With 100000 dials and 7 7 on every line, every dial contributes one choice. The algorithm performs exactly 100000 iterations and prints 1, without attempting to construct any combinations.

Finally, the number of combinations can be much larger than a standard 32-bit integer. Ten dials with the range 1 100 have (100^{10}=10^{20}) possible combinations. Python's arbitrary-precision integers allow the product to be computed exactly, so no special overflow handling is required.