CF 1845A - Forbidden Integer

Rating: 800
Tags: constructive algorithms, implementation, math, number theory
Solve time: 5m 13s
Verified: no

Solution

Problem Understanding

We are asked to determine whether a target sum n can be formed using any number of integers from 1 to k, excluding a single forbidden integer x. If it is possible, we must construct one valid sequence of integers whose sum is exactly n. Each test case provides three integers: n (the target sum), k (the largest available integer), and x (the forbidden integer).

The constraints are small: n, k, and x are at most 100, and there can be up to 100 test cases. This allows for algorithms with roughly O(n²) complexity, but simpler O(n) or O(1) constructive approaches are preferable. Because we are allowed unlimited copies of each allowed number, the problem reduces to finding a multiset of numbers from the set [1, k] \ {x} that sums to n.

Edge cases appear when the forbidden integer x is either very small or very large. For instance, if x = 1 and n = 1, we cannot make the sum because the smallest allowed number is 2. Similarly, if x = k = 1, the set of usable integers is empty, and any positive n is impossible. Another subtlety is that if n is smaller than the smallest allowed number (e.g., n = 2 but 1 is forbidden), it is impossible to reach the target.

Approaches

The brute-force solution would try all combinations of integers from 1 to k excluding x. This could be implemented recursively or with dynamic programming. Its time complexity is high, potentially O(n·k), and unnecessary for small n.

A key observation simplifies the problem: since we can use unlimited copies of allowed integers, we only need to check whether n can be formed using repeated copies of the smallest allowed integer. If 1 is not forbidden, we can always form n by taking n copies of 1. If 1 is forbidden but 2 is allowed, n must be at least 2 and can be written as a sum of twos (or a mix of twos and threes if they are available). The constructive nature allows us to output a simple greedy sequence: fill as much as possible with the smallest allowed integer, and then adjust with the next smallest if necessary.

This insight reduces the problem to a simple check on the smallest allowed integer and then constructing the sum directly, yielding O(n) time per test case, which is more than fast enough.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n·k)	O(n)	Works but unnecessary
Greedy/Constructive	O(n)	O(n)	Accepted

Algorithm Walkthrough

Identify the smallest allowed integer a. This is either 1 (if x != 1) or 2 (if x = 1 and k >= 2). If x = 1 and k = 1, no integers are allowed, and the answer is "NO".
Determine whether n is reachable using a. If n < a, print "NO" because we cannot form n with integers greater than n. Otherwise, proceed to construct the sequence.
Construct the sequence greedily. If a = 1, take n copies of 1. If a = 2, and n is even, take n / 2 copies of 2. If n is odd and k >= 3 and 3 is allowed, take one 3 and (n - 3)/2 copies of 2. Otherwise, it is impossible.
Output "YES", the total count of integers in the sequence, and the sequence itself.

Why it works: the smallest allowed integers dominate the sum construction because they give the most flexibility. Any sum n that is at least the smallest allowed integer can be constructed using multiples of that integer, with minor adjustments using the next smallest allowed integer if n is not divisible by the smallest integer.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, k, x = map(int, input().split())
    
    if k == 1 and x == 1:
        print("NO")
        continue
    
    if x != 1:
        # 1 is allowed
        print("YES")
        print(n)
        print("1 " * n)
    else:
        # 1 is forbidden, use 2 if possible
        if n == 1:
            print("NO")
        elif n % 2 == 0:
            print("YES")
            print(n // 2)
            print("2 " * (n // 2))
        elif k >= 3:
            print("YES")
            count = 1 + (n - 3) // 2
            print(count)
            print("3 " + "2 " * ((n - 3) // 2))
        else:
            print("NO")

The solution first handles the degenerate case when no numbers are allowed. Then it checks whether 1 is usable, allowing an immediate construction. If 1 is forbidden, the solution uses 2 for even n and 3 + 2 for odd n when possible. This matches the constructive logic described above.

Worked Examples

Example 1: n = 10, k = 3, x = 2

Step	Allowed Integers	Sequence Construction	Sequence
Identify smallest	1	1 is allowed	-
Construct sequence	n=10	Take ten 1s	1 1 1 1 1 1 1 1 1 1

Output:

YES
10
1 1 1 1 1 1 1 1 1 1

Example 2: n = 5, k = 2, x = 1

Step	Allowed Integers	Sequence Construction	Sequence
Identify smallest	2	1 forbidden, use 2	n=5 odd, k<3 cannot adjust
Result	impossible	-	-

Output:

NO

These traces show that the algorithm correctly handles both possible and impossible cases, including small sums and forbidden integers.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Sequence construction may need up to n integers in worst case
Space	O(n) per test case	Storing the sequence for output

With n <= 100 and t <= 100, the worst-case time is 10,000 operations, which easily fits in the 2-second limit. Memory is also minimal.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    exec(open('solution.py').read())  # assume solution is in solution.py
    return out.getvalue().strip()

# provided samples
assert run("5\n10 3 2\n5 2 1\n4 2 1\n7 7 3\n6 1 1\n") == \
"""YES
10
1 1 1 1 1 1 1 1 1 1
NO
YES
2
2 2
YES
1
7
NO""", "Sample 1"

# custom cases
assert run("1\n1 1 1\n") == "NO", "Minimum input, forbidden 1"
assert run("1\n2 2 1\n") == "YES\n1\n2", "n even, 1 forbidden, 2 allowed"
assert run("1\n3 3 1\n") == "YES\n2\n3 2", "n odd, 1 forbidden, 2 and 3 allowed"
assert run("1\n5 5 2\n") == "YES\n5\n1 1 1 1 1", "1 allowed, simple sum"

Test input	Expected output	What it validates
1 1 1	NO	No integers available
2 2 1	YES 2	Using 2 when 1 is forbidden, even n
3 3 1	YES 2	Using 3 + 2 when 1 is forbidden, odd n
5 5 2	YES 5	Using smallest allowed integer directly

Edge Cases

If the only allowed number is 2 and n is odd, the algorithm correctly outputs "NO" because there is no way to sum an odd number using only 2s. For example, n = 5, k = 2, x = 1 yields "NO". If 1 is allowed, even large n are trivially constructed using n copies of 1. If n equals the forbidden integer but other integers are available, the algorithm selects the smallest allowed integer, ensuring the sum can be reached without using the forbidden number.