23:I[344796,["/_next/static/chunks/083cs_iw_ie0y.js","/_next/static/chunks/2sk2f47durg8o.js","/_next/static/chunks/00mfjk9z2ewie.js","/_next/static/chunks/2q9x6spyz4qxv.js","/_next/static/chunks/1wo5i8je1ygp5.js","/_next/static/chunks/1h4l04ysuholp.js","/_next/static/chunks/2qcnia-og0qh0.js","/_next/static/chunks/40webkzfr6rcp.js"],"default"] 29:I[27201,["/_next/static/chunks/083cs_iw_ie0y.js","/_next/static/chunks/2sk2f47durg8o.js","/_next/static/chunks/00mfjk9z2ewie.js","/_next/static/chunks/2q9x6spyz4qxv.js","/_next/static/chunks/1wo5i8je1ygp5.js","/_next/static/chunks/1h4l04ysuholp.js","/_next/static/chunks/2qcnia-og0qh0.js"],"IconMark"] 24:Td2d,

Neel partied hard last night, forgetting there was a test in school today. The first question asked him to write down a set of "Special" numbers. Having no idea what a Special number was, he glanced at his neighbour's paper and managed to catch only a partial glimpse. Taking a chance, he guessed the remaining digits and submitted his answers.

Naturally, he got the numbers wrong! For each number Neel wrote, determine how far off the nearest Special number is from his answer.

A positive integer p is a Special number if dividing p by the sum of its digits yields a prime number. That division must be exact (no remainder), and the resulting quotient must be prime.

For example, 18 is Special because 18 / (1 + 8) = 2, and 2 is prime.

For each number x in Neel's list, find the Special number s closest to x (measured by absolute difference |s - x|), and report the signed offset s - x.

Input Format

The first line contains a single integer n, the count of numbers Neel wrote.
The second line contains n space-separated integers, Neel's numbers.

Output Format

Print, on a single line, n space-separated integers. The i-th value is s - x for the i-th number x, where s is the Special number nearest to x.

If two Special numbers are equally far from x (one below and one above), choose the lower Special number.

Constraints

1 <= n <= 2000
10 < each number <= 1000

Example 1

Input

3
17 19 20

Output

1 -1 1

Explanation

The Special number nearest to 17 is 18, so 18 - 17 = 1.
The Special number nearest to 19 is 18, so 18 - 19 = -1.
The Special number nearest to 20 is 21, so 21 - 20 = 1.

Example 2

Input

4
43 44 116 115

Output

-1 1 1 -1

Explanation

Nearest to 43 is 42 (42 / (4 + 2) = 7, prime), so 42 - 43 = -1.
Nearest to 44 is 45 (45 / (4 + 5) = 5, prime), so 45 - 44 = 1.
Nearest to 116 is 117 (117 / (1 + 1 + 7) = 13, prime), so 117 - 116 = 1.
Nearest to 115 is 114 (114 / (1 + 1 + 4) = 19, prime), so 114 - 115 = -1.

Notes

The offset is 0 when the number is itself Special.
Special numbers may lie below or above x; search outward in both directions until the closest one is found. The smallest Special number is 18.

25:T44c,#include using namespace std; static bool isPrime(long long x){ if(x < 2) return false; if(x < 4) return true; if(x % 2 == 0) return false; for(long long i = 3; i * i <= x; i += 2) if(x % i == 0) return false; return true; } static int digitSum(int p){ int s = 0; while(p > 0){ s += p % 10; p /= 10; } return s; } static bool isSpecial(int p){ if(p < 1) return false; int s = digitSum(p); if(s == 0 || p % s != 0) return false; return isPrime(p / s); } static int nearestDiff(int x){ if(isSpecial(x)) return 0; for(int d = 1; ; ++d){ if(x - d >= 1 && isSpecial(x - d)) return -d; if(isSpecial(x + d)) return d; } } void printResult(int n, vector arr){ string out; for(int i = 0; i < n; ++i){ if(i) out += ' '; out += to_string(nearestDiff(arr[i])); } printf("%s\n", out.c_str()); } int main(){ int n; if(scanf("%d", &n) != 1) return 0; vector arr(n); for(int i = 0; i < n; ++i) scanf("%d", &arr[i]); printResult(n, arr); return 0; } 26:T441,import sys def is_prime(x): if x < 2: return False if x < 4: return True if x % 2 == 0: return False i = 3 while i * i <= x: if x % i == 0: return False i += 2 return True def digit_sum(p): s = 0 while p > 0: s += p % 10 p //= 10 return s def is_special(p): if p < 1: return False s = digit_sum(p) if s == 0 or p % s != 0: return False return is_prime(p // s) def nearest_diff(x): if is_special(x): return 0 d = 1 while True: lo = x - d hi = x + d lo_ok = lo >= 1 and is_special(lo) hi_ok = is_special(hi) if lo_ok: # tie or lower-only -> choose lower return -d if hi_ok: return d d += 1 def main(): data = sys.stdin.read().split() if not data: return n = int(data[0]) arr = list(map(int, data[1:1 + n])) out = [str(nearest_diff(x)) for x in arr] print(" ".join(out)) if __name__ == "__main__": main() 27:T96b,# Nearest Special Number Offsets Neel partied hard last night, forgetting there was a test in school today. The first question asked him to write down a set of "Special" numbers. Having no idea what a Special number was, he glanced at his neighbour's paper and managed to catch only a partial glimpse. Taking a chance, he guessed the remaining digits and submitted his answers. Naturally, he got the numbers wrong! For each number Neel wrote, determine how far off the **nearest** Special number is from his answer. A positive integer `p` is a **Special number** if dividing `p` by the sum of its digits yields a prime number. That division must be exact (no remainder), and the resulting quotient must be prime. For example, `18` is Special because `18 / (1 + 8) = 2`, and `2` is prime. For each number `x` in Neel's list, find the Special number `s` closest to `x` (measured by absolute difference `|s - x|`), and report the signed offset `s - x`. ### Input Format - The first line contains a single integer `n`, the count of numbers Neel wrote. - The second line contains `n` space-separated integers, Neel's numbers. ### Output Format Print, on a single line, `n` space-separated integers. The `i`-th value is `s - x` for the `i`-th number `x`, where `s` is the Special number nearest to `x`. If two Special numbers are equally far from `x` (one below and one above), choose the **lower** Special number. ### Constraints - `1 <= n <= 2000` - `10 < each number <= 1000` ### Examples #### Example 1 **Input** ``` 3 17 19 20 ``` **Output** ``` 1 -1 1 ``` **Explanation** - The Special number nearest to `17` is `18`, so `18 - 17 = 1`. - The Special number nearest to `19` is `18`, so `18 - 19 = -1`. - The Special number nearest to `20` is `21`, so `21 - 20 = 1`. #### Example 2 **Input** ``` 4 43 44 116 115 ``` **Output** ``` -1 1 1 -1 ``` **Explanation** - Nearest to `43` is `42` (`42 / (4 + 2) = 7`, prime), so `42 - 43 = -1`. - Nearest to `44` is `45` (`45 / (4 + 5) = 5`, prime), so `45 - 44 = 1`. - Nearest to `116` is `117` (`117 / (1 + 1 + 7) = 13`, prime), so `117 - 116 = 1`. - Nearest to `115` is `114` (`114 / (1 + 1 + 4) = 19`, prime), so `114 - 115 = -1`. ### Notes - The offset is `0` when the number is itself Special. - Special numbers may lie below or above `x`; search outward in both directions until the closest one is found. The smallest Special number is `18`.28:T927,# Nearest Special Number Offsets — Editorial ## Overview For each queried number, we must find the Special number closest to it and report the signed gap. A number is Special when it divides evenly by its digit sum and the resulting quotient is prime. ## Key Insight Special numbers are dense: within the input range every number has a Special number only a small distance away (never more than a couple dozen steps). So instead of precomputing anything global, we can simply expand a symmetric search window around each query until we land on the first Special number. ## Approach Define a predicate `isSpecial(p)`: - Compute the digit sum `s`. - Check that `p` is divisible by `s`. - Check that `p / s` is prime (trial division up to `sqrt(p / s)`). For a query `x`: 1. First test `x` itself (offset `0`). 2. Otherwise increase a radius `d` from `1` upward. 3. At each `d`, examine `x - d` before `x + d`. - Testing the lower candidate first automatically implements the tie‑break rule — when both are Special, the lower one is returned. 4. Return the signed difference `s - x` for the first Special value found. ## Why This Works Scanning `d = 1, 2, 3, ...` visits candidates in non‑decreasing order of distance, so the first Special number encountered is genuinely the nearest. Because the lower candidate at each radius is inspected first, an exact tie always resolves to the smaller number, exactly as the problem requires. The quotient being prime guarantees `p / s` is an integer greater than `1`, so single‑digit values (whose quotient would be `1`) are never Special. ## Complexity - Each primality test is `O(sqrt(p))`. - The search radius per query is bounded by a small constant for the given range. - Digit sum computation is `O(log p)`. - With `n` up to `2000` the total work is comfortably within limits. ## Common Pitfalls - Forgetting the exact‑division requirement and treating `p / s` as a truncating quotient marks non‑Special numbers as Special. - Treating `1` as prime wrongly classifies numbers whose quotient is `1`. - Reporting the absolute distance instead of the signed offset `s - x` loses the direction. - Breaking ties toward the higher number, or checking the upper candidate first, violates the tie‑break rule. - Skipping the offset‑`0` case when the number is itself Special.1c:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"learningResourceType\":\"Coding Question\",\"name\":\"Nearest Special Number Offsets\",\"description\":\"Neel partied hard last night, forgetting there was a test in school today. 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