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A14075. GCD Sum

编程题 普及/提高-
知识点

题目描述

The $\text{$ gcdSum $}$ of a positive integer is the $gcd$ of that integer with its sum of digits. Formally, $\text{$ gcdSum $}(x) = gcd(x, \text{ sum of digits of } x)$ for a positive integer $x$ . $gcd(a, b)$ denotes the greatest common divisor of $a$ and $b$ — the largest integer $d$ such that both integers $a$ and $b$ are divisible by $d$ .

For example: $\text{$ gcdSum $}(762) = gcd(762, 7 + 6 + 2)=gcd(762,15) = 3$ .

Given an integer $n$ , find the smallest integer $x \ge n$ such that $\text{$ gcdSum $}(x) > 1$ .

输入格式

The first line of input contains one integer $t$ $(1 \le t \le 10^4)$ — the number of test cases.

Then $t$ lines follow, each containing a single integer $n$ $(1 \le n \le 10^{18})$ .

All test cases in one test are different.

输出格式

Output $t$ lines, where the $i$ -th line is a single integer containing the answer to the $i$ -th test case.

输入输出样例

输入 #1
3
11
31
75
输出 #1
12
33
75

说明/提示

Let us explain the three test cases in the sample.

Test case 1: $n = 11$ :

$\text{$ gcdSum $}(11) = gcd(11, 1 + 1) = gcd(11,\ 2) = 1$ .

$\text{$ gcdSum $}(12) = gcd(12, 1 + 2) = gcd(12,\ 3) = 3$ .

So the smallest number $\ge 11$ whose $gcdSum$ $> 1$ is $12$ .

Test case 2: $n = 31$ :

$\text{$ gcdSum $}(31) = gcd(31, 3 + 1) = gcd(31,\ 4) = 1$ .

$\text{$ gcdSum $}(32) = gcd(32, 3 + 2) = gcd(32,\ 5) = 1$ .

$\text{$ gcdSum $}(33) = gcd(33, 3 + 3) = gcd(33,\ 6) = 3$ .

So the smallest number $\ge 31$ whose $gcdSum$ $> 1$ is $33$ .

Test case 3: $\ n = 75$ :

$\text{$ gcdSum $}(75) = gcd(75, 7 + 5) = gcd(75,\ 12) = 3$ .

The $\text{$ gcdSum $}$ of $75$ is already $> 1$ . Hence, it is the answer.
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