Codeforces Round 879 (Div. 2)
B
Maximum Strength
Codeforces
1000
原题
← AC →
翻译中…

Fedya is playing a new game called "The Legend of Link", in which one of the character's abilities is to combine two materials into one weapon. Each material has its own strength, which can be represented by a positive integer $x$. The strength of the resulting weapon is determined as the sum of the absolute differences of the digits in the decimal representation of the integers at each position.

翻译中…

Formally, let the first material have strength $X = \overline{x_{1}x_{2} \ldots x_{n}}$, and the second material have strength $Y = \overline{y_{1}y_{2} \ldots y_{n}}$. Then the strength of the weapon is calculated as $|x_{1} - y_{1}| + |x_{2} - y_{2}| + \ldots + |x_{n} - y_{n}|$. If the integers have different lengths, then the shorter integer is padded with leading zeros.

翻译中…

Fedya has an unlimited supply of materials with all possible strengths from $L$ to $R$, inclusive. Help him find the maximum possible strength of the weapon he can obtain.

翻译中…

An integer $C = \overline{c_{1}c_{2} \ldots c_{k}}$ is defined as an integer obtained by sequentially writing the digits $c_1, c_2, \ldots, c_k$ from left to right, i.e. $10^{k-1} \cdot c_1 + 10^{k-2} \cdot c_2 + \ldots + c_k$.

翻译中…

Input

翻译中…

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

翻译中…

The first line of each test case contains two integers $L$ and $R$ ($1 \le L \le R \lt 10^{100}$) — the decimal representation of the integers representing the minimum and maximum strength of the materials that Fedya has. It is guaranteed that the integers $L$ and $R$ do not contain leading zeros.

翻译中…

Note that the input data may not fit into standard $32$-bit or $64$-bit integer data types.

翻译中…

Output

翻译中…

For each test case print one integer — the maximum possible strength of the weapon that Fedya can obtain from the given materials.

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Note

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In the first test case, the weapon made from materials with strengths $53$ and $57$ will have the maximum possible strength: $|5 - 5| + |3 - 7| = 4$.

翻译中…

In the second test case, the maximum strength is achieved with materials with strengths $190$ and $209$: $|1 - 2| + |9 - 0| + |0 - 9| = 19$.

翻译中…

In the fourth test case, there is only one valid strength, so the answer is $0$.

翻译中…

In the sixth test case, the maximum strength is achieved with materials with strengths $1909$ and $90$: $|1 - 0| + |9 - 0| + |0 - 9| + |9 - 0| = 28$. Note that the shorter integer was padded with leading zeros.

翻译中…

样例

Input #1
6
53 57
179 239
13 37
132228 132228
54943329752812629795 55157581939688863366
88 1914
Output #1
4
19
11
0
163
28
InputCopy OutputCopy
题目信息
题号
1834B
时间限制
1 s
内存限制
256 MB
标签
贪心
数学

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。