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chore: update lc problem: No.0115 (doocs#1282)
* chore: update lc problem: No.0115 * fix: add skip lc problem
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solution/0100-0199/0115.Distinct Subsequences/README.md

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<p><strong>示例&nbsp;1:</strong></p>
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<pre>
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<strong>输入:</strong>s = "rabbbit", t = "rabbit"<code>
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<strong>输出</strong></code><strong>:</strong><code>3
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</code><strong>解释:</strong>
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如下所示, 有 3 种可以从 s 中得到 <code>"rabbit" 的方案</code>
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<code><strong><u>rabb</u></strong>b<strong><u>it</u></strong></code>
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<code><strong><u>ra</u></strong>b<strong><u>bbit</u></strong></code>
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<code><strong><u>rab</u></strong>b<strong><u>bit</u></strong></code></pre>
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<strong>输入:</strong>s = "rabbbit", t = "rabbit"
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<strong>输出</strong><strong>:</strong>3
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<strong>解释:</strong>
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如下所示, 有 3 种可以从 s 中得到 "rabbit" 的方案。
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<strong><u>rabb</u></strong>b<strong><u>it</u></strong>
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<strong><u>ra</u></strong>b<strong><u>bbit</u></strong>
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<strong><u>rab</u></strong>b<strong><u>bit</u></strong></pre>
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<p><strong>示例&nbsp;2:</strong></p>
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<pre>
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<strong>输入:</strong>s = "babgbag", t = "bag"
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<code><strong>输出</strong></code><strong>:</strong><code>5
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</code><strong>解释:</strong>
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如下所示, 有 5 种可以从 s 中得到 <code>"bag" 的方案</code>
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<code><strong><u>ba</u></strong>b<u><strong>g</strong></u>bag</code>
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<code><strong><u>ba</u></strong>bgba<strong><u>g</u></strong></code>
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<code><u><strong>b</strong></u>abgb<strong><u>ag</u></strong></code>
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<code>ba<u><strong>b</strong></u>gb<u><strong>ag</strong></u></code>
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<code>babg<strong><u>bag</u></strong></code>
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<strong>输出</strong><strong>:</strong>5
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<strong>解释:</strong>
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如下所示, 有 5 种可以从 s 中得到 "bag" 的方案。
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<strong><u>ba</u></strong>b<u><strong>g</strong></u>bag
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<strong><u>ba</u></strong>bgba<strong><u>g</u></strong>
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<u><strong>b</strong></u>abgb<strong><u>ag</u></strong>
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ba<u><strong>b</strong></u>gb<u><strong>ag</strong></u>
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babg<strong><u>bag</u></strong>
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</pre>
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<p>&nbsp;</p>

solution/config.py

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