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    <title>Div3 on shift&#39;s Blog</title>
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      <title>Codeforces Round 929 (Div. 3)</title>
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      <pubDate>Wed, 28 Feb 2024 12:45:36 +0800</pubDate>
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      <description>A. Turtle Puzzle: Rearrange and Negate 因为可以任意排列这个数组，并将一段区间内的数乘 $-1$，我们可以排序后把所有负数变为正数，即可 $code:$ void solve() { int n; std::cin &amp;gt;&amp;gt; n; int ans = 0; for(int i = 0, x; i &amp;lt; n; i ++ ) { std::cin &amp;gt;&amp;gt; x; ans += std::abs(x); } std::cout &amp;lt;&amp;lt; ans &amp;lt;&amp;lt; &amp;#39;\n&amp;#39;; } B. Turtle Math: Fast Three Task 分类讨论，$\sum_{i=1}^na_i\mod 3$ 等于 $0$ 的话不需要操作，等于 $1$ 就看有</description>
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