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2026-07-13 07:46:06 +00:00 · 2023-11-09 05:13:54 +08:00
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@@ -3370,7 +3370,7 @@
 <p class="admonition-title">Question</p>
 <p>给定 <span class="arithmatex">\(n\)</span> 个物品，第 <span class="arithmatex">\(i\)</span> 个物品的重量为 <span class="arithmatex">\(wgt[i-1]\)</span>、价值为 <span class="arithmatex">\(val[i-1]\)</span> ，和一个容量为 <span class="arithmatex">\(cap\)</span> 的背包。每个物品只能选择一次，<strong>但可以选择物品的一部分，价值根据选择的重量比例计算</strong>，问在不超过背包容量下背包中物品的最大价值。</p>
 </div>
-<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="分数背包问题的示例数据" src="../fractional_knapsack_problem.assets/fractional_knapsack_example.png" /></a></p>
+<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="分数背包问题的示例数据" class="animation-figure" src="../fractional_knapsack_problem.assets/fractional_knapsack_example.png" /></a></p>
 <p align="center"> 图 15-3 &nbsp; 分数背包问题的示例数据 </p>

 <p>分数背包和 0-1 背包整体上非常相似，状态包含当前物品 <span class="arithmatex">\(i\)</span> 和容量 <span class="arithmatex">\(c\)</span> ，目标是求不超过背包容量下的最大价值。</p>
@@ -3379,7 +3379,7 @@
 <li>对于物品 <span class="arithmatex">\(i\)</span> ，它在单位重量下的价值为 <span class="arithmatex">\(val[i-1] / wgt[i-1]\)</span> ，简称为单位价值。</li>
 <li>假设放入一部分物品 <span class="arithmatex">\(i\)</span> ，重量为 <span class="arithmatex">\(w\)</span> ，则背包增加的价值为 <span class="arithmatex">\(w \times val[i-1] / wgt[i-1]\)</span> 。</li>
 </ol>
-<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="物品在单位重量下的价值" src="../fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png" /></a></p>
+<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="物品在单位重量下的价值" class="animation-figure" src="../fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png" /></a></p>
 <p align="center"> 图 15-4 &nbsp; 物品在单位重量下的价值 </p>

 <h3 id="1">1. &nbsp; 贪心策略确定<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
@@ -3389,7 +3389,7 @@
 <li>遍历所有物品，<strong>每轮贪心地选择单位价值最高的物品</strong>。</li>
 <li>若剩余背包容量不足，则使用当前物品的一部分填满背包即可。</li>
 </ol>
-<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="分数背包的贪心策略" src="../fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png" /></a></p>
+<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="分数背包的贪心策略" class="animation-figure" src="../fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png" /></a></p>
 <p align="center"> 图 15-5 &nbsp; 分数背包的贪心策略 </p>

 <h3 id="2">2. &nbsp; 代码实现<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
@@ -3811,7 +3811,7 @@
 <p>现在从背包中拿出单位重量的任意物品，并替换为单位重量的物品 <span class="arithmatex">\(x\)</span> 。由于物品 <span class="arithmatex">\(x\)</span> 的单位价值最高，因此替换后的总价值一定大于 <code>res</code> 。<strong>这与 <code>res</code> 是最优解矛盾，说明最优解中必须包含物品 <span class="arithmatex">\(x\)</span></strong> 。</p>
 <p>对于该解中的其他物品，我们也可以构建出上述矛盾。总而言之，<strong>单位价值更大的物品总是更优选择</strong>，这说明贪心策略是有效的。</p>
 <p>如图 15-6 所示，如果将物品重量和物品单位价值分别看作一个 2D 图表的横轴和纵轴，则分数背包问题可被转化为“求在有限横轴区间下的最大围成面积”。这个类比可以帮助我们从几何角度理解贪心策略的有效性。</p>
-<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="分数背包问题的几何表示" src="../fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png" /></a></p>
+<p><a class="glightbox" href="../fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="分数背包问题的几何表示" class="animation-figure" src="../fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png" /></a></p>
 <p align="center"> 图 15-6 &nbsp; 分数背包问题的几何表示 </p>

 <!-- Source file information -->
@@ -3392,7 +3392,7 @@
 <p>给定 <span class="arithmatex">\(n\)</span> 种硬币，第 <span class="arithmatex">\(i\)</span> 种硬币的面值为 <span class="arithmatex">\(coins[i - 1]\)</span> ，目标金额为 <span class="arithmatex">\(amt\)</span> ，每种硬币可以重复选取，问能够凑出目标金额的最少硬币个数。如果无法凑出目标金额则返回 <span class="arithmatex">\(-1\)</span> 。</p>
 </div>
 <p>本题的贪心策略如图 15-1 所示。给定目标金额，<strong>我们贪心地选择不大于且最接近它的硬币</strong>，不断循环该步骤，直至凑出目标金额为止。</p>
-<p><a class="glightbox" href="../greedy_algorithm.assets/coin_change_greedy_strategy.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="零钱兑换的贪心策略" src="../greedy_algorithm.assets/coin_change_greedy_strategy.png" /></a></p>
+<p><a class="glightbox" href="../greedy_algorithm.assets/coin_change_greedy_strategy.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="零钱兑换的贪心策略" class="animation-figure" src="../greedy_algorithm.assets/coin_change_greedy_strategy.png" /></a></p>
 <p align="center"> 图 15-1 &nbsp; 零钱兑换的贪心策略 </p>

 <p>实现代码如下所示。你可能会不由地发出感叹：So Clean ！贪心算法仅用十行代码就解决了零钱兑换问题。</p>
@@ -3648,7 +3648,7 @@
 <li><strong>反例 <span class="arithmatex">\(coins = [1, 20, 50]\)</span></strong>：假设 <span class="arithmatex">\(amt = 60\)</span> ，贪心算法只能找到 <span class="arithmatex">\(50 + 1 \times 10\)</span> 的兑换组合，共计 <span class="arithmatex">\(11\)</span> 枚硬币，但动态规划可以找到最优解 <span class="arithmatex">\(20 + 20 + 20\)</span> ，仅需 <span class="arithmatex">\(3\)</span> 枚硬币。</li>
 <li><strong>反例 <span class="arithmatex">\(coins = [1, 49, 50]\)</span></strong>：假设 <span class="arithmatex">\(amt = 98\)</span> ，贪心算法只能找到 <span class="arithmatex">\(50 + 1 \times 48\)</span> 的兑换组合，共计 <span class="arithmatex">\(49\)</span> 枚硬币，但动态规划可以找到最优解 <span class="arithmatex">\(49 + 49\)</span> ，仅需 <span class="arithmatex">\(2\)</span> 枚硬币。</li>
 </ul>
-<p><a class="glightbox" href="../greedy_algorithm.assets/coin_change_greedy_vs_dp.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="贪心无法找出最优解的示例" src="../greedy_algorithm.assets/coin_change_greedy_vs_dp.png" /></a></p>
+<p><a class="glightbox" href="../greedy_algorithm.assets/coin_change_greedy_vs_dp.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="贪心无法找出最优解的示例" class="animation-figure" src="../greedy_algorithm.assets/coin_change_greedy_vs_dp.png" /></a></p>
 <p align="center"> 图 15-2 &nbsp; 贪心无法找出最优解的示例 </p>

 <p>也就是说，对于零钱兑换问题，贪心算法无法保证找到全局最优解，并且有可能找到非常差的解。它更适合用动态规划解决。</p>
@@ -3292,7 +3292,7 @@
 <!-- Page content -->
 <h1 id="15">第 15 章 &nbsp; 贪心<a class="headerlink" href="#15" title="Permanent link">&para;</a></h1>
 <div class="center-table">
-<p><a class="glightbox" href="../assets/covers/chapter_greedy.jpg" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="贪心" src="../assets/covers/chapter_greedy.jpg" width="600" /></a></p>
+<p><a class="glightbox" href="../assets/covers/chapter_greedy.jpg" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="贪心" class="cover-image" src="../assets/covers/chapter_greedy.jpg" /></a></p>
 </div>
 <div class="admonition abstract">
 <p class="admonition-title">Abstract</p>
@@ -3372,7 +3372,7 @@
 <p>容器的容量等于高度和宽度的乘积（即面积），其中高度由较短的隔板决定，宽度是两个隔板的数组索引之差。</p>
 <p>请在数组中选择两个隔板，使得组成的容器的容量最大，返回最大容量。</p>
 </div>
-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大容量问题的示例数据" src="../max_capacity_problem.assets/max_capacity_example.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大容量问题的示例数据" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_example.png" /></a></p>
 <p align="center"> 图 15-7 &nbsp; 最大容量问题的示例数据 </p>

 <p>容器由任意两个隔板围成，<strong>因此本题的状态为两个隔板的索引，记为 <span class="arithmatex">\([i, j]\)</span></strong> 。</p>
@@ -3383,16 +3383,16 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
 <p>设数组长度为 <span class="arithmatex">\(n\)</span> ，两个隔板的组合数量（即状态总数）为 <span class="arithmatex">\(C_n^2 = \frac{n(n - 1)}{2}\)</span> 个。最直接地，<strong>我们可以穷举所有状态</strong>，从而求得最大容量，时间复杂度为 <span class="arithmatex">\(O(n^2)\)</span> 。</p>
 <h3 id="1">1. &nbsp; 贪心策略确定<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
 <p>这道题还有更高效率的解法。如图 15-8 所示，现选取一个状态 <span class="arithmatex">\([i, j]\)</span> ，其满足索引 <span class="arithmatex">\(i &lt; j\)</span> 且高度 <span class="arithmatex">\(ht[i] &lt; ht[j]\)</span> ，即 <span class="arithmatex">\(i\)</span> 为短板、<span class="arithmatex">\(j\)</span> 为长板。</p>
-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_initial_state.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="初始状态" src="../max_capacity_problem.assets/max_capacity_initial_state.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_initial_state.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="初始状态" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_initial_state.png" /></a></p>
 <p align="center"> 图 15-8 &nbsp; 初始状态 </p>

 <p>如图 15-9 所示，<strong>若此时将长板 <span class="arithmatex">\(j\)</span> 向短板 <span class="arithmatex">\(i\)</span> 靠近，则容量一定变小</strong>。</p>
 <p>这是因为在移动长板 <span class="arithmatex">\(j\)</span> 后，宽度 <span class="arithmatex">\(j-i\)</span> 肯定变小；而高度由短板决定，因此高度只可能不变（ <span class="arithmatex">\(i\)</span> 仍为短板）或变小（移动后的 <span class="arithmatex">\(j\)</span> 成为短板）。</p>
-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_moving_long_board.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="向内移动长板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_long_board.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_moving_long_board.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="向内移动长板后的状态" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_moving_long_board.png" /></a></p>
 <p align="center"> 图 15-9 &nbsp; 向内移动长板后的状态 </p>

 <p>反向思考，<strong>我们只有向内收缩短板 <span class="arithmatex">\(i\)</span> ，才有可能使容量变大</strong>。因为虽然宽度一定变小，<strong>但高度可能会变大</strong>（移动后的短板 <span class="arithmatex">\(i\)</span> 可能会变长）。例如在图 15-10 中，移动短板后面积变大。</p>
-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_moving_short_board.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="向内移动短板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_short_board.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_moving_short_board.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="向内移动短板后的状态" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_moving_short_board.png" /></a></p>
 <p align="center"> 图 15-10 &nbsp; 向内移动短板后的状态 </p>

 <p>由此便可推出本题的贪心策略：初始化两指针分裂容器两端，每轮向内收缩短板对应的指针，直至两指针相遇。</p>
@@ -3406,31 +3406,31 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
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-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大容量问题的贪心过程" src="../max_capacity_problem.assets/max_capacity_greedy_step1.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大容量问题的贪心过程" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step1.png" /></a></p>
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+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step2" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step2.png" /></a></p>
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+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step3" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step3.png" /></a></p>
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+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step4.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step4" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step4.png" /></a></p>
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+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step5.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step5" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step5.png" /></a></p>
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+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step6.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step6" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step6.png" /></a></p>
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+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step7.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step7" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step7.png" /></a></p>
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-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step8.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step8" src="../max_capacity_problem.assets/max_capacity_greedy_step8.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step8.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step8" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step8.png" /></a></p>
 </div>
 <div class="tabbed-block">
-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step9.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step9" src="../max_capacity_problem.assets/max_capacity_greedy_step9.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_greedy_step9.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="max_capacity_greedy_step9" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_greedy_step9.png" /></a></p>
 </div>
 </div>
 </div>
@@ -3707,7 +3707,7 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
 <div class="arithmatex">\[
 cap[i, i+1], cap[i, i+2], \dots, cap[i, j-2], cap[i, j-1]
 \]</div>
-<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_skipped_states.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="移动短板导致被跳过的状态" src="../max_capacity_problem.assets/max_capacity_skipped_states.png" /></a></p>
+<p><a class="glightbox" href="../max_capacity_problem.assets/max_capacity_skipped_states.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="移动短板导致被跳过的状态" class="animation-figure" src="../max_capacity_problem.assets/max_capacity_skipped_states.png" /></a></p>
 <p align="center"> 图 15-12 &nbsp; 移动短板导致被跳过的状态 </p>

 <p>观察发现，<strong>这些被跳过的状态实际上就是将长板 <span class="arithmatex">\(j\)</span> 向内移动的所有状态</strong>。而在第二步中，我们已经证明内移长板一定会导致容量变小。也就是说，被跳过的状态都不可能是最优解，<strong>跳过它们不会导致错过最优解</strong>。</p>
@@ -3370,7 +3370,7 @@
 <p class="admonition-title">Question</p>
 <p>给定一个正整数 <span class="arithmatex">\(n\)</span> ，将其切分为至少两个正整数的和，求切分后所有整数的乘积最大是多少。</p>
 </div>
-<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_definition.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大切分乘积的问题定义" src="../max_product_cutting_problem.assets/max_product_cutting_definition.png" /></a></p>
+<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_definition.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大切分乘积的问题定义" class="animation-figure" src="../max_product_cutting_problem.assets/max_product_cutting_definition.png" /></a></p>
 <p align="center"> 图 15-13 &nbsp; 最大切分乘积的问题定义 </p>

 <p>假设我们将 <span class="arithmatex">\(n\)</span> 切分为 <span class="arithmatex">\(m\)</span> 个整数因子，其中第 <span class="arithmatex">\(i\)</span> 个因子记为 <span class="arithmatex">\(n_i\)</span> ，即</p>
@@ -3393,13 +3393,13 @@ n &amp; \geq 4
 \]</div>
 <p>如图 15-14 所示，当 <span class="arithmatex">\(n \geq 4\)</span> 时，切分出一个 <span class="arithmatex">\(2\)</span> 后乘积会变大，<strong>这说明大于等于 <span class="arithmatex">\(4\)</span> 的整数都应该被切分</strong>。</p>
 <p><strong>贪心策略一</strong>：如果切分方案中包含 <span class="arithmatex">\(\geq 4\)</span> 的因子，那么它就应该被继续切分。最终的切分方案只应出现 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span>、<span class="arithmatex">\(3\)</span> 这三种因子。</p>
-<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="切分导致乘积变大" src="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer1.png" /></a></p>
+<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="切分导致乘积变大" class="animation-figure" src="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer1.png" /></a></p>
 <p align="center"> 图 15-14 &nbsp; 切分导致乘积变大 </p>

 <p>接下来思考哪个因子是最优的。在 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span>、<span class="arithmatex">\(3\)</span> 这三个因子中，显然 <span class="arithmatex">\(1\)</span> 是最差的，因为 <span class="arithmatex">\(1 \times (n-1) &lt; n\)</span> 恒成立，即切分出 <span class="arithmatex">\(1\)</span> 反而会导致乘积减小。</p>
 <p>如图 15-15 所示，当 <span class="arithmatex">\(n = 6\)</span> 时，有 <span class="arithmatex">\(3 \times 3 &gt; 2 \times 2 \times 2\)</span> 。<strong>这意味着切分出 <span class="arithmatex">\(3\)</span> 比切分出 <span class="arithmatex">\(2\)</span> 更优</strong>。</p>
 <p><strong>贪心策略二</strong>：在切分方案中，最多只应存在两个 <span class="arithmatex">\(2\)</span> 。因为三个 <span class="arithmatex">\(2\)</span> 总是可以被替换为两个 <span class="arithmatex">\(3\)</span> ，从而获得更大乘积。</p>
-<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最优切分因子" src="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer2.png" /></a></p>
+<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最优切分因子" class="animation-figure" src="../max_product_cutting_problem.assets/max_product_cutting_greedy_infer2.png" /></a></p>
 <p align="center"> 图 15-15 &nbsp; 最优切分因子 </p>

 <p>总结以上，可推出以下贪心策略。</p>
@@ -3671,7 +3671,7 @@ n = 3 a + b
 </div>
 </div>
 </div>
-<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_greedy_calculation.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大切分乘积的计算方法" src="../max_product_cutting_problem.assets/max_product_cutting_greedy_calculation.png" /></a></p>
+<p><a class="glightbox" href="../max_product_cutting_problem.assets/max_product_cutting_greedy_calculation.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="最大切分乘积的计算方法" class="animation-figure" src="../max_product_cutting_problem.assets/max_product_cutting_greedy_calculation.png" /></a></p>
 <p align="center"> 图 15-16 &nbsp; 最大切分乘积的计算方法 </p>

 <p><strong>时间复杂度取决于编程语言的幂运算的实现方法</strong>。以 Python 为例，常用的幂计算函数有三种。</p>