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chapter_dynamic_programming/unbounded_knapsack_problem/index.html
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@@ -4684,7 +4684,7 @@ dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)
<div class="arithmatex">\[
dp[i, a] = dp[i-1, a] + dp[i, a - coins[i-1]]
\]</div>
<p>当目标金额为 <span class="arithmatex">\(0\)</span> 时，无需选择任何硬币即可凑出目标金额，因此应将首列所有 <span class="arithmatex">\(dp[i, 0]\)</span> 都初始化为 <span class="arithmatex">\(1\)</span> 。当无硬币时，无法凑出任何 <span class="arithmatex">\(&gt;0\)</span> 的目标金额，因此首行所有 <span class="arithmatex">\(dp[0, a]\)</span> 都等于 <span class="arithmatex">\(0\)</span> 。</p>
<p>当目标金额为 <span class="arithmatex">\(0\)</span> 时，无须选择任何硬币即可凑出目标金额，因此应将首列所有 <span class="arithmatex">\(dp[i, 0]\)</span> 都初始化为 <span class="arithmatex">\(1\)</span> 。当无硬币时，无法凑出任何 <span class="arithmatex">\(&gt;0\)</span> 的目标金额，因此首行所有 <span class="arithmatex">\(dp[0, a]\)</span> 都等于 <span class="arithmatex">\(0\)</span> 。</p>
<h3 id="2_2">2. &nbsp; 代码实现<a class="headerlink" href="#2_2" title="Permanent link">&para;</a></h3>
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