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<h2 id="Brief"><a href="#Brief" class="headerlink" title="Brief"></a>Brief</h2><p>本章節的題目比較沒有什麼特定演算法，根據題目敘述去解即可</p>
<a id="more"></a>
<h2 id="Your-Ride-Is-Here"><a href="#Your-Ride-Is-Here" class="headerlink" title="Your Ride Is Here"></a><a href="http://cerberus.delos.com:790/usacoprob2?a=23kbfvJXJab&S=ride" target="_blank" rel="noopener">Your Ride Is Here</a></h2><p>將兩個input都分別轉換成數字，最後再用mod去比較看看是否相等即可</p>
<h2 id="Greedy-Gift-Givers"><a href="#Greedy-Gift-Givers" class="headerlink" title="Greedy Gift Givers"></a><a href="http://cerberus.delos.com:790/usacoprob2?a=23kbfvJXJab&S=gift1" target="_blank" rel="noopener">Greedy Gift Givers</a></h2><p>將所有人的金錢按照規則分配下去，即可得到答案</p>
<h2 id="Friday-the-Thirteenth"><a href="#Friday-the-Thirteenth" class="headerlink" title="Friday the Thirteenth"></a><a href="http://cerberus.delos.com:790/usacoprob2?a=23kbfvJXJab&S=friday" target="_blank" rel="noopener">Friday the Thirteenth</a></h2><p>題目規定輸出的時候，必須要按照<code>"Saturday, Sunday, Monday, Tuesday, ..., Friday,"</code>這邊是有涵義的，<br>因為 1900/01/13 第一個十三號就是星期六，以此為條件在計算上會更方便。</p>
<p>使用一個陣列，先記住每個月份的天數<br><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">int</span> days[<span class="number">12</span>]={<span class="number">31</span>,<span class="number">28</span>,<span class="number">31</span>,<span class="number">30</span>,<span class="number">31</span>,<span class="number">30</span>,<span class="number">31</span>,<span class="number">31</span>,<span class="number">30</span>,<span class="number">31</span>,<span class="number">30</span>,<span class="number">31</span>};</span><br></pre></td></tr></table></figure></p>
<p>接者根據題目要求的年數來迴圈計算，每次計算時要先判斷該年是不是閏年<br><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">if</span>( ( <span class="number">0</span> == i%<span class="number">4</span> && <span class="number">0</span> !=i%<span class="number">100</span>) || (i+<span class="number">1900</span>)%<span class="number">1000</span>==<span class="number">0</span>) </span><br><span class="line"> leap = <span class="number">1</span>;</span><br><span class="line">``` </span><br><span class="line">接下來則是計算每個月份的天數，使用day這個變數紀錄十三號是星期幾，使用ans這個<span class="built_in">array</span>來紀錄每個星期出現多少次十三號，如果是閏年的二月份，則記得要再多加一天，</span><br><span class="line">``` c</span><br><span class="line"><span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">0</span>; j<<span class="number">12</span>;++j){</span><br><span class="line"> ans[day]++;</span><br><span class="line"> addDays(day,days[j]);</span><br><span class="line"> <span class="keyword">if</span>(j==<span class="number">1</span> && leap)</span><br><span class="line"> addDays(day,<span class="number">1</span>);</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
<p>最後就按照順序把結果輸出即可</p>
<h2 id="Broken-Necklace"><a href="#Broken-Necklace" class="headerlink" title="Broken Necklace"></a><a href="http://cerberus.delos.com:790/usacoprob2?a=23kbfvJXJab&S=beads" target="_blank" rel="noopener">Broken Necklace</a></h2><p>這題可以使用暴力法，每個點都往左右兩邊去試試看，此方法就很直覺，不再多說。<br>若為了效率，我們可以採用DP的方法來計算答案，<br>首先，為了計算方便，先將input給複製一份並串起來<br>我們的DP的概念是紀錄第<code>I</code>點能夠往左與往右的最大可能性，紅珠跟藍珠要分開來計算<br>左邊部分，我們計算不包含I以前的往左最大值<br><figure class="highlight"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">r_left[i] = r_left[i-1]+1 if input[i-1] == 'red' or input[i-1] == 'white'</span><br><span class="line"> 0 if input[i-1] == 'blue'</span><br><span class="line">b_left[i] = b_left[i-1]+1 if input[i-1] == 'blue' or input[i-1] == 'white'</span><br><span class="line"> 0 if input[i-1] == 'red'</span><br></pre></td></tr></table></figure></p>
<p>右邊部分則是計算包含I後往右的最大值。<br><figure class="highlight"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">r_right[i] = r_right[i+1]+1 if input[i] =='red' or input[i] =='white'</span><br><span class="line"> 0 if input[i] =='blue'</span><br><span class="line">b_right[i] = b_right[i+1]+1 if input[i] =='blue' or input[i] =='white'</span><br><span class="line"> 0 if input[i] =='red'</span><br></pre></td></tr></table></figure></p>
<p>當全部都計算完畢後，就取最大值，每個點的最大值就是左邊最大加上右邊最大。<br><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<num*<span class="number">2</span>;i++){</span><br><span class="line"> max_ans = max(max_ans,max(r_left[i],b_left[i])+max(r_right[i],b_right[i]));</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
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