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        <div class="chatlog__reply-symbol"></div><img class="chatlog__avatar" src="https://cdn.discordapp.com/avatars/269801719796334592/c7ac354c0870623258accfe97d7b8ffa.png?size=512" alt="Avatar" loading="lazy">
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          <div class="chatlog__reply-author" style="color:rgb(231,76,60)" title="tarunon">tarunon</div>
          <div class="chatlog__reply-content"><span class="chatlog__reply-link" onclick="scrollToMessage(event,'877206900578586674')"> <span class="chatlog__markdown-spoiler chatlog__markdown-spoiler--hidden" onclick="showSpoiler(event, this)">[A0,A1]×[[X,Y],[Z,W]]=[A1,A2]となる行列XYZWを求める。この時、mod100は最終項の下2桁を取れば良いので無視する。A2=A0+A1のとき、行列は[[0,1],[1,1]]なので、答えは([A0,A1]×[[0,1],[1,1]]^N)[0]%100となる。計算量はO(logN)</span> </span> <span class="chatlog__reply-edited-timestamp" title="Tuesday, August 17, 2021 3:08 PM">(edited)</span></div>
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        <div class="chatlog__header"><span class="chatlog__author" style="color:rgb(241,196,15)" title="t.ae" data-user-id="269801719796334592">t.ae</span> <a href="/channels/763427634179735592?category=main&amp;channel=atcoder&amp;message_id=877416575785914418"><span class="chatlog__timestamp" title="Wednesday, August 18, 2021 5:00 AM"></span></a><a href="#chatlog__message-container-877416575785914418">8/18/2021 5:00 AM</a></div>
        <div class="chatlog__content chatlog__markdown"><span class="chatlog__markdown-preserve">これ対角化すれば行列の累乗計算いらないですね。 <code class="chatlog__markdown-pre chatlog__markdown-pre--inline">pow((1-sqrt(5))/2, N)</code> が出てきますがこれって計算量どうなるんですっけ</span></div>
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