https://111qqz.com/en/tags/%E5%89%A9%E4%BD%99%E7%B3%BB/Recent content in 剩余系 on 111qqz's blogHugo -- gohugo.ioen© 2015-2026 111qqzMon, 10 Oct 2016 12:08:41 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-10-10-e4ba8ce6aca1e589a9e4bd99efbc88cipollas-algorithmefbc89e5ada6e4b9a0e7ac94e8aeb0/Mon, 10 Oct 2016 12:08:41 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-10-10-e4ba8ce6aca1e589a9e4bd99efbc88cipollas-algorithmefbc89e5ada6e4b9a0e7ac94e8aeb0/<p>先放资料。</p> <hr> <h3 class="relative group"><strong>前置技能点：</strong> <div id="前置技能点" class="anchor"></div> <span class="absolute top-0 w-6 transition-opacity opacity-0 -start-6 not-prose group-hover:opacity-100 select-none"> <a class="text-primary-300 dark:text-neutral-700 !no-underline" href="#%e5%89%8d%e7%bd%ae%e6%8a%80%e8%83%bd%e7%82%b9" aria-label="Anchor">#</a> </span> </h3> <p><a href="http://baike.baidu.com/link?url=7nEL8QRUFHmQQor4elIUVpLVJG0i63Q3iRRXhrzzr1p2bhJ4qq4d-whxKrZhx6O8GEBf_iv_wsM2bWH1NJaJDVyLM8Mft0PCPAhMnYj6eNG2zh4lTNt5YvGDn1R4vrYE" target="_blank" rel="noreferrer">剩余系</a></p>https://111qqz.com/en/post/acm-icpc/2016/2016-02-29-poj2356/Mon, 29 Feb 2016 13:06:00 +0000https://111qqz.com/en/post/acm-icpc/2016/2016-02-29-poj2356/<p><a href="http://poj.org/problem?id=2356" target="_blank" rel="noreferrer">http://poj.org/problem?id=2356</a></p> <p>题意：有 n 个数，从中选取若干个（1..n），和能被 n 整除。问是否有解，无解输出 0，有解的话，<strong>输出个数以及选择的 <code>a[i]</code>（不是 i）</strong>。</p>https://111qqz.com/en/post/acm-icpc/2015/2015-08-21-poj3370halloweentreatse589a9e4bd99e7b1bbe68abde5b189e58e9fe79086/Fri, 21 Aug 2015 05:36:00 +0000https://111qqz.com/en/post/acm-icpc/2015/2015-08-21-poj3370halloweentreatse589a9e4bd99e7b1bbe68abde5b189e58e9fe79086/<p>昨天那道签到的数学题没搞出来不开心.</p> <p>是时候刷一波数学了</p> <p>这题题意是说,从n个数中任选m个,使得m个的和为c的倍数.</p>