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2019/08/14/题解 P5497 【[LnOI2019SP]龟速单项式变换(SMT)】/index.html
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| <meta property="og:description" content="这题不是大水题吗，我随便口胡就做出来了 ——来自周子衡的嘲讽 首先考虑连续一段$[l,r]$的和 设$s[i]$为前$i$个数的和 则有$sum(l,r)=s[r]-s[l-1]$ 若$s[r]\equiv s[l-1] \pmod{m}$，则$sum(l,r)$为$m$的倍数 所以若$n\geq m$，则$s[0],s[1]…s[n]$共$n+1$个数，由容斥原理可知，必有两个数$\pmod{m"> | ||
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| <meta name="twitter:description" content="这题不是大水题吗，我随便口胡就做出来了 ——来自周子衡的嘲讽 首先考虑连续一段$[l,r]$的和 设$s[i]$为前$i$个数的和 则有$sum(l,r)=s[r]-s[l-1]$ 若$s[r]\equiv s[l-1] \pmod{m}$，则$sum(l,r)$为$m$的倍数 所以若$n\geq m$，则$s[0],s[1]…s[n]$共$n+1$个数，由容斥原理可知，必有两个数$\pmod{m"> | ||
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| 题解 P5497 【[LnOI2019SP]龟速单项式变换(SMT)】 | ||
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| <p>这题不是大水题吗，我随便口胡就做出来了 ——来自<a href="https://www.luogu.org/space/show?uid=112794" target="_blank" rel="noopener">周子衡</a>的嘲讽</p> | ||
| <p>首先考虑连续一段$[l,r]$的和</p> | ||
| <p>设$s[i]$为前$i$个数的和</p> | ||
| <p>则有$sum(l,r)=s[r]-s[l-1]$</p> | ||
| <p>若$s[r]\equiv s[l-1] \pmod{m}$，则$sum(l,r)$为$m$的倍数</p> | ||
| <p>所以若$n\geq m$，则$s[0],s[1]…s[n]$共$n+1$个数，<br>由容斥原理可知，必有两个数$\pmod{m}$同余，即这个序列为”司$m$序列”</p> | ||
| <p>若$n<m$，则可以令$a[i]=1,i\in [0,n]$，此时$s[0]…s[n]$<br>$\pmod{m}$的余数均不相等，此时序列不为”司$m$序列”</p> | ||
| <p>放代码</p> | ||
| <figure class="highlight cpp"><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></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="keyword">long</span> <span class="keyword">long</span> x,y;</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%lld%lld"</span>,&x,&y);</span><br><span class="line"> <span class="keyword">if</span>(x>=y) <span class="built_in">printf</span>(<span class="string">"YES"</span>);</span><br><span class="line"> <span class="keyword">else</span> <span class="built_in">printf</span>(<span class="string">"NO"</span>);</span><br><span class="line">}</span><br></pre></td></tr></table></figure> | ||
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