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<?xml version="1.0" encoding="utf-8"?>
<search>
<entry>
<title><![CDATA[\[PAT乙级]1013 组个最小数]]></title>
<url>/2018/02/07/%5BPAT%E4%B9%99%E7%BA%A7%5D1013%20%E7%BB%84%E4%B8%AA%E6%9C%80%E5%B0%8F%E6%95%B0/</url>
<content type="html"><![CDATA[<p>####题目描述<br>给定数字0-9各若干个。你可以以任意顺序排列这些数字，但必须全部使用。目标是使得最后得到的数尽可能小（注意0不能做首位）。例如：给定两个0，两个1，三个5，一个8，我们得到的最小的数就是10015558。<br>现给定数字，请编写程序输出能够组成的最小的数。 </p>
<p>####输入描述:<br>每个输入包含1个测试用例。每个测试用例在一行中给出10个非负整数，顺序表示我们拥有数字0、数字1、……数字9的个数。整数间用一个空格分隔。10个数字的总个数不超过50，且至少拥有1个非0的数字。</p>
<p>####输出描述:<br>在一行中输出能够组成的最小的数。</p>
<p>####输入例子:<br>2 2 0 0 0 3 0 0 1 0</p>
<p>####输出例子:<br>10015558</p>
<p>####代码实现：<br><figure class="highlight plain"><figcaption><span><stdio.h></stdio.h></span></figcaption><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> int num[10],i;</span><br><span class="line"> for(i=0;i<10;i++)</span><br><span class="line"> scanf("%d",&num[i]);</span><br><span class="line"> if(num[0]!=0){</span><br><span class="line"> i=1;</span><br><span class="line"> while(num[i]==0&i<10)</span><br><span class="line"> i++;</span><br><span class="line"> if(i!=10){</span><br><span class="line"> printf("%d",i);</span><br><span class="line"> num[i]--;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> for(i=0;i<10;i++)</span><br><span class="line"> while(num[i]!=0){</span><br><span class="line"> printf("%d",i);</span><br><span class="line"> num[i]--;</span><br><span class="line"> }</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
]]></content>
</entry>
<entry>
<title><![CDATA[\[PAT乙级]1011 D进制的A+B]]></title>
<url>/2018/02/07/%5BPAT%E4%B9%99%E7%BA%A7%5D1011%20D%E8%BF%9B%E5%88%B6%E7%9A%84A+B/</url>
<content type="html"><![CDATA[<p>####题目描述<br>输入两个非负10进制整数A和B(<=230-1)，输出A+B的D (1 < D <= 10)进制数。</p>
<p>####输入描述:<br>输入在一行中依次给出3个整数A、B和D。</p>
<p>####输出描述:<br>输出A+B的D进制数。</p>
<p>####输入例子:<br>123 456 8</p>
<p>####输出例子:<br>1103</p>
<p>####实现代码：<br><figure class="highlight plain"><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><span class="line">#include <stdio.h></span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> long long int A,B,C=0;</span><br><span class="line"> int D,out[40];</span><br><span class="line"> scanf("%lld %lld %d",&A,&B,&D);</span><br><span class="line"> C=A+B;</span><br><span class="line"> int i=0;</span><br><span class="line"> while(C!=0){</span><br><span class="line"> out[i]=C%D;</span><br><span class="line"> i++;</span><br><span class="line"> C/=D;</span><br><span class="line"> }</span><br><span class="line"> i--;</span><br><span class="line"> while(i>=0){</span><br><span class="line"> printf("%d",out[i]);</span><br><span class="line"> i--;</span><br><span class="line"> }</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
]]></content>
</entry>
<entry>
<title><![CDATA[\[PAT乙级]1011 个位数统计]]></title>
<url>/2018/02/07/%5BPAT%E4%B9%99%E7%BA%A7%5D1011%20%E4%B8%AA%E4%BD%8D%E6%95%B0%E7%BB%9F%E8%AE%A1/</url>
<content type="html"><![CDATA[<p>####题目描述<br>给定一个k位整数N = dk-1<em>10k-1 + … + d1</em>101 + d0(0<=di<=9, i=0,…,k-1, dk-1>0)，请编写程序统计每种不同的个位数字出现的次数。例如：给定N = 100311，则有2个0，3个1，和1个3。</p>
<p>####输入描述:<br>每个输入包含1个测试用例，即一个不超过1000位的正整数N。</p>
<p>####输出描述:<br>对N中每一种不同的个位数字，以D:M的格式在一行中输出该位数字D及其在N中出现的次数M。要求按D的升序输出。</p>
<p>####输入例子:<br>100311</p>
<p>####输出例子:<br>0:2<br>1:3<br>3:1 </p>
<p>####tips:<br>为了方便统计，直接采用字符串的输入形式。</p>
<p>####实现代码：<br><figure class="highlight plain"><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br></pre></td><td class="code"><pre><span class="line">#include <stdio.h></span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> char num[1001];</span><br><span class="line"> int i,con[10]={0,};</span><br><span class="line"> scanf("%s",num);</span><br><span class="line"> for(i=0;num[i]!='\0';i++)</span><br><span class="line"> switch (num[i]) {</span><br><span class="line"> case('0'):{</span><br><span class="line"> con[0]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '1':{</span><br><span class="line"> con[1]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '2':{</span><br><span class="line"> con[2]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '3':{</span><br><span class="line"> con[3]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '4':{</span><br><span class="line"> con[4]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '5':{</span><br><span class="line"> con[5]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '6':{</span><br><span class="line"> con[6]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '7':{</span><br><span class="line"> con[7]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '8':{</span><br><span class="line"> con[8]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> case '9':{</span><br><span class="line"> con[9]++;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> for(i=0;i<10;i++){</span><br><span class="line"> if(con[i]==0)</span><br><span class="line"> continue;</span><br><span class="line"> else{</span><br><span class="line"> printf("%d:%d\n",i,con[i]);</span><br><span class="line"> con[i]=0;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
]]></content>
</entry>
<entry>
<title><![CDATA[\[PAT乙级]1010 月饼]]></title>
<url>/2018/02/07/%5BPAT%E4%B9%99%E7%BA%A7%5D1010%20%E6%9C%88%E9%A5%BC/</url>
<content type="html"><![CDATA[<p>####题目描述<br>月饼是中国人在中秋佳节时吃的一种传统食品，不同地区有许多不同风味的月饼。现给定所有种类月饼的库存量、总售价、以及市场的最大需求量，请你计算可以获得的最大收益是多少。<br>注意：销售时允许取出一部分库存。样例给出的情形是这样的：假如我们有3种月饼，其库存量分别为18、15、10万吨，总售价分别为75、72、45亿元。如果市场的最大需求量只有20万吨，那么我们最大收益策略应该是卖出全部15万吨第2种月饼、以及5万吨第3种月饼，获得72 + 45/2 = 94.5（亿元）。</p>
<p>####输入描述:<br>每个输入包含1个测试用例。每个测试用例先给出一个不超过1000的正整数N表示月饼的种类数、以及不超过500（以万吨为单位）的正整数D表示市场最大需求量。随后一行给出N个正数表示每种月饼的库存量（以万吨为单位）；最后一行给出N个正数表示每种月饼的总售价（以亿元为单位）。数字间以空格分隔。</p>
<p>####输出描述:<br>对每组测试用例，在一行中输出最大收益，以亿元为单位并精确到小数点后2位。</p>
<p>####输入例子:<br>3 20<br>18 15 10<br>75 72 45 </p>
<p>####输出例子:<br>94.50</p>
<p>####tips：<br>1.月饼的总售价需要转换成月饼单价，如果需要有最大收益，则需要尽量卖出单价高的月饼。<br>2.每次都卖出单价最高的月饼，然后将单价最高的月饼排出排序范围。我用的方法是将当次单价最高的月饼放到下次排序的数组前面一个地方，然后将还未排序的月饼放入当次排序最高的地方。（交换）</p>
<p>####实现代码：<br><figure class="highlight plain"><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br></pre></td><td class="code"><pre><span class="line">#include <stdio.h></span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> int N,kind[1000],all;</span><br><span class="line"> float price[1000],ap=0,pt[1000];</span><br><span class="line"> scanf("%d %d",&N,&all);</span><br><span class="line"> for(int i=0;i<N;i++)</span><br><span class="line"> scanf("%d",&kind[i]);</span><br><span class="line"> for(int i=0;i<N;i++){</span><br><span class="line"> scanf("%f",&price[i]);</span><br><span class="line"> pt[i]=price[i]/kind[i];</span><br><span class="line"> }</span><br><span class="line"> while(all!=0){</span><br><span class="line"> int i,j,tem;</span><br><span class="line"> for(i=0;i<N;i++){</span><br><span class="line"> tem=i;</span><br><span class="line"> for(j=i+1;j<N;j++)</span><br><span class="line"> if(pt[j]>pt[tem])</span><br><span class="line"> tem=j;</span><br><span class="line"> if(all>=kind[tem]){</span><br><span class="line"> ap+=price[tem];</span><br><span class="line"> all=all-kind[tem];</span><br><span class="line"> }</span><br><span class="line"> else{</span><br><span class="line"> ap+=(price[tem]*all/kind[tem]);</span><br><span class="line"> all=0;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> price[tem]=price[i];</span><br><span class="line"> kind[tem]=kind[i];</span><br><span class="line"> pt[tem]=pt[i];</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> printf("%.2f",ap);</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
]]></content>
</entry>
<entry>
<title><![CDATA[\[PAT乙级]1009 数字黑洞]]></title>
<url>/2018/02/03/%5BPAT%E4%B9%99%E7%BA%A7%5D1009%20%E6%95%B0%E5%AD%97%E9%BB%91%E6%B4%9E/</url>
<content type="html"><![CDATA[<p>###题目描述<br>给定任一个各位数字不完全相同的4位正整数，如果我们先把4个数字按非递增排序，再按非递减排序，然后用第1个数字减第2个数字，将得到一个新的数字。一直重复这样做，我们很快会停在有“数字黑洞”之称的6174，这个神奇的数字也叫Kaprekar常数。例如，我们从6767开始，将得到<br> 7766 - 6677 = 1089<br> 9810 - 0189 = 9621<br> 9621 - 1269 = 8352<br> 8532 - 2358 = 6174<br> 7641 - 1467 = 6174<br> … …<br> 现给定任意4位正整数，请编写程序演示到达黑洞的过程。</p>
<p>###输入描述:<br>输入给出一个(0, 10000)区间内的正整数N。</p>
<p>###输出描述:<br>如果N的4位数字全相等，则在一行内输出“N - N = 0000”；否则将计算的每一步在一行内输出，直到6174作为差出现，输出格式见样例,每行中间没有空行。注意每个数字按4位数格式输出。</p>
<p>###输入例子:<br>6767</p>
<p>###输出例子:<br>7766 - 6677 = 1089<br>9810 - 0189 = 9621<br>9621 - 1269 = 8352<br>8532 - 2358 = 6174</p>
<p>###tips：<br>%d：输出一个整数<br>%4d：输出一个占四位的整数<br>%04d：输出一个占四位的整数，不足部分前面补0 </p>
<p>###代码解答<br><figure class="highlight plain"><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br></pre></td><td class="code"><pre><span class="line">#include <stdio.h></span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> int num,a=0,b=0,c,i,in[4];</span><br><span class="line"> scanf("%d",&num);</span><br><span class="line"> c=num;</span><br><span class="line"> for(i=0;i<4;i++){</span><br><span class="line"> in[i]=num%10;</span><br><span class="line"> num/=10;</span><br><span class="line"> }</span><br><span class="line"> if(in[0]==in[1]&&in[1]==in[2]&&in[2]==in[3])</span><br><span class="line"> printf("%04d - %04d = 0000",c,c);</span><br><span class="line"> else</span><br><span class="line"> while(1){</span><br><span class="line"> for(i=0;i<4;i++)</span><br><span class="line"> for(int j=i+1;j<4;j++)</span><br><span class="line"> if(in[i]>in[j]){</span><br><span class="line"> int tem;</span><br><span class="line"> tem = in[j];</span><br><span class="line"> in[j]=in[i];</span><br><span class="line"> in[i]=tem;</span><br><span class="line"> }</span><br><span class="line"> for(i=0;i<4;i++){</span><br><span class="line"> b=b*10+in[i];</span><br><span class="line"> a=a*10+in[3-i];</span><br><span class="line"> }</span><br><span class="line"> c=a-b;</span><br><span class="line"> printf("%04d - %04d = %4d\n",a,b,c);</span><br><span class="line"> if(c==6174)</span><br><span class="line"> break;</span><br><span class="line"> num=c;</span><br><span class="line"> for(i=0;i<4;i++){</span><br><span class="line"> in[i]=num%10;</span><br><span class="line"> num/=10;</span><br><span class="line"> }</span><br><span class="line"> a=0;b=0;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
]]></content>
</entry>
<entry>
<title><![CDATA[\[PAT乙级]1003 数素数]]></title>
<url>/2018/02/03/%5BPAT%E4%B9%99%E7%BA%A7%5D1003%20%E6%95%B0%E7%B4%A0%E6%95%B0/</url>
<content type="html"><![CDATA[<h3 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述:"></a>题目描述:</h3><p>令Pi表示第i个素数。现任给两个正整数M <= N <= 10000，请输出PM到PN的所有素数。</p>
<h3 id="输入描述"><a href="#输入描述" class="headerlink" title="输入描述:"></a>输入描述:</h3><p>输入在一行中给出M和N，其间以空格分隔。</p>
<h3 id="输出描述"><a href="#输出描述" class="headerlink" title="输出描述:"></a>输出描述:</h3><p>输出从PM到PN的所有素数，每10个数字占1行，其间以空格分隔，但行末不得有多余空格。</p>
<h3 id="输入例子"><a href="#输入例子" class="headerlink" title="输入例子:"></a>输入例子:</h3><p>5 27</p>
<h3 id="输出例子"><a href="#输出例子" class="headerlink" title="输出例子:"></a>输出例子:</h3><p>11 13 17 19 23 29 31 37 41 43<br>47 53 59 61 67 71 73 79 83 89<br>97 101 103</p>
<h3 id="解答代码："><a href="#解答代码：" class="headerlink" title="解答代码："></a>解答代码：</h3><figure class="highlight plain"><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line">#include <stdio.h></span><br><span class="line">#include <math.h></span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> int M,N,i,flag=1,num=0;</span><br><span class="line"> double now=2,k=2;</span><br><span class="line"> scanf("%d %d",&N,&M);</span><br><span class="line"> for(i=0;i<M;i++){</span><br><span class="line"> flag=1;</span><br><span class="line"> k=2;</span><br><span class="line"> while(flag & k<=sqrt(now)){</span><br><span class="line"> for(k=2;k<=sqrt(now);k++)</span><br><span class="line"> if(int(now)%int(k)==0){</span><br><span class="line"> flag=0;</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> if(flag==0)</span><br><span class="line"> i--;</span><br><span class="line"> else if(i>=N-1){</span><br><span class="line"> printf("%.0f",now);</span><br><span class="line"> num++;</span><br><span class="line"> if(num%10==0)</span><br><span class="line"> printf("\n");</span><br><span class="line"> else{</span><br><span class="line"> if(i==M-1)</span><br><span class="line"> break;</span><br><span class="line"> printf(" ");</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> now++;</span><br><span class="line"> }</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
]]></content>
</entry>
<entry>
<title><![CDATA[\[PAT乙级]1002 数字分类]]></title>
<url>/2018/02/03/%5BPAT%E4%B9%99%E7%BA%A7%5D1002%20%E6%95%B0%E5%AD%97%E5%88%86%E7%B1%BB/</url>
<content type="html"><![CDATA[<h3 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h3><p>给定一系列正整数，请按要求对数字进行分类，并输出以下5个数字：<br>A1 = 能被5整除的数字中所有偶数的和；<br>A2 = 将被5除后余1的数字按给出顺序进行交错求和，即计算n1-n2+n3-n4…；<br>A3 = 被5除后余2的数字的个数；<br>A4 = 被5除后余3的数字的平均数，精确到小数点后1位；<br>A5 = 被5除后余4的数字中最大数字。</p>
<h3 id="输入描述"><a href="#输入描述" class="headerlink" title="输入描述:"></a>输入描述:</h3><p>每个输入包含1个测试用例。每个测试用例先给出一个不超过1000的正整数N，随后给出N个不超过1000的待分类的正整数。数字间以空格分隔。</p>
<h3 id="输出描述"><a href="#输出描述" class="headerlink" title="输出描述:"></a>输出描述:</h3><p>对给定的N个正整数，按题目要求计算A1~A5并在一行中顺序输出。数字间以空格分隔，但行末不得有多余空格。若其中某一类数字不存在，则在相应位置输出“N”。</p>
<h3 id="输入例子"><a href="#输入例子" class="headerlink" title="输入例子:"></a>输入例子:</h3><p>13 1 2 3 4 5 6 7 8 9 10 20 16 18</p>
<h3 id="输出例子"><a href="#输出例子" class="headerlink" title="输出例子:"></a>输出例子:</h3><p>30 11 2 9.7 9</p>
<h3 id="解答代码："><a href="#解答代码：" class="headerlink" title="解答代码："></a>解答代码：</h3><figure class="highlight plain"><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br></pre></td><td class="code"><pre><span class="line">#include <stdio.h></span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> int N,num,i,flag[5]={0,},nums[5]={0,},now=1;</span><br><span class="line"> double n=0;</span><br><span class="line"> scanf("%d",&N);</span><br><span class="line"> for(i=0;i<N;i++)</span><br><span class="line"> {</span><br><span class="line"> scanf("%d",&num);</span><br><span class="line"> if(num%10==0)</span><br><span class="line"> {</span><br><span class="line"> nums[0]+=num;</span><br><span class="line"> flag[0]=1;</span><br><span class="line"> }</span><br><span class="line"> else if(num%5==1)</span><br><span class="line"> {</span><br><span class="line"> flag[1]=1;</span><br><span class="line"> nums[1]+=now*num;</span><br><span class="line"> now*=-1;</span><br><span class="line"> }</span><br><span class="line"> else if(num%5==2)</span><br><span class="line"> {</span><br><span class="line"> flag[2]=1;</span><br><span class="line"> nums[2]++;</span><br><span class="line"> }</span><br><span class="line"> else if(num%5==3)</span><br><span class="line"> {</span><br><span class="line"> nums[3]+=num;</span><br><span class="line"> flag[3]=1;</span><br><span class="line"> n++;</span><br><span class="line"> }</span><br><span class="line"> else if(num%5==4)</span><br><span class="line"> {</span><br><span class="line"> flag[4]=1;</span><br><span class="line"> if(num>nums[4])</span><br><span class="line"> nums[4]=num;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> for(i=0;i<4;i++)</span><br><span class="line"> {</span><br><span class="line"> if(i==3)</span><br><span class="line"> {</span><br><span class="line"> if(flag[i]==0)</span><br><span class="line"> printf("N ");</span><br><span class="line"> else</span><br><span class="line"> printf("%.1f ",nums[3]/n);</span><br><span class="line"> }</span><br><span class="line"> else</span><br><span class="line"> {</span><br><span class="line"> if(flag[i]==0)</span><br><span class="line"> printf("N ");</span><br><span class="line"> else</span><br><span class="line"> printf("%d ",nums[i]);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> if(flag[i]==0)</span><br><span class="line"> printf("N");</span><br><span class="line"> else</span><br><span class="line"> printf("%d",nums[i]);</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
]]></content>
</entry>
<entry>
<title><![CDATA[/[PAT乙级]1001 A+B和C]]></title>
<url>/2018/02/03/%5BPAT%E4%B9%99%E7%BA%A7%5D1001%20A+B%E5%92%8CC/</url>
<content type="html"><![CDATA[<h3 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述:"></a>题目描述:</h3><p>给定区间[-2的31次方, 2的31次方]内的3个整数A、B和C，请判断A+B是否大于C。</p>
<h3 id="输入描述"><a href="#输入描述" class="headerlink" title="输入描述:"></a>输入描述:</h3><p>输入第1行给出正整数T(<=10)，是测试用例的个数。随后给出T组测试用例，每组占一行，顺序给出A、B和C。整数间以空格分隔。</p>
<h3 id="输出描述"><a href="#输出描述" class="headerlink" title="输出描述:"></a>输出描述:</h3><p>对每组测试用例，在一行中输出“Case #X: true”如果A+B>C，否则输出“Case #X: false”，其中X是测试用例的编号（从1开始）。</p>
<h3 id="输入例子"><a href="#输入例子" class="headerlink" title="输入例子:"></a>输入例子:</h3><p>4<br>1 2 3<br>2 3 4<br>2147483647 0 2147483646<br>0 -2147483648 -2147483647 </p>
<h3 id="输出例子"><a href="#输出例子" class="headerlink" title="输出例子:"></a>输出例子:</h3><p>Case #1: false<br>Case #2: true<br>Case #3: true<br>Case #4: false </p>
<h3 id="解答代码："><a href="#解答代码：" class="headerlink" title="解答代码："></a>解答代码：</h3><figure class="highlight plain"><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><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line">#include <iostream></span><br><span class="line">int main()</span><br><span class="line">{</span><br><span class="line"> double a,b,c;</span><br><span class="line"> int i,num;</span><br><span class="line"> scanf("%d",&num);</span><br><span class="line"> for(i=0;i<num;i++)</span><br><span class="line"> {</span><br><span class="line"> scanf("%lf %lf %lf",&a,&b,&c);</span><br><span class="line"> if(a+b>c)</span><br><span class="line"> printf("Case #%d: true\n",i+1);</span><br><span class="line"> else</span><br><span class="line"> printf("Case #%d: false\n",i+1);</span><br><span class="line"> }</span><br><span class="line"> return 0;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
]]></content>
</entry>
<entry>
<title><![CDATA[PAT乙级题目笔记全解目录]]></title>
<url>/2018/02/03/PAT%E4%B9%99%E7%BA%A7%E9%A2%98%E7%9B%AE%E7%AC%94%E8%AE%B0%E5%85%A8%E8%A7%A3%E7%9B%AE%E5%BD%95/</url>
<content type="html"><![CDATA[<p>记于2018年2月，目标，十天内刷完PAT乙级然后晋击甲级。因时间有限，只刷真题。 </p>
<h4 id="目录："><a href="#目录：" class="headerlink" title="目录："></a>目录：</h4><p>1001 A+B和C(15)<br>1002 数字分类(20)<br>1003 数素数(20)<br>1004 福尔摩斯的约会(20)<br>1005 德才论(25)<br>1006 部分A+B(15)<br>1007 A除以B<br>1008 锤子剪刀布<br>1009 数字黑洞<br>1010 月饼 </p>
]]></content>
</entry>
<entry>
<title><![CDATA[masscan源码跟随笔记]]></title>
<url>/2018/01/24/masscan%E6%BA%90%E7%A0%81%E8%B7%9F%E9%9A%8F%E7%AC%94%E8%AE%B0/</url>
<content type="html"><![CDATA[<h3 id="入口进入："><a href="#入口进入：" class="headerlink" title="入口进入："></a>入口进入：</h3><pre><code>masscan_command_line 命令行入口
</code></pre><h3 id="头文件："><a href="#头文件：" class="headerlink" title="头文件："></a>头文件：</h3><pre><code>masscan.h 定义Masscan，用于接收命令行参数
</code></pre>]]></content>
</entry>
<entry>
<title><![CDATA[安装笔记]]></title>
<url>/2018/01/17/*%E5%AE%89%E8%A3%85%E7%AC%94%E8%AE%B0/</url>
<content type="html"><![CDATA[<h4 id="brew安装："><a href="#brew安装：" class="headerlink" title="brew安装："></a>brew安装：</h4><p>地址：/usr/local/Cellar<br>–python2.7<br>–python3.5</p>
<h4 id="git安装："><a href="#git安装：" class="headerlink" title="git安装："></a>git安装：</h4><p>地址：/Users/seventailcat/git_things<br>–masscan & make</p>
]]></content>
</entry>
<entry>
<title><![CDATA[Mac安装Homebrew]]></title>
<url>/2018/01/17/Mac%E5%AE%89%E8%A3%85Homebrew/</url>
<content type="html"><![CDATA[<p>系统：Mac OS High Sierra 10.13.2</p>
<p> Mac上官配的python是python2.7，然而在学校的做课程设计的时候，因为要配置环境，电脑环境混乱，刷机后决定用homebrew重新安装python。</p>
<h4 id="安装方法："><a href="#安装方法：" class="headerlink" title="安装方法："></a>安装方法：</h4><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">ruby -e <span class="string">"<span class="variable">$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/master/install)</span>"</span></span><br></pre></td></tr></table></figure>
<p>Homebrew官网：<a href="https://brew.sh/" target="_blank" rel="noopener">https://brew.sh/</a></p>
<h4 id="使用："><a href="#使用：" class="headerlink" title="使用："></a>使用：</h4><pre><code>安装：brew install wget
</code></pre>]]></content>
</entry>
<entry>
<title><![CDATA[Hello World]]></title>
<url>/2018/01/15/hello-world/</url>
<content type="html"><![CDATA[<p>Welcome to <a href="https://hexo.io/" target="_blank" rel="noopener">Hexo</a>! This is your very first post. Check <a href="https://hexo.io/docs/" target="_blank" rel="noopener">documentation</a> for more info. If you get any problems when using Hexo, you can find the answer in <a href="https://hexo.io/docs/troubleshooting.html" target="_blank" rel="noopener">troubleshooting</a> or you can ask me on <a href="https://github.com/hexojs/hexo/issues" target="_blank" rel="noopener">GitHub</a>.</p>
<h2 id="Quick-Start"><a href="#Quick-Start" class="headerlink" title="Quick Start"></a>Quick Start</h2><h3 id="Create-a-new-post"><a href="#Create-a-new-post" class="headerlink" title="Create a new post"></a>Create a new post</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo new <span class="string">"My New Post"</span></span><br></pre></td></tr></table></figure>
<p>More info: <a href="https://hexo.io/docs/writing.html" target="_blank" rel="noopener">Writing</a></p>
<h3 id="Run-server"><a href="#Run-server" class="headerlink" title="Run server"></a>Run server</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo server</span><br></pre></td></tr></table></figure>
<p>More info: <a href="https://hexo.io/docs/server.html" target="_blank" rel="noopener">Server</a></p>
<h3 id="Generate-static-files"><a href="#Generate-static-files" class="headerlink" title="Generate static files"></a>Generate static files</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo generate</span><br></pre></td></tr></table></figure>
<p>More info: <a href="https://hexo.io/docs/generating.html" target="_blank" rel="noopener">Generating</a></p>
<h3 id="Deploy-to-remote-sites"><a href="#Deploy-to-remote-sites" class="headerlink" title="Deploy to remote sites"></a>Deploy to remote sites</h3><figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">$ hexo deploy</span><br></pre></td></tr></table></figure>
<p>More info: <a href="https://hexo.io/docs/deployment.html" target="_blank" rel="noopener">Deployment</a></p>
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