<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[大了七八个的博客]]></title><description><![CDATA[全部大残]]></description><link>https://blog.youkuaiyun.com/Ychaoding</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Ychaoding]]></copyright><item><title><![CDATA[ERROR: Could not find a version that satisfies the requirement]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/126469021</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/126469021</guid><author>Ychaoding</author><pubDate>Mon, 22 Aug 2022 16:42:25 +0800</pubDate><description><![CDATA[ERROR: Could not find a version that satisfies the requirement]]></description><category></category></item><item><title><![CDATA[1030 Travel Plan (30分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106574291</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106574291</guid><author>Ychaoding</author><pubDate>Fri, 05 Jun 2020 17:02:40 +0800</pubDate><description><![CDATA[1030 Travel Plan (30分)
分析
采用迪杰斯特拉算法，选取最短路径的同时记录前驱节点，若到同一点的最短路径不止一条，取花费最少的一条
代码
#include&lt;iostream&gt;
#include&lt;algorithm&gt;
#include&lt;queue&gt;
#include&lt;vector&gt;
using namespace std;
int n, m, s, d, a, b, l, co, distTo[510], pre[510], marked[5]]></description><category></category></item><item><title><![CDATA[1029 Median (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106568827</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106568827</guid><author>Ychaoding</author><pubDate>Fri, 05 Jun 2020 13:47:00 +0800</pubDate><description><![CDATA[1029 Median (25分)
分析
为了节省时间，先将第一个数组全部输入，而第二个数组一边输入一边找中位数，cnt记录在中位数之前一共有多少个数字，cnt_i用于计数在中位数之前，有多少个数字是属于第一个数组的，cnt_j用于计数在中位数之前有多少个数字是属于第二个数组的，当cnt_i+cnt_j=cnt时，表明中位数已经找到。
当cnt_i为n或者cnt_j为m，则表示第一个数组或者第二个数组中的元素全部位于中位数之前。
当cnt_i+cnt_j=cnt时，若cnt_i为n，则中位数为b，若cnt_]]></description><category></category></item><item><title><![CDATA[1028 List Sorting (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106568071</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106568071</guid><author>Ychaoding</author><pubDate>Fri, 05 Jun 2020 12:36:34 +0800</pubDate><description><![CDATA[1028 List Sorting (25分)
分析
注意比较函数和结构体应尽量有助于缩短排序时间
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
#include&lt;string&gt;
#include&lt;string.h&gt;
#include&lt;algorithm&gt;
using namespace std;
int n, a;
struct node {
 int id, score;
 string name;
 node(]]></description><category></category></item><item><title><![CDATA[1027 Colors in Mars (20分)（氵）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106565367</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106565367</guid><author>Ychaoding</author><pubDate>Fri, 05 Jun 2020 10:37:46 +0800</pubDate><description><![CDATA[1027 Colors in Mars (20分)
代码
#include&lt;iostream&gt;
using namespace std;
int main() {
 int a;
 char v[13] = { '0','1','2','3','4','5','6','7','8','9','A','B','C' };
 printf("#");
 for (int i = 0; i &lt; 3; ++i) {
  scanf("%d", &amp;a);
  printf("%c%c", v]]></description><category></category></item><item><title><![CDATA[1025 PAT Ranking (25分)（水题）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106565055</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106565055</guid><author>Ychaoding</author><pubDate>Fri, 05 Jun 2020 10:26:47 +0800</pubDate><description><![CDATA[1025 PAT Ranking (25分)
分析
先按组排序，后所有记录排序即可
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
#include&lt;algorithm&gt;
#include&lt;string&gt;
#include&lt;string.h&gt;
using namespace std;
struct node {
 string id;
 int score, group_id, l_rank, g_rank;
 nod]]></description><category></category></item><item><title><![CDATA[1022 Digital Library (30分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106544595</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106544595</guid><author>Ychaoding</author><pubDate>Thu, 04 Jun 2020 13:39:18 +0800</pubDate><description><![CDATA[1022 Digital Library (30分)
分析
在输入时，将id插入到对应的特征下即可
注意输出id时要按照%07d的格式输出
代码
#include&lt;iostream&gt;
#include&lt;map&gt;
#include&lt;set&gt;
#include&lt;string&gt;
using namespace std;
map&lt;string, set&lt;int&gt; &gt; rcd[6];//1:title,2:author,3:keywords,4:]]></description><category></category></item><item><title><![CDATA[1021 Deepest Root (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106543946</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106543946</guid><author>Ychaoding</author><pubDate>Thu, 04 Jun 2020 12:36:06 +0800</pubDate><description><![CDATA[1021 Deepest Root (25分)
分析
在深度优先遍历的同时，计算连通分量的个数，以及最大深度的节点，加入set中。
第二次选第一次遍历得到的节点中的其中一个，开始深度优先遍历，得到的深度最深的节点也加入set中，从而得到所有的可以作为高度最大的树的根节点的节点。
提供一个测试用例
5
1 2
2 3
2 4
4 5
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
#include&lt;set&gt;
using namespace st]]></description><category></category></item><item><title><![CDATA[1023 Have Fun with Numbers (20分)（一个字水）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106543352</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106543352</guid><author>Ychaoding</author><pubDate>Thu, 04 Jun 2020 11:46:24 +0800</pubDate><description><![CDATA[1023 Have Fun with Numbers (20分)
分析
考察大整数的处理方法，对字符串进行乘二操作，并设置进位标记tag
代码
#include&lt;iostream&gt;
#include&lt;string&gt;
using namespace std;
int cnt[10];
string s;
int main() {
 cin &gt;&gt; s;
 bool tag = false;
 for (int i = s.size() - 1; i &gt;= 0; --i)]]></description><category></category></item><item><title><![CDATA[1024 Palindromic Number (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106543013</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106543013</guid><author>Ychaoding</author><pubDate>Thu, 04 Jun 2020 11:36:50 +0800</pubDate><description><![CDATA[1024 Palindromic Number (25分)
分析
大整数加法，需要处理字符串相加，设置一个tag来标记进位
代码
#include&lt;iostream&gt;
#include&lt;string&gt;
#include&lt;string.h&gt;
#include&lt;algorithm&gt;
using namespace std;
string getsum(string x, string y) {
 bool tag = false;
 for (int i = x.]]></description><category></category></item><item><title><![CDATA[1020 Tree Traversals (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106541643</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106541643</guid><author>Ychaoding</author><pubDate>Thu, 04 Jun 2020 11:07:49 +0800</pubDate><description><![CDATA[1020 Tree Traversals (25分)
分析
给出二叉树的后序遍历序列和中序遍历序列，可唯一确定二叉树，从而可输出层序遍历序列
代码
#include&lt;iostream&gt;
#include&lt;queue&gt;
using namespace std;
int n, post_odr[33], in_odr[33], level_odr[33], cnt = 0;
typedef pair&lt;pair&lt;int, int&gt;, pair&lt;int, int&gt]]></description><category></category></item><item><title><![CDATA[1010 Radix (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106527694</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106527694</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 16:50:55 +0800</pubDate><description><![CDATA[1010 Radix (25分)
分析
二分法搜索目标进制，不失一般性，假设tag=1，初始的左边界设为b各位中最小数字加一，右边界设为a
代码
#include&lt;iostream&gt;
#include&lt;string&gt;
using namespace std;
long long int getnum(string s, int d) {
 long long int a = 0, z = 1;
 for (int i = s.size() - 1; i &gt;= 0; --i, z]]></description><category></category></item><item><title><![CDATA[1018 Public Bike Management (30分)（不长，不水）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106527399</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106527399</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 16:36:08 +0800</pubDate><description><![CDATA[1018 Public Bike Management (30分)
分析
求最短路的同时，要求带出的车子最少，再要求带回的车子最少
迪杰斯特拉算法是贪心算法的一种，迪杰斯特拉算法能得出最优解，要求问题满足最优子结构条件。若携带车辆数目与带回车辆数目这两个尺度加到其中的话，满足最优子结构条件，即不能保证当前步骤选择的节点满足路径最短、带出车子最少、带回车子最少的条件时，之后的步骤也满足全局最优的条件
反例如下：
10 4 4 5
6 7 5 0
0 1 1
0 2 1
1 3 1
2 3 1
3 4 1
因此]]></description><category></category></item><item><title><![CDATA[1019 General Palindromic Number (20分)（一个字水）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106527191</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106527191</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 16:24:13 +0800</pubDate><description><![CDATA[1019 General Palindromic Number (20分)
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
using namespace std;
int main() {
 int a, b;
 vector&lt;int&gt;v;
 scanf("%d %d", &amp;a, &amp;b);
 while (a != 0) {
  v.push_back(a % b);
  a /= b;
 }
 bool tag = tru]]></description><category></category></item><item><title><![CDATA[1017 Queueing at Bank (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106527120</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106527120</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 16:22:06 +0800</pubDate><description><![CDATA[1017 Queueing at Bank (25分)
分析
在输入记录时，将五点之后到达的筛除
利用priority_queue的特性，每次挑选出结束时间最早的窗口，记录等待时间，出队，并添加新的结束时间
代码
#include&lt;iostream&gt;
#include&lt;queue&gt;
#include&lt;vector&gt;
using namespace std;
int n, m, hh, mm, ss, serving_time, cnt = 0, wait_time = 0]]></description><category></category></item><item><title><![CDATA[1016 Phone Bills (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106525376</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106525376</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 15:38:21 +0800</pubDate><description><![CDATA[1016 Phone Bills (25分)
分析
将所有记录按照名字以及记录时间排序，将两条相邻的前后同名且前为on后为off的记录提取为一对用于计算通话时长的有效记录，并保存在map中对应的用户的记录中，最后遍历map计算每一对记录的时长、费用，以及总费用即可
注意时长以及费用的计算可以从00：00：00开始计算然后相减
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
#include&lt;algorithm&gt;
#include&lt;str]]></description><category></category></item><item><title><![CDATA[1015 Reversible Primes (20分)（一个字水）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106525237</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106525237</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 15:26:30 +0800</pubDate><description><![CDATA[1015 Reversible Primes (20分)
分析
打印素数表，判断给定的数以及其在给定进制下的转置数是否为素数
代码
#include&lt;iostream&gt;
using namespace std;
int n, d, nonprime[100010];
int getreverse(int n,int d) {
 int s = 0;
 while (n != 0) {
  s = s * d + n % d;
  n /= d;
 }
 return s;
}
int main(]]></description><category></category></item><item><title><![CDATA[1014 Waiting in Line (30分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106524691</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106524691</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 15:11:12 +0800</pubDate><description><![CDATA[1014 Waiting in Line (30分)
分析
按照时间遍历，记录每位顾客的结束时间即可
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
#include&lt;queue&gt;
using namespace std;
int n, m, k, q, a;
vector&lt;int&gt; f, service_time;
vector&lt;vector&lt;pair&lt;int,int&gt; &gt; &gt;w;//窗口，记]]></description><category></category></item><item><title><![CDATA[1013 Battle Over Cities (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106524384</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106524384</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 14:56:36 +0800</pubDate><description><![CDATA[1013 Battle Over Cities (25分)
分析
依次将每个点标为已访问状态，计算连通分量个数
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
using namespace std;
vector&lt;vector&lt;int&gt; &gt;g;
int n, m, k, a, b, marked[1010];
int main() {
 scanf("%d %d %d", &amp;n, &amp;m, &amp;k);
 g.]]></description><category></category></item><item><title><![CDATA[1012 The Best Rank (25分)（一个字短！）]]></title><link>https://blog.youkuaiyun.com/Ychaoding/article/details/106523943</link><guid>https://blog.youkuaiyun.com/Ychaoding/article/details/106523943</guid><author>Ychaoding</author><pubDate>Wed, 03 Jun 2020 14:53:38 +0800</pubDate><description><![CDATA[1012 The Best Rank (25分)
分析
依次对各个成绩排序，并按照优先级选取最高排名
代码
#include&lt;iostream&gt;
#include&lt;vector&gt;
#include&lt;algorithm&gt;
#include&lt;map&gt;
using namespace std;
vector&lt;pair&lt;int, vector&lt;int&gt; &gt; &gt;rcd;
map&lt;int, vector&lt;int&gt; &g]]></description><category></category></item></channel></rss>