<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[privilage的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/privilage</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; privilage]]></copyright><item><title><![CDATA[A1001 A+B Format]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/89041455</link><guid>https://blog.youkuaiyun.com/privilage/article/details/89041455</guid><author>privilage</author><pubDate>Fri, 05 Apr 2019 11:27:29 +0800</pubDate><description><![CDATA[思路：

数字范围可以用int型， 先进行计算，然后利用string型进行处理； 最前方部分特殊处理，之后部分按逗号+数字的格式输出。




#include &lt;iostream&gt;
#include &lt;string&gt;
using namespace std;

int main(){
	int a, b;
	cin &gt;&gt; a &gt;&gt; b;
	i...]]></description><category></category></item><item><title><![CDATA[1808 Problem  C：字符串的查找删除]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79978918</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79978918</guid><author>privilage</author><pubDate>Tue, 17 Apr 2018 18:39:40 +0800</pubDate><description><![CDATA[题目描述给定一个短字符串（不含空格），再给定若干字符串，在这些字符串中删除所含有的短字符串。输入输入只有1组数据。输入一个短字符串（不含空格），再输入若干字符串直到文件结束为止。输出删除输入的短字符串(不区分大小写)并去掉空格,输出。样例输入in
#include 
int main()
{

printf(&quot; Hi &quot;);
}样例输出#clude
tma()
{

prtf(&quot;Hi&quot;);
}提示...]]></description><category></category></item><item><title><![CDATA[1805 Problem  B：首字母大写]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79972464</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79972464</guid><author>privilage</author><pubDate>Tue, 17 Apr 2018 11:38:27 +0800</pubDate><description><![CDATA[题目描述对一个字符串中的所有单词，如果单词的首字母不是大写字母，则把单词的首字母变成大写字母。在字符串中，单词之间通过空白符分隔，空白符包括：空格(' ')、制表符('\t')、回车符('\r')、换行符('\n')。输入输入一行：待处理的字符串（长度小于100）。输出可能有多组测试数据，对于每组数据，输出一行：转换后的字符串。样例输入if so, you already have a googl...]]></description><category></category></item><item><title><![CDATA[1785 Problem  A：字符串连接]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79971222</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79971222</guid><author>privilage</author><pubDate>Tue, 17 Apr 2018 10:35:48 +0800</pubDate><description><![CDATA[题目描述不借用任何字符串库函数实现无冗余地接受两个字符串，然后把它们无冗余的连接起来。输入每一行包括两个字符串，长度不超过100。输出可能有多组测试数据，对于每组数据，不借用任何字符串库函数实现无冗余地接受两个字符串，然后把它们无冗余的连接起来。输出连接后的字符串。样例输入abc def样例输出abcdef思路：本题直接用scanf将两个字符串存储到二维数组中，按顺序输出即可；代码如下：#incl...]]></description><category></category></item><item><title><![CDATA[B1009. 说反话 (20)]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79967214</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79967214</guid><author>privilage</author><pubDate>Mon, 16 Apr 2018 21:54:13 +0800</pubDate><description><![CDATA[题目描述：给定一句英语，要求你编写程序，将句中所有单词的顺序颠倒输出。输入格式：测试输入包含一个测试用例，在一行内给出总长度不超过80的字符串。字符串由若干单词和若干空格组成，其中单词是由英文字母（大小写有区分）组成的字符串，单词之间用1个空格分开，输入保证句子末尾没有多余的空格。输出格式：每个测试用例的输出占一行，输出倒序后的句子。输入样例：Hello World Here I Come
输出样...]]></description><category></category></item><item><title><![CDATA[5901 Problem  I:【字符串】回文串]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79966788</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79966788</guid><author>privilage</author><pubDate>Mon, 16 Apr 2018 21:25:49 +0800</pubDate><description><![CDATA[题目描述读入一串字符，判断是否是回文串。“回文串”是一个正读和反读都一样的字符串，比如“level”或者“noon”等等就是回文串。输入一行字符串，长度不超过255。输出如果是回文串，输出“YES”，否则输出“NO”。样例输入12321
样例输出YES思路：假设字符串str从下标0开始，i位对应len - 1 - i位，只要存在不同，即不是回文数；因为回文数是正反读都一样的字符串，所以只需要遍历字...]]></description><category></category></item><item><title><![CDATA[1944 Problem  D：八进制]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79965519</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79965519</guid><author>privilage</author><pubDate>Mon, 16 Apr 2018 20:01:13 +0800</pubDate><description><![CDATA[题目描述输入一个整数，将其转换成八进制数输出。输入输入包括一个整数N(0&amp;lt;=N&amp;lt;=100000)。输出可能有多组测试数据，对于每组数据，输出N的八进制表示数。样例输入9
8
5样例输出11
10
5思路：用int型读取n，直接除基取余即可；代码如下：#include &amp;lt;cstdio&amp;gt;

int main()
{
	int n, ans[40];
	while (~ scan...]]></description><category></category></item><item><title><![CDATA[1943 Problem  C：进制转换]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79961809</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79961809</guid><author>privilage</author><pubDate>Mon, 16 Apr 2018 16:08:05 +0800</pubDate><description><![CDATA[题目描述将一个长度最多为30位数字的十进制非负整数转换为二进制数输出。输入多组数据，每行为一个长度不超过30位的十进制非负整数。（注意是10进制数字的个数可能有30个，而非30bits的整数）输出每行输出对应的二进制数。样例输入985
211
1126样例输出1111011001
11010011
10001100110思路：本题中数字可能达到30位，远超于longlong型，所以只能用字符串对数...]]></description><category></category></item><item><title><![CDATA[1942 Problem  B：数制转换]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79959279</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79959279</guid><author>privilage</author><pubDate>Mon, 16 Apr 2018 13:17:27 +0800</pubDate><description><![CDATA[题目描述求任意两个不同进制非负整数的转换（2进制～16进制），所给整数在long所能表达的范围之内。不同进制的表示符号为（0，1，...，9，a，b，...，f）或者（0，1，...，9，A，B，...，F）。输入输入只有一行，包含三个整数a，n，b。a表示其后的n 是a进制整数，b表示欲将a进制整数n转换成b进制整数。a，b是十进制整数，2 =&amp;lt; a，b &amp;lt;= 16。输出可能有多组测...]]></description><category></category></item><item><title><![CDATA[1941 Problem  A：又一版 A+B]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79957503</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79957503</guid><author>privilage</author><pubDate>Mon, 16 Apr 2018 10:42:31 +0800</pubDate><description><![CDATA[题目描述输入两个不超过整型定义的非负10进制整数A和B(&amp;lt;=231-1)，输出A+B的m (1 &amp;lt; m &amp;lt;10)进制数。输入输入格式：测试输入包含若干测试用例。每个测试用例占一行，给出m和A，B的值。当m为0时输入结束。输出输出格式：每个测试用例的输出占一行，输出A+B的m进制数。样例输入2 4 58 123 4560样例输出10011103提示注意输入的两个数相加后的结果可能会...]]></description><category></category></item><item><title><![CDATA[Codeup 导航]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79954347</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79954347</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 22:22:46 +0800</pubDate><description><![CDATA[1932    《算法笔记》3.4小节1928    1929    1931    2026    2063《算法笔记》3.5小节1941    1942    1943    1944《算法笔记》3.6小节1785    1805    1808    5901...]]></description><category></category></item><item><title><![CDATA[PAT (Advanced Level) Practice 导航]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79954327</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79954327</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 22:21:39 +0800</pubDate><description><![CDATA[A1002 A1006 A1009 A1011 A1019 A1027 A1031 A1036 A1042 A1058 A1065]]></description><category></category></item><item><title><![CDATA[PAT(Basic Level)  导航]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79954154</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79954154</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 22:10:08 +0800</pubDate><description><![CDATA[B1004 B1008 B1010 B1012 B1018 B1016 B1022 B1026 B1027 B1028 B1032 B1036 B1037 B1041 B1046 ]]></description><category></category></item><item><title><![CDATA[B1001. 害死人不偿命的(3n+1)猜想 (15)]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79953986</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79953986</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 22:00:55 +0800</pubDate><description><![CDATA[卡拉兹(Callatz)猜想：对任何一个自然数n，如果它是偶数，那么把它砍掉一半；如果它是奇数，那么把(3n+1)砍掉一半。这样一直反复砍下去，最后一定在某一步得到n=1。卡拉兹在1950年的世界数学家大会上公布了这个猜想，传说当时耶鲁大学师生齐动员，拼命想证明这个貌似很傻很天真的命题，结果闹得学生们无心学业，一心只证(3n+1)，以至于有人说这是一个阴谋，卡拉兹是在蓄意延缓美国数学界教学与科研的...]]></description><category></category></item><item><title><![CDATA[A1058. A+B in Hogwarts (20)]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79953193</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79953193</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 21:01:27 +0800</pubDate><description><![CDATA[If you are a fan of Harry Potter, you would know the world of magic has its own currency system -- as Hagrid explained it to Harry, &quot;Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Si...]]></description><category></category></item><item><title><![CDATA[A1027. Colors in Mars (20)]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79952603</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79952603</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 20:19:03 +0800</pubDate><description><![CDATA[People in Mars represent the colors in their computers in a similar way as the Earth people. That is, a color is represented by a 6-digit number, where the first 2 digits are for Red, the middle 2 dig...]]></description><category></category></item><item><title><![CDATA[A1019. General Palindromic Number (20)]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79950474</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79950474</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 16:39:38 +0800</pubDate><description><![CDATA[A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers...]]></description><category></category></item><item><title><![CDATA[B1037. 在霍格沃茨找零钱（20）]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79948626</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79948626</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 13:35:25 +0800</pubDate><description><![CDATA[题目描述：如果你是哈利·波特迷，你会知道魔法世界有它自己的货币系统 —— 就如海格告诉哈利的：“十七个银西可(Sickle)兑一个加隆(Galleon)，二十九个纳特(Knut)兑一个西可，很容易。”现在，给定哈利应付的价钱P和他实付的钱A，你的任务是写一个程序来计算他应该被找的零钱。输入格式：输入在1行中分别给出P和A，格式为“Galleon.Sickle.Knut”，其间用1个空格分隔。这里G...]]></description><category></category></item><item><title><![CDATA[B1022. D进制的A+B (20)]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79948051</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79948051</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 12:21:00 +0800</pubDate><description><![CDATA[题目描述：输入两个非负10进制整数A和B(&amp;lt;=230-1)，输出A+B的D (1 &amp;lt; D &amp;lt;= 10)进制数。输入格式：输入在一行中依次给出3个整数A、B和D。输出格式：输出A+B的D进制数。输入样例：123 456 8
输出样例：1103    思路：有题值A+B最大为2^31-2，int型可以满足；之后用除基除余法即可对十进制进行转换；（1）十进制x转换为Q进制y：采用“除基...]]></description><category></category></item><item><title><![CDATA[2063 Problem  E：日期累加]]></title><link>https://blog.youkuaiyun.com/privilage/article/details/79943244</link><guid>https://blog.youkuaiyun.com/privilage/article/details/79943244</guid><author>privilage</author><pubDate>Sun, 15 Apr 2018 10:52:38 +0800</pubDate><description><![CDATA[题目描述设计一个程序能计算一个日期加上若干天后是什么日期。输入输入第一行表示样例个数m，接下来m行每行四个整数分别表示年月日和累加的天数。输出输出m行，每行按yyyy-mm-dd的个数输出。样例输入1
2008 2 3 100样例输出2008-05-13 思路：通过判断是否为闰年，修改二月份的天数； 在年份增加时也许判别是否为闰年；代码如下：#include &amp;lt;cstdio&amp;gt;

int...]]></description><category></category></item></channel></rss>