<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[wzs1997的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/wzs1997</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; wzs1997]]></copyright><item><title><![CDATA[Binary String Matching]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79645718</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79645718</guid><author>wzs1997</author><pubDate>Wed, 21 Mar 2018 20:49:49 +0800</pubDate><description><![CDATA[Binary String Matching
时间限制：3000 ms  |  内存限制：65535 KB
难度：3
描述
Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring ...]]></description><category></category></item><item><title><![CDATA[大数加法]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79577210</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79577210</guid><author>wzs1997</author><pubDate>Fri, 16 Mar 2018 09:45:50 +0800</pubDate><description><![CDATA[//大数加法2  
#include&amp;lt;stdio.h&amp;gt;  
#include&amp;lt;string.h&amp;gt;  
#define M 100005  
char s1[M],s2[M];  
int a[M],b[M],c[M];  
int main()  
{  
    int i,j,k,n,m;  
    while(scanf(&quot;%s%s&quot;,s1,s2))  
    {...]]></description><category></category></item><item><title><![CDATA[string]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79573468</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79573468</guid><author>wzs1997</author><pubDate>Thu, 15 Mar 2018 20:29:01 +0800</pubDate><description><![CDATA[c++中 string 用于字符串的处理//	string str(&quot;sssss&quot;);// 输出sssss 
       string s=&quot;hello&quot; ;// 输出hello 
    //string s(str);//将str中的字符串 保存到s中 
//string s(10,'c'); 将10个字符'c' 保存到s中 
  //    char cs[]=&quot;12345&quot;;
    s...]]></description><category></category></item><item><title><![CDATA[素数打表]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79552826</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79552826</guid><author>wzs1997</author><pubDate>Wed, 14 Mar 2018 12:14:43 +0800</pubDate><description><![CDATA[美素数　小明对数的研究比较热爱，一谈到数，脑子里就涌现出好多数的问题，今天，小明想考考你对素数的认识。  　　问题是这样的：一个十进制数，如果是素数，而且它的各位数字和也是素数，则称之为“美素数”，如29，本身是素数，而且2+9 = 11也是素数，所以它是美素数。  　　给定一个区间，你能计算出这个区间内有多少个美素数吗？             Input        第一行输入一个正整数T，...]]></description><category></category></item><item><title><![CDATA[sort 排序]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79510176</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79510176</guid><author>wzs1997</author><pubDate>Sat, 10 Mar 2018 17:46:50 +0800</pubDate><description><![CDATA[#include &amp;lt;iostream&amp;gt;     // std::cout
#include &amp;lt;algorithm&amp;gt;    // std::sort
using namespace std;
int main()
{
	 
	int a[5];
	for(int i=0;i&amp;lt;5;i++)
	cin&amp;gt;&amp;gt;a[i];
	sort(a,a+5);
	for(int ...]]></description><category></category></item><item><title><![CDATA[库函数]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79501759</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79501759</guid><author>wzs1997</author><pubDate>Fri, 09 Mar 2018 19:16:03 +0800</pubDate><description><![CDATA[pow()函数：计算a^b#include&amp;lt;stdio.h&amp;gt;
#include&amp;lt;math.h&amp;gt;
int main()
{
	double a=2,b=3;
	printf(&quot;%f\n&quot;,pow(a,b));
	return 0;
	
 } pow(x,y);//其作用是计算x的y次方。x、y及函数值都是double型 要加入头文件 math.h 与字符串有关的函数1. st...]]></description><category></category></item><item><title><![CDATA[快速幂]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79501683</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79501683</guid><author>wzs1997</author><pubDate>Fri, 09 Mar 2018 19:11:49 +0800</pubDate><description><![CDATA[取模运算(a*b)%c=(a%c)*(b%c)%c  计算a^b%cint quick(int a,int b,int c)  {      int ans=1;   //记录结果      a=a%c;   //预处理，使得a处于c的数据范围之下      while(b!=0)      {          if(b&amp;amp;1) ans=(ans*a)%c;   //如果b的二进制位不是0...]]></description><category></category></item><item><title><![CDATA[输入问题]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79489281</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79489281</guid><author>wzs1997</author><pubDate>Thu, 08 Mar 2018 20:13:26 +0800</pubDate><description><![CDATA[当输入n为0时则不进行处理，可以这样写while(~scanf(&quot;%d&quot;,&amp;amp;n)&amp;amp;&amp;amp;n!=0)，也可以直接写while(~scanf(&quot;%d&quot;,&amp;amp;n)&amp;amp;&amp;amp;n)，进而变成while(scanf(&quot;%d&quot;,&amp;amp;n)，n)。while (~scanf(&quot;%d&quot;,&amp;amp;n),n) 当n为0时中止循环while (~s]]></description><category></category></item><item><title><![CDATA[if语句判断积偶   与   位运算]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79489086</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79489086</guid><author>wzs1997</author><pubDate>Thu, 08 Mar 2018 20:06:49 +0800</pubDate><description><![CDATA[1.if(n&amp;amp;1)n为奇数  // 奇数2进制末位为1，1与1进行与运算结果为1；2.n&amp;gt;&amp;gt;=1   右移1位 除以2；n&amp;lt;&amp;lt;=1  左移1位  乘以2；]]></description><category></category></item><item><title><![CDATA[memset 函数的使用]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79479074</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79479074</guid><author>wzs1997</author><pubDate>Thu, 08 Mar 2018 07:55:44 +0800</pubDate><description><![CDATA[C语言中memset函数头文件为&amp;lt;memory.h&amp;gt; 或 &amp;lt;string.h&amp;gt;C++   &amp;lt;cstring&amp;gt;#include&amp;lt;stdio.h&amp;gt;#include&amp;lt;memory.h&amp;gt; main(){	char a[]=&quot;abcd&quot;;	printf(&quot;%s\n&quot;,a);	memset(a,'#',sizeof(a));	printf(&]]></description><category></category></item><item><title><![CDATA[最小公倍数 简单代码]]></title><link>https://blog.youkuaiyun.com/wzs1997/article/details/79476527</link><guid>https://blog.youkuaiyun.com/wzs1997/article/details/79476527</guid><author>wzs1997</author><pubDate>Wed, 07 Mar 2018 20:42:13 +0800</pubDate><description><![CDATA[#include&amp;lt;stdio.h&amp;gt;int gcd(int a,int b){    return b==0?a:gcd(b,a%b); }     int main() { 	int a,b; 	scanf(&quot;%d%d&quot;,&amp;amp;a,&amp;amp;b); 	printf(&quot;%d&quot;,gcd(a,b));   //最大公约数       printf(&quot;%d&quot;,a*b/gcd(a,b)); ...]]></description><category></category></item></channel></rss>