<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[GUOYUNTING的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/GUOYUNTING</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; GUOYUNTING]]></copyright><item><title><![CDATA[南阳acm士兵杀敌系列]]></title><link>https://blog.youkuaiyun.com/GUOYUNTING/article/details/76222507</link><guid>https://blog.youkuaiyun.com/GUOYUNTING/article/details/76222507</guid><author>GUOYUNTING</author><pubDate>Thu, 27 Jul 2017 22:16:20 +0800</pubDate><description><![CDATA[士兵杀敌（一）
#include 
int main()
{
int M,N;
scanf("%d%d",&N,&M);
int *a;
a=new int[1000000];
int i,j;int start,end,sum;
for(i=0;i
{scanf("%d",&a[i]);}
for(i=0;i
{
sum=0;
scanf("%d%d",&start,]]></description><category></category></item><item><title><![CDATA[7.25南阳理工acm评测系统数据结构]]></title><link>https://blog.youkuaiyun.com/GUOYUNTING/article/details/76052084</link><guid>https://blog.youkuaiyun.com/GUOYUNTING/article/details/76052084</guid><author>GUOYUNTING</author><pubDate>Tue, 25 Jul 2017 10:18:18 +0800</pubDate><description><![CDATA[2

Binary String Matching
首先理解题意和案例输出 
难度在于比如A为11 B为111 那么就有两个字符串 
思路 先记录A的长度 然后从B 一个一个开始判断其后面的是不是符合 一旦发现不符合 就下一个 符合就+1 
#include 
#include 
#include 
using namespace std;
int main()
{
int]]></description><category></category></item><item><title><![CDATA[7.14 vector 南阳理工测评的数据结构题]]></title><link>https://blog.youkuaiyun.com/GUOYUNTING/article/details/75096711</link><guid>https://blog.youkuaiyun.com/GUOYUNTING/article/details/75096711</guid><author>GUOYUNTING</author><pubDate>Fri, 14 Jul 2017 09:58:08 +0800</pubDate><description><![CDATA[vector array;  
    array.push_back(1);  
    array.push_back(6);  
    array.push_back(3);  
    array.push_back(6);  
    array.push_back(6);  
    array.push_back(2); 
vector::iterator itor;]]></description><category></category></item></channel></rss>