<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[pcccccccccc的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/pcccccccccc</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; pcccccccccc]]></copyright><item><title><![CDATA[STL by pcccc]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/78429550</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/78429550</guid><author>pcccccccccc</author><pubDate>Thu, 02 Nov 2017 21:23:18 +0800</pubDate><description><![CDATA[前言 
所有常用操作都在代码里 
STL可能会很慢，时间允许的话可替代操作还是手写好 
建议对每一种都盲敲2~3遍代码，不要眼高手低#include<iostream>
#include<cstdio>
#include<vector>//！！！头文件
using namespace std;
vector<int> a[100];
int main()
{
    for(int i=1;i<=2]]></description><category></category></item><item><title><![CDATA[vector]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/78327549</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/78327549</guid><author>pcccccccccc</author><pubDate>Tue, 24 Oct 2017 11:16:51 +0800</pubDate><description><![CDATA[#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
vector<int> a[100];
int main()
{
    for(int i=1;i<=200;i++)
    {
        a[1].push_back(i);
    }
    a[1].pop_back();
    co]]></description><category></category></item><item><title><![CDATA[priority queue]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/78327536</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/78327536</guid><author>pcccccccccc</author><pubDate>Tue, 24 Oct 2017 11:16:18 +0800</pubDate><description><![CDATA[//大根堆 
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
priority_queue<int>q;
int main()
{
    q.push(1);
    q.push(2);
    q.push(3);
    while(q.size())
    {
        cout<<q]]></description><category></category></item><item><title><![CDATA[bitset]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/78327500</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/78327500</guid><author>pcccccccccc</author><pubDate>Tue, 24 Oct 2017 11:14:43 +0800</pubDate><description><![CDATA[//http://blog.youkuaiyun.com/chaiwenjun000/article/details/71154235
#include<iostream>
#include<cstdio>
#include<bitset>
using namespace std;
bitset<10> a;
bitset<10> b;
int main()
{
    a.set();
    cout<<a]]></description><category></category></item><item><title><![CDATA[map用法]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/78326657</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/78326657</guid><author>pcccccccccc</author><pubDate>Tue, 24 Oct 2017 10:22:40 +0800</pubDate><description><![CDATA[#include <map>
#include <iostream>
using namespace std;
int main()
{   
    map< string,int > a; 
    cout<<"插入:"<<endl;
    a["April"]=112;//插入
    cout<<a["April"]<<endl;
    cout<<endl;    cout<<"查询]]></description><category></category></item><item><title><![CDATA[spfa+位运算 2017.10.9 t3]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/78188477</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/78188477</guid><author>pcccccccccc</author><pubDate>Tue, 10 Oct 2017 08:25:30 +0800</pubDate><description><![CDATA[#include<iostream>
#include<cstdio>
#include<queue>
#define Maxn (1<<20)+200005
#define Maxm 400001
using namespace std;
int n,m,to[Maxn],nxt[Maxn],head[Maxn],e,val[Maxn],a[Maxm],x,y,dis[Maxn];
bool us]]></description><category></category></item><item><title><![CDATA[RMQ求LCA]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/77887331</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/77887331</guid><author>pcccccccccc</author><pubDate>Thu, 07 Sep 2017 21:40:35 +0800</pubDate><description><![CDATA[落谷模板#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#define MAXN 1010001
using namespace std;
int n,m,to[MAXN],nxt[MAXN],s,head[MAXN],e,pos[MAXN],dfn[MAXN],u,]]></description><category></category></item><item><title><![CDATA[SPFA模板]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/77019047</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/77019047</guid><author>pcccccccccc</author><pubDate>Thu, 10 Aug 2017 07:59:40 +0800</pubDate><description><![CDATA[模板题
  代码
#include<iostream>
#include<cstdio>
int n,m,to[100001],nxt[100001],head[100001],e,hed,tail,dis[100001],stack[100001];
int cost[100001],x,y;
bool in[100001];
using namespace std;
inline void ad]]></description><category></category></item><item><title><![CDATA[（方法）枚举2^k所有情况]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/76944033</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/76944033</guid><author>pcccccccccc</author><pubDate>Tue, 08 Aug 2017 21:03:54 +0800</pubDate><description><![CDATA[#include<iostream>
#include<cstdio>
using namespace std;
int t,n,m,a[101][101],k,ans;
int s1,s2,s;
void get()
{
    int pc=min(k+1,m);ans=2e9;
    for(int i=1;i<=pc;i++)
    {
        s=0;
        for(]]></description><category></category></item><item><title><![CDATA[线段树]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/76671282</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/76671282</guid><author>pcccccccccc</author><pubDate>Fri, 04 Aug 2017 11:18:00 +0800</pubDate><description><![CDATA[贾比奇的博客带lazytag的线段树模板/*
第一行包含两个整数N、M，分别表示该数列数字的个数和操作的总个数。
第二行包含N个用空格分隔的整数，其中第i个数字表示数列第i项的初始值。
接下来M行每行包含3或4个整数，表示一个操作，具体如下：
操作1： 格式：1 x y k 含义：将区间[x,y]内每个数加上k
操作2： 格式：2 x y 含义：输出区间[x,y]内每个数的和
*/
#include]]></description><category></category></item><item><title><![CDATA[排列组合]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/76598513</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/76598513</guid><author>pcccccccccc</author><pubDate>Wed, 02 Aug 2017 21:06:19 +0800</pubDate><description><![CDATA[小比琪的屁屁
  排列 
  
  
  组合 
  
排列有顺序，组合没顺序；就是这么简单！！公式 
很好理解的 
1：从m中选n个，假设m个中有一个是x，所有的情况就可分为选x和不选x，对于选x的情况，就相当于从m-1个里选n-1个，对于不选x的情况，就相当于从m-1里选n个 
2：c(n,m)=c(m-n,m)（假设选为涂红色，不选为涂白色） 
从m个里选n个的涂红色总情况数等于从m个里选m-]]></description><category></category></item><item><title><![CDATA[归并排序求逆序对]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/76495253</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/76495253</guid><author>pcccccccccc</author><pubDate>Tue, 01 Aug 2017 09:44:33 +0800</pubDate><description><![CDATA[某人的讲解 
归并排序好懂 
求逆序对不好懂qwq 
将2段序列合并时，如果后面序列的一个数比前面序列的一个数x小，他必然比第一个序列x后的所有数小，tot+=mid-i+1 
（很简单是不是） 
裸题#include<iostream>
#include<cstdio>
using namespace std;
int a[1000001],n,q[10000001],l,r,mid,m,i,j,]]></description><category></category></item><item><title><![CDATA[蒜头君的树]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/76458704</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/76458704</guid><author>pcccccccccc</author><pubDate>Mon, 31 Jul 2017 17:29:07 +0800</pubDate><description><![CDATA[蒜头君的树题解
  先存图，再dfs，dfs的过程中统计以x点为子树的点的总数cnt[x]，这样可以求出来x和fa[x]连的这条边一共被使用了(n-cnt[x])*cnt[x]次 
  再求出来最短路之和tot（因为是树，所以最短路好求），对于每次修改，只需要将将差乘上使用次数就好了
#include<iostream>
#include<cstdio>
using namespace std;
l]]></description><category></category></item><item><title><![CDATA[强连通分量]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/76262497</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/76262497</guid><author>pcccccccccc</author><pubDate>Fri, 28 Jul 2017 15:30:39 +0800</pubDate><description><![CDATA[讲解一道模板题牛的舞会The Cow Prom#include<iostream>
#include<cstdio>using namespace std;
int n,m,to[1000001],nxt[100001],head[1000001];
int edge,dfn[100001],low[100001],xx,yy,ans,top,time;
bool in[1000001];
int]]></description><category></category></item><item><title><![CDATA[DP动态规划（持续更新）]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/76101700</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/76101700</guid><author>pcccccccccc</author><pubDate>Wed, 26 Jul 2017 09:07:47 +0800</pubDate><description><![CDATA[最长不下降子序列
  不用我说了吧 
  有一个O(nlogn)的做法  
最长公共子序列
  smartoj1212
#include<stdio.h>
using namespace std;
char a[30001],b[30001];
int dp[3001][3001];
int main()
{
    scanf("%s",a+1);scanf("%s",b+1);
    for(]]></description><category></category></item><item><title><![CDATA[数论积累]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/75950049</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/75950049</guid><author>pcccccccccc</author><pubDate>Mon, 24 Jul 2017 08:11:56 +0800</pubDate><description><![CDATA[辗转相除法排列组合翔翔的排列组合]]></description><category></category></item><item><title><![CDATA[AtCoder Grand Contest 018]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/75950017</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/75950017</guid><author>pcccccccccc</author><pubDate>Mon, 24 Jul 2017 07:47:43 +0800</pubDate><description><![CDATA[A - Getting Difference
  Time limit : 2sec / Memory limit : 256MB
  
  Score : 300 points
  
  Problem Statement There is a box containing N balls. The i-th ball has 
  the integer Ai written on it. Sn]]></description><category></category></item><item><title><![CDATA[NOIP模板大全]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/75453818</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/75453818</guid><author>pcccccccccc</author><pubDate>Thu, 20 Jul 2017 09:00:58 +0800</pubDate><description><![CDATA[我也不知道为什么会有这种东西？？1 数据结构
1.1 线段树
#include <iostream>
#include <cstdio>
using namespace std;
const int N=1e5;
int n,m,tot,root;
struct Tree{
    int l,r,ls,rs,sum,max,tag;
}t[N*2];
void push_up(int x)
{]]></description><category></category></item><item><title><![CDATA[SPFA模板]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/75453498</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/75453498</guid><author>pcccccccccc</author><pubDate>Thu, 20 Jul 2017 07:47:22 +0800</pubDate><description><![CDATA[洛谷2951#include<iostream>
#include<cstdio>
using namespace std;
int n,m,d[200001],x,y,edge,head[500001],maxn,biaohao,sum;
struct node{int to,cost,nxt;}a[500001];
int team[1000001],hed,tail;
bool v[50000]]></description><category></category></item><item><title><![CDATA[矩阵快速幂（以斐波那契数列为例）]]></title><link>https://blog.youkuaiyun.com/pcccccccccc/article/details/73862754</link><guid>https://blog.youkuaiyun.com/pcccccccccc/article/details/73862754</guid><author>pcccccccccc</author><pubDate>Wed, 28 Jun 2017 18:28:11 +0800</pubDate><description><![CDATA[小 M 玩数列【问题描述】 
小 W 发现了一个神奇的数列： 
() = ( − 1) + ( − 2) ｛ ≥ 3, (1) = 1, (2) = 1｝ ，这就是著名的 Fibonacci 
Sequence = =!。 
众所周知，小 M 的数学超级超级好，于是给小 W 出了一道题： 
给小 W 两个数 X,Y,其中 X ≤ Y≤ 2^31−1。 
小 W 任务就是求出 Fib]]></description><category></category></item></channel></rss>