https://blog.youkuaiyun.com/Paradise_for_whyzh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/Paradise_for_why/article/details/6022952https://blog.youkuaiyun.com/Paradise_for_why/article/details/6022952Paradise_for_whySat, 20 Nov 2010 10:02:00 +0800本文的说明依据STM32参考手册（RM0008）第10版：英文：http://www.st.com/stonline/products/literature/rm/13902.pdf 中译文：http://www.stmicroelectronics.com.cn/stonline/mcu/images/STM32_RM_CH_V10_1.pdf 在STM32参考手册的第13、14章中，都有一张定时器的框图，下面是第14章中定时器框图的局部，图中黄色框所示的是auto-reload regist]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/6022940https://blog.youkuaiyun.com/Paradise_for_why/article/details/6022940Paradise_for_whySat, 20 Nov 2010 09:56:00 +0800STM32的定时器输入通道都有一个滤波单元，分别位于每个输入通路上(下图中的黄色框)和外部触发输入通路上(下图中的兰色框)，它们的作用是滤除输入信号上的高频干扰。

具体操作原理如下：
在TIMx_CR1中的CKD[1:0]可以由用户设置对输入信号的采样频率基准，有三种选择：
1）采样频率基准fDTS=定时器输入频率fCK_INT
2）采样频率基准fDTS=定时器输入频率fCK_INT/2
3）采样频率基准fDTS=定时器输入频率fCK]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/6022937https://blog.youkuaiyun.com/Paradise_for_why/article/details/6022937Paradise_for_whySat, 20 Nov 2010 09:54:00 +0800https://blog.youkuaiyun.com/Paradise_for_why/article/details/5938627https://blog.youkuaiyun.com/Paradise_for_why/article/details/5938627Paradise_for_whyWed, 13 Oct 2010 16:16:00 +0800https://blog.youkuaiyun.com/Paradise_for_why/article/details/5809577https://blog.youkuaiyun.com/Paradise_for_why/article/details/5809577Paradise_for_whyFri, 13 Aug 2010 13:54:00 +0800/*PKU 1273 Drainage Ditches
  Edmonds-Karp Maximum Flow  O(VE^2)*/
#include
#include
using namespace std;
const int N = 210;
int ad[N][N],flow[N][N];
int pre[N],inc[N];
int n,m;
bool BFS(]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5807450https://blog.youkuaiyun.com/Paradise_for_why/article/details/5807450Paradise_for_whyThu, 12 Aug 2010 17:06:00 +0800先来DP的代码O(∩_∩)O~
 
/*自顶向下的分析，自底向上的求解*/
#include
int n,k;
int map[101][101],dp[101][101];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int DFS(int x,int y)
{
 int i,j,nx,ny,t,max=0;
 if(dp[x][y]!]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5805097https://blog.youkuaiyun.com/Paradise_for_why/article/details/5805097Paradise_for_whyWed, 11 Aug 2010 19:14:00 +0800Floyd算法：
如果要求所有两结点间的最短路径，则可以使用dijkstra算法n次完成，时间复杂度是
O(n^3)，Floyd还提出过另一个算法，同样是O(n^3)的复杂度，但形式上更简单些。
　　要求的解是一个矩阵Ｓ[N][N]，其中S[i][j]表示结点i到j的最短路径，算法很像动态
规划算法，甚至更简单些，不同的是这里规划的是一个矩阵，而不是简单的数组。
Floyd算法过程描述如下：
（在整个循环结束后，S[i][j]肯定是i和]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5802400https://blog.youkuaiyun.com/Paradise_for_why/article/details/5802400Paradise_for_whyTue, 10 Aug 2010 19:56:00 +0800#include
using namespace std;
int n,i,j,k,q,t,map[200][200];
int main()
{
    cin >> n;
    while (n != 0)
    {
      for (i = 1;i <= n;i++)
       for (j = 1;j <= n;j++)
          map[i][]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5798776https://blog.youkuaiyun.com/Paradise_for_why/article/details/5798776Paradise_for_whyMon, 09 Aug 2010 14:34:00 +0800这个题并不难，就是并查集的运用，下面的程序是正确的~
但是里面的红色代码部分应该可以用break来进行优化剪枝的，这样运行时间会短一些，但是不知为什么如果加上break运行结果就不对了的，如果有哪位高手知道了的话请告诉我一下哟~先谢谢喽O(∩_∩)O~
 
 
#include
using namespace std;
const int Max = 1005;
struct data
{
https://blog.youkuaiyun.com/Paradise_for_why/article/details/5797131https://blog.youkuaiyun.com/Paradise_for_why/article/details/5797131Paradise_for_whySun, 08 Aug 2010 18:03:00 +0800#include
#include
using namespace std;
const int Max = 30050;
int pa[Max], num[Max], under[Max]; 
  // num[x]用于根节点x记录集合的元素数量，集合其他元素的num[]以后不会用到。
  // under[x]用于记录集合元素x与根节点元素的位置差值，由于根节点的under[]为0，
https://blog.youkuaiyun.com/Paradise_for_why/article/details/5796820https://blog.youkuaiyun.com/Paradise_for_why/article/details/5796820Paradise_for_whySun, 08 Aug 2010 15:09:00 +0800#include
#include
using namespace std;
#define SIZN 1001
#define SIZM 20001
typedef struct node{
    int u,v,w;
}edge;
int p[SIZN];
edge e[SIZM];//边集数组
int rank[SIZN];
int to]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5793802https://blog.youkuaiyun.com/Paradise_for_why/article/details/5793802Paradise_for_whyFri, 06 Aug 2010 17:19:00 +0800#include
using namespace std;
const long MAXN = 100000;
inline long& len(long* a)
{
 return a[MAXN-1];
}
void copy(long* to, long* scr)
{
 len(to) = len(scr);
 memcpy(to,scr,sizeof(long)*le]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5793337https://blog.youkuaiyun.com/Paradise_for_why/article/details/5793337Paradise_for_whyFri, 06 Aug 2010 15:08:00 +0800#include
#include
int DP[100003][11];
int max(int a,int b,int c)
{
 int m ;
 m=a ;
 m=b>m?b:m ;
 m=c>m?c:m ;
 return m ;
}
int main()
{
    int i,j,n,t,sum,x ;
https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792733https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792733Paradise_for_whyFri, 06 Aug 2010 11:22:00 +0800Robberies http://acm.hdu.edu.cn/showproblem.php?pid=2955
    背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和… 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);
   
    正确的方程是:f[j]=max(f[j],f[j-]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792610https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792610Paradise_for_whyFri, 06 Aug 2010 10:53:00 +0800#include
#include
int a[21][191];
int main()
{
 int i,j,k,sum,p,q,n;
 a[0][0]=1;
 a[1][0]=1;
 a[2][0]=1;
 a[2][1]=1;
 for(i=3;i<=20;i++)
 {
  for(j=0;j  {
https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792606https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792606Paradise_for_whyFri, 06 Aug 2010 10:52:00 +0800#include
#include
#include
typedef struct
{
 int time;
 int x;
 int y;
}node;
typedef struct
{node pos_mes[100];
 int front,rear;
}sequeue;
sequeue *q;
https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792594https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792594Paradise_for_whyFri, 06 Aug 2010 10:50:00 +0800这个一定要注意输出格式的问题，尤其是回车符的输出~
 
#include
#include
#include
char a[1001],b[1001];
int c[1001],d[1001],jin[1001],res[1002];
int main()
{
 int i,j,k,T,lena,lenb,max,min,flag,u=0;
https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792586https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792586Paradise_for_whyFri, 06 Aug 2010 10:49:00 +0800#include
#include
int R,C,max,a[100][100]={10001},b[100][100];
int DFS(int sx,int sy)
{
 int i,x,y,tmp,dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
 if(b[sx][sy])
  return b[sx][sy];
 b[sx][sy]=1;<]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792585https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792585Paradise_for_whyFri, 06 Aug 2010 10:47:00 +0800#include
#include
int R,C,max,a[100][100]={10001},b[100][100];
int DFS(int sx,int sy)
{
 int i,x,y,tmp,dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
 if(b[sx][sy])
  return b[sx][sy];
 b[sx][sy]=1;<]]>https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792574https://blog.youkuaiyun.com/Paradise_for_why/article/details/5792574Paradise_for_whyFri, 06 Aug 2010 10:45:00 +0800#include
#include
int m,n,N,A[10000];
int cmp(const void *a ,const void *b )
{
 return *(int *)b-*(int *)a;
}
long int f()
{
 int k,t;
 long int sum=0,sum1=0;
 qsort(A,n,siz]]>