<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[xlp的acm专栏]]></title><description><![CDATA[Littlefriends]]></description><link>https://blog.youkuaiyun.com/u010863471</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; u010863471]]></copyright><item><title><![CDATA[ORACLE EBS系统中FORM的编译]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/47426045</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/47426045</guid><author>u010863471</author><pubDate>Tue, 11 Aug 2015 20:10:24 +0800</pubDate><description><![CDATA[cd $AU_TOP forms/US

frmcmp_batch Module=/d01/oracle/VIS/apps/apps_st/appl/au/12.0.0/forms/US/SYFFOLDERTAB.fmb Userid=apps/apps Output_File=/d01/oracle/VIS/apps/apps_st/appl/au/12.0.0/forms/US/SYFFOLD]]></description><category></category></item><item><title><![CDATA[ORA-00600: [kcratr1_lastbwr] 报错处理办法]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/47344249</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/47344249</guid><author>u010863471</author><pubDate>Fri, 07 Aug 2015 20:16:07 +0800</pubDate><description><![CDATA[强制关机造成的数据库无法打开


处理办法：


shutdown

startup mount

recover database

alter database open;]]></description><category></category></item><item><title><![CDATA[manacher算法总结]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/45126379</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/45126379</guid><author>u010863471</author><pubDate>Sun, 19 Apr 2015 09:02:19 +0800</pubDate><description><![CDATA[所谓回文串，简单来说就是正着读和反着读都是一样的字符串，比如abba，noon等等，一个字符串的最长回文子串即为这个字符串的子串中，是回文串的最长的那个。

计算字符串的最长回文字串最简单的算法就是枚举该字符串的每一个子串，并且判断这个子串是否为回文串，这个算法的时间复杂度为O(n^3)的，显然无法令人 满意，稍微优化的一个算法是枚举回文串的中点，这里要分为两种情况，一种是回文串长度是奇数的情]]></description><category></category></item><item><title><![CDATA[HDU 4950 Monster]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38581135</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38581135</guid><author>u010863471</author><pubDate>Fri, 15 Aug 2014 10:18:07 +0800</pubDate><description><![CDATA[分情况讨论
#include
using namespace std;
int main()
{
    long long h,a,b,k;
    int kase=1;
    while(scanf("%I64d%I64d%I64d%I64d",&h,&a,&b,&k)==4)
    {
        if(h==0&&a==0&&b==0&&k==0)
            br]]></description><category></category></item><item><title><![CDATA[TopCoder SRM629 CandyMaking]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38532543</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38532543</guid><author>u010863471</author><pubDate>Wed, 13 Aug 2014 10:00:31 +0800</pubDate><description><![CDATA[将所有密度遍历一遍求最小即可
#include 
using namespace std;
struct CandyMaking {
    double findSuitableDensity(vector  containerVolume, vector  desiredWeight) {
        vectorv;
        double Min=50000000;
     ]]></description><category></category></item><item><title><![CDATA[TopCoder SRM629  RectangleCoveringEasy]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38532375</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38532375</guid><author>u010863471</author><pubDate>Wed, 13 Aug 2014 09:51:08 +0800</pubDate><description><![CDATA[只要完全覆盖且不完全相同即可
#include 
using namespace std;
struct RectangleCoveringEasy {
    int solve(int holeH, int holeW, int boardH, int boardW) {
        if((boardH>=holeH&&boardW>holeW)||(boardH>holeH&&boa]]></description><category></category></item><item><title><![CDATA[Codeforces 455A Boredom (dp)]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38531955</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38531955</guid><author>u010863471</author><pubDate>Wed, 13 Aug 2014 09:20:22 +0800</pubDate><description><![CDATA[很裸的dp 
状态转移方程 dp[i]=max(dp[i-1],dp[i-2]+dp[i]*i)

#include
using namespace std;
long long dp[100020];
int main()
{
    int n,a;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d]]></description><category></category></item><item><title><![CDATA[HiHo 1000 A+B]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38516857</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38516857</guid><author>u010863471</author><pubDate>Tue, 12 Aug 2014 16:06:34 +0800</pubDate><description><![CDATA[#include
using namespace std;
int main()
{
   int A,B;
   while(scanf("%d%d",&A,&B)==2)
       printf("%d\n",A+B);
   return 0;
}]]></description><category></category></item><item><title><![CDATA[HiHo 1014 Trie树]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38516759</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38516759</guid><author>u010863471</author><pubDate>Tue, 12 Aug 2014 16:04:03 +0800</pubDate><description><![CDATA[Trie树模板
#include
using namespace std;
struct Trie
{
    int word[100020<<2][26];
    int ex[100020<<2];
    int sz;
    Trie()
    {
        sz=1;
        memset(word,0,sizeof(word));
        memset(]]></description><category></category></item><item><title><![CDATA[HiHo 1037 数字三角形 (dp)]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38516713</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38516713</guid><author>u010863471</author><pubDate>Tue, 12 Aug 2014 15:58:19 +0800</pubDate><description><![CDATA[经典动态规划数塔问题]]></description><category></category></item><item><title><![CDATA[HiHo 1038 01背包]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38516687</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38516687</guid><author>u010863471</author><pubDate>Tue, 12 Aug 2014 15:54:22 +0800</pubDate><description><![CDATA[#include
using namespace std;
int need[1020],value[1020],dp[100020];
int main()
{
    int N,M;
    while(scanf("%d%d",&N,&M)==2)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=N;i++)
  ]]></description><category></category></item><item><title><![CDATA[LA 6437 Power Plant (prim最小生成树)]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38512189</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38512189</guid><author>u010863471</author><pubDate>Tue, 12 Aug 2014 10:53:18 +0800</pubDate><description><![CDATA[还是裸的最小生成树
#include
using namespace std;
int T,N,M,P,K,a,b,c;
int dist[1020],m[1020][1020];
bool p[1020];
void prim()
{
    for(int i=1;i<=N;i++)
    {
        if(!p[i])
        {
            int Min=]]></description><category></category></item><item><title><![CDATA[Ural 1982 Electrification Plan (prim最小生成树)]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38511691</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38511691</guid><author>u010863471</author><pubDate>Tue, 12 Aug 2014 10:33:06 +0800</pubDate><description><![CDATA[很明显的最小生成树模板题
#include
using namespace std;
int n,k,a;
int dist[120],m[120][120];
bool p[120];
void prim()
{
    for(int i=1;i<=n;i++)
    {
        if(p[i]==false)
        {
            int Min=10002]]></description><category></category></item><item><title><![CDATA[BNU 1082 人品问题]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38495571</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38495571</guid><author>u010863471</author><pubDate>Mon, 11 Aug 2014 17:14:02 +0800</pubDate><description><![CDATA[这个题我是1A 突然觉得自己好叼
#include
int main()
{
    printf("25\n");
    return 0;
}]]></description><category></category></item><item><title><![CDATA[Codeforces 458B Distributed Join]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38489101</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38489101</guid><author>u010863471</author><pubDate>Mon, 11 Aug 2014 11:10:50 +0800</pubDate><description><![CDATA[题目没看懂 数据库什么的。。
照着样例直接贪心居然又AC了。。
#include
using namespace std;
long long a[100020],b[100020];
int main()
{
    int m,n;
    scanf("%d%d",&m,&n);
    long long sum1=0,sum2=0,s1=0,s2=0,Max1=0,Max2=0;
 ]]></description><category></category></item><item><title><![CDATA[Codeforces 458A Golden System]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38489061</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38489061</guid><author>u010863471</author><pubDate>Mon, 11 Aug 2014 11:09:12 +0800</pubDate><description><![CDATA[比较两个串计算后的大小 主要是q^2=q+1 明显是一个斐波那契数列
100000位肯定超LL 我在每一位只取到两个以内 居然ac了
#include
using namespace std;
int a[100020];
char s1[100020],s2[100020];
int main()
{
    scanf("%s%s",s1,s2);
    int len1=strlen(]]></description><category></category></item><item><title><![CDATA[MemSQL Start[c]UP 2.0 - Round 2 - Online Round]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38488909</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38488909</guid><author>u010863471</author><pubDate>Mon, 11 Aug 2014 11:05:53 +0800</pubDate><description><![CDATA[今天凌晨的cf题目都没看明白]]></description><category></category></item><item><title><![CDATA[HDU 4932 Miaomiao's Geometry]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38488527</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38488527</guid><author>u010863471</author><pubDate>Mon, 11 Aug 2014 10:45:09 +0800</pubDate><description><![CDATA[答案只可能有两种情况 要么是已知点的距离 要么是已知点距离的一半
那么就枚举每个点之间的距离 和距离的一半
先把所有点按照升序排序 然后用枚举的值贪心
对于点A[i] 如果能放[[A[i]-now,A[i]]就放 否则就放[A[i],A[i]+now]
#include
using namespace std;
double a[120],b[120];
int main()
{
    ]]></description><category></category></item><item><title><![CDATA[HDU 4931 Happy Three Friends]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38488439</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38488439</guid><author>u010863471</author><pubDate>Mon, 11 Aug 2014 10:40:18 +0800</pubDate><description><![CDATA[为了让Grandpa Shawn赢 Dong-Hao肯定让他拿zu]]></description><category></category></item><item><title><![CDATA[NBUT 1225 NEW RDSP MODE I]]></title><link>https://blog.youkuaiyun.com/u010863471/article/details/38417775</link><guid>https://blog.youkuaiyun.com/u010863471/article/details/38417775</guid><author>u010863471</author><pubDate>Thu, 07 Aug 2014 14:14:24 +0800</pubDate><description><![CDATA[找出循环周期就行了
#include
using namespace std;
int N,M,X;
int time(int x,int y,int z)
{
    if(z!=0&&x==1)
        return z;
    if(x%2==0)
        x/=2;
    else
        x=(x+y+1)/2;
    return time(x,y,z+]]></description><category></category></item></channel></rss>