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oracle性能检测sql语句
1. 监控事例的等待 select event,sum(decode(wait_Time,0,0,1)) "Prev",sum(decode(wait_Time,0,1,0)) "Curr",count(*) "Tot" from v$session_Wait group by event orde...
分类:数据库   时间:2014-06-27 19:18:03    阅读次数:273
Broken Sword II The Smoking Mirror v2.0.09 Android-DeBTPDA 美式冒险解谜大作
Broken Sword II The Smoking Mirror v2.0.09 Android-DeBTPDA美式冒险解谜大作 Broken Sword II The Smoking Mirror非常经典的一款美式冒险解谜大作。游戏精美的画面和细节上的精雕细琢,给人非常华丽的感觉;引人入...
分类:移动开发   时间:2014-06-27 18:41:29    阅读次数:327
BZOJ 3155: Preprefix sum
大意:给一个数组,先求出SUM[I],然后动态的求出1-I的SUM[I]的和, 这题得化公式: 树状数组维护两个和:SUM(A[I])(1#include#include#includeusing namespace std;typedef long long ll;const int N=...
分类:其他好文   时间:2014-06-27 17:53:38    阅读次数:141
LeetCode:Triangle
Given a triangle, find the minimum path sum from top to bottom. Each step you maymove to adjacent numbers on the row below.For example, given the fol....
分类:其他好文   时间:2014-06-27 16:29:58    阅读次数:188
数据结构与算法一
题目一:求1!+2!+…..+n! 的和的后6位,(注意n的范围)#include using namespace std;const int MAX = 1000000;int getResu(int n){ int sum=0; int temp= 1; for(int i=1; i >n) {...
分类:其他好文   时间:2014-06-27 13:47:37    阅读次数:161
Leetcode Combination Sum
Given a set of candidate numbers (C) and a target number (T), find all unique combinations inCwhere the candidate numbers sums toT.Thesamerepeated num...
分类:其他好文   时间:2014-06-26 15:50:35    阅读次数:176
Leetcode Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inCwhere the candidate numbers sums toT.Each number ...
分类:其他好文   时间:2014-06-26 15:45:49    阅读次数:163
Leetcode Maximum Subarray
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.For example, given the array[?2,1,?3,4,?1,2,1,...
分类:其他好文   时间:2014-06-26 15:10:47    阅读次数:168
BZOJ:3209: 花神的数论题
今天居然没参考任何资料解决了这道数位DP,事先只是搞一道数论题练练;思路:求SUM[1]-SUM[N]的二进制的乘积mod1000000007; N#include#include#includeusing namespace std;#define N 10000007typedef long.....
分类:其他好文   时间:2014-06-26 15:09:30    阅读次数:221
uva10487
#include #include using namespace std; int n,m,k,cases,sum,arr[1005],S[1000000]; void calsum(){ k=0; for (int i=0;i<n-1;i++){ for (int j=i+1;j<n;j++){ S[k++]=arr[i]+arr[j]; } } sort(S,S...
分类:其他好文   时间:2014-06-26 10:51:01    阅读次数:198
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