If you runfsck, the filesystem check and repair
command, it might find data fragments that are not referenced anywhere in the
filesystem. In particula...
分类：
其他好文 时间：
2014-05-13 16:59:54
阅读次数：
276
Cut the Sequence
Time Limit: 2000MS
Memory Limit: 131072K
Total Submissions: 8764
Accepted: 2576
Description
Given an integer sequence { an } of length N, you are to ...
分类：
其他好文 时间：
2014-05-13 16:04:29
阅读次数：
267
A：A. Points and Segments (easy)
题目看了n久，开始觉得尼玛这是div2的题目么，题目还标明了easy。。
意思是给你一n个点，m个区间，在n个点上放蓝球或者红球，然后让你找一种选择方案使得m个区间内的蓝球和红球数量之差不超过1.
开始想过用dfs，不过这只是div2的A题而已。。
然后想了下，直接输出010101序列不就可以么。
交了一发，发现...
分类：
其他好文 时间：
2014-05-13 15:42:51
阅读次数：
296
Pat1044代码
题目描述：
Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can...
分类：
其他好文 时间：
2014-05-13 14:33:08
阅读次数：
437
原题：
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of th...
分类：
其他好文 时间：
2014-05-13 14:12:41
阅读次数：
342
本文为senlie原创，转载请保留此地址：http://blog.csdn.net/zhengsenlie
Best Time to Buy and Sell Stock II
Total Accepted: 13030 Total
Submissions: 36467
Say you have an array for which the ith ...
分类：
其他好文 时间：
2014-05-13 14:08:39
阅读次数：
414
K-th Number
Time Limit: 20000MS
Memory Limit: 65536K
Total Submissions: 35704
Accepted: 11396
Case Time Limit: 2000MS
Description
You are working for Macrohard com...
分类：
其他好文 时间：
2014-05-13 14:07:41
阅读次数：
203
You are given a sequence of integers, A1,A2,...,An. And you are allowed a manipulation on the sequence to transform the origin sequence into another sequence B1,B2,...,Bn(Maybe the two sequences are s...
分类：
其他好文 时间：
2014-05-13 13:45:10
阅读次数：
327
Prime Ring Problem
Problem Description
A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent...
分类：
其他好文 时间：
2014-05-13 11:54:17
阅读次数：
444
中心节点就是树的中心，2遍dfs求到树的直径，而中心一定在直径上，顺着直径找到中心就够了。
然后可以一遍树形DP找到最小值或者二分+判断是否访问到叶子节点。
#include
#include
#include
#include
using namespace std;
struct node
{
int next;
int power;
int length...
分类：
其他好文 时间：
2014-05-13 11:28:38
阅读次数：
306
