30
06/2015
《算法竞赛入门经典》第一章
1.PI除了宏定义外，更好的办法可以通过下面获得：#include <math.h>
const double pi=acos(-1.0);2.通过程序计时：#include <time.h>
int main(){
...
...
分类： 读书笔记, 《算法竞赛入门经典》 | 浏览 (3614) | 评论 (1) | 查看全文
26
06/2015
我要长高！！坚持就是胜利！（2）
增高是很多个子不高的朋友的梦想，如何能够快速增高呢？小编今天教你一个能够轻松有效增高的秘诀。那就是增高瑜伽——增延脊柱式。它的前屈位不仅锻炼了我们的腹部肌肉，更重要的是拉伸了我们的脊柱。通过坚持做这个动作，经常锻炼我们的脊柱，能够有效地把它拉伸，把身高“拉长”哦！1）双脚并拢，抬举双手并高举过头，呼气，双手打开与肩同宽，带动身体，身体往前，往下。2）指尖触地，保持抬头状态，双脚绷直。3）然后双手抓...
26
06/2015
我要长高！！坚持就是胜利！（1）
成年人的骨骺已经闭合，不可能再长高。这是个让人很烦恼的问题，但瑜伽增高可以助你一臂之力！瑜伽增高，瑜伽中有很多的增高动作，同时也能使自身的韧带得到史无前例的充分放松。瑜伽增高的原理是：基本上每个人的脊柱都会或多或少有些不正，不正的程度决定您是否含胸驼背。通过瑜伽体式的练习，以正确的方向地不断舒展，伸展、扩宽骨间隙，增加身高的长度，让脊柱直起来。平时弯腰驼背，坐姿不正，谈话含胸，长时间久坐的伽人，通...
25
06/2015
[LeetCode] Jump Game II
Jump Game IIGiven an array of non-negative integers, you are initially positioned at the first index of the array.Each element in the array represents your maximum jump length at that position.Your go...
25
06/2015
[LeetCode] Wildcard Matching
Wildcard MatchingImplement wildcard pattern matching with support for '?' and '*'.'?' Matches any single character.
'*' Matches...
24
06/2015
[LeetCode] Regular Expression Matching
Regular Expression MatchingImplement regular expression matching with support for '.' and '*'.'.' Matches any single character.
'*'&...
23
06/2015
[LeetCode] Search a 2D Matrix
Search a 2D MatrixWrite an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:Integers in each row are sorted from left to r...
22
06/2015
[LeetCode] Basic Calculator II
Basic Calculator IIImplement a basic calculator to evaluate a simple expression string.The expression string contains only non-negative integers, +, -, *, and / operators. The integer divisi...
19
06/2015
[LeetCode] Single Number II
Single Number IIGiven an array of integers, every element appears three times except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you impl...
18
06/2015
[LeetCode] Single Number
Single NumberGiven an array of integers, every element appears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it ...