Dream It Possible-Delacey

很喜欢的一首歌，共勉！！！
I will run, I will climb, I will soar
I‘m undefeated
Jumpiing out of my skin, pull the chord
Yeah I believe it
The past, is everything we were
don‘t make us who we are
So I‘ll dream, until I make it real,
and all I see is stars
Its not until you fall that you fly
When your dreams come alive you‘re unstoppable
Take a shot, chase the sun, find the beautiful
We will glow in the dark turning dust to gold
And we‘ll dream it possible
possible
And we‘ll dream it possible
I will chase, I will reach, I will fly
Until I‘m breaking, until I‘m breaking
Out of my cage, like a bird in the night
I know I‘m changing, I know I‘m changing
In, into something big, better than before
And if it takes, takes a thousand lives
Then it‘s worth fighting for
Its not until you fall that you fly
When your dreams come alive you‘re unstoppable
Take a shot, chase the sun, find the beautiful
We will glow in the dark turning dust to gold
And we‘ll dream it possible
it possible
From the bottom to the top
We‘re sparking wild fire‘s
Never quit and never stop
The rest of our lives
From the bottom to the top
We‘re sparking wild fire‘s
Never quit and never stop
Its not until you fall that you fly
When your dreams come alive you‘re unstoppable
Take a shot, chase the sun, find the beautiful
We will glow in the dark turning dust to gold
And we‘ll dream it possible
possible
And we‘ll dream it possible

时间： 2024-10-10 10:03:59

Dream It Possible-Delacey的相关文章

[Daily Life]百首好歌

百首歌曲已经整理完成√PS1:风格以"中国情歌 + 欧美电音 + 励志歌曲 + 日系歌曲"为主PS2:如果符合PS1的话保证每首歌都很好听都很经典PS3:即使风格不大相同也可以看一看指不定能找到你喜欢的歌PS4:百首歌曲纯手打,不喜勿喷PS5:格式为"序列号 + 歌曲 + 歌手(不一定是原唱,有些是翻唱)"PS6:序列号仅仅是表面意思,不代表个人心目中的排名(因为太多了太累了)PS7:之所以发这个的原因主要是为了想跟大家分享一些好听的歌曲,并且,当自己难受的时候,听

POJ2411(Mondriaan's Dream)

题目链接:传送门题目大意:用1*2大小的砖块去铺满n*m大小的地面,有多少种方案题目思路:因为1<=n,m<=11,并且砖块是1*2,故可以用二进制思想,也就是状态压缩DP,其中矩阵中为0的元素表示当前位置竖着放一块砖,而连着两个1表示横着放一块砖(状态压缩真的很奇妙) #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <

poj 2411 Mondriaan's Dream（状压DP）

Mondriaan's Dream Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 12232 Accepted: 7142 Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series

POJ 2411 Mondriaan's Dream

题目链接:http://poj.org/problem?id=2411 状态压缩Dynamic Programming. For each row, at ith position, 1 means that there is a block placed at this row and next row (vertically). otherwise, its 0. For the example in question, the state of For the example in que

【poj2411】 Mondriaan's Dream

http://poj.org/problem?id=2411 (题目链接) 题意一个$n*m$的网格,用$1*2$的方块填满有多少种方案. Solution 轮廓线dp板子.按格dp,对上方和左方的格子的占用情况进行讨论转移.0表示已放置,1表示未放置. 细节 LL,滚动清空数组. 代码 // poj2411 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring>

Dream of boy, quadrotor helicopter

no.7 Hey,guys,how's everything going? This is Jack from Shanghai. Today's topic is about an significant project in my undergraduate period -- quadrotor helicopter. Quadcopter is a kind of helicopter with 4 propellers. Because of its simple structure,

poj 2411 Mondriaan's Dream（状态压缩+dp）

题意:用1*2砖块铺满n*m的房间. 思路转自:http://www.cnblogs.com/scau20110726/archive/2013/03/14/2960448.html 因为这道题输入范围在11*11之间,所以可以先打表直接输出.......... 状态压缩DP 经典覆盖问题,输入n和m表示一个n*m的矩形,用1*2的方块进行覆盖,不能重叠,不能越出矩形边界,问完全覆盖完整个矩形有多少种不同的方案其中n和m均为奇数的话,矩形面积就是奇数,可知是不可能完全覆盖的.接着我们来看

sdutoj Thrall’s Dream

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2604 Thrall’s Dream Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 We never paid any heed to the ancient prophecies, like fools we clung to the old hatreds, and fought

ZOJ 3211 Dream City (J) DP

Dream City Time Limit: 1 Second Memory Limit: 32768 KB JAVAMAN is visiting Dream City and he sees a yard of gold coin trees. There are n trees in the yard. Let's call them tree 1, tree 2 ...and tree n. At the first day, each tree i has ai coins

POJ 2411 Mondriaan's Dream(状压DP）

http://poj.org/problem?id=2411 求一个n*m矩阵用1*2方块去填满的情况有几种思路:状压dp,先预处理那些状态之间能互相到达,情况就几种,上一个两个1,下一个状态也两个1,上一个为0,下一个必须为1,还有一种是上一个为1,下一个为0的情况然后就一层层往后递推即可代码: #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int