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Codeforces Round #299 (Div. 2) A. Tavas and Nafas
题目链接:http://codeforces.com/problemset/problem/535/A#include #include using namespace std;int main() { string s1[10]={"zero","one","two","three","four","five","six","seven","eigh...
分类:其他好文   时间:2015-06-24 16:34:49    阅读次数:111
Project Euler:Problem 50 Consecutive prime sum
The prime 41, can be written as the sum of six consecutive primes: 41 = 2 + 3 + 5 + 7 + 11 + 13 This is the longest sum of consecutive primes that adds to a prime below one-hundred. The longest s...
分类:其他好文   时间:2015-06-07 17:35:13    阅读次数:129
20150606-six minutes
In Greece, people with disabilities are often heavily stigmatized. The BBC has gained access to one state-run care home where some disabled children a...
分类:其他好文   时间:2015-06-06 13:27:33    阅读次数:91
梦殇 chapter six
如行尸走肉般的坐在教室,看著书上密密麻麻的字,心却久久不能平静。 自己已经不再是当初那个懵懂,无知的小女孩了。仰望天空45度,让眼泪流回眼睛。 是该告别彷徨,告别懵懂的年纪了。 早些年许下的诺言,如今都已化作纷飞的落叶,飘散在风中。 曾经一心追逐的梦想,都被残酷的现实所吞噬,俨然没有再去追逐梦想的....
分类:其他好文   时间:2015-06-05 10:01:27    阅读次数:108
wake,awake,waken,awaken的区别
wake,awake,waken,awaken的区别这4个单词都有"醒来,唤醒,弄醒"的意意思,但又有不同之处。 1:wake,作及物动词和不及物动词的时候都常与up连用。a:I wake up at six every morning。我每天早上六点醒来。b:Please wake me up a...
分类:其他好文   时间:2015-06-02 23:20:18    阅读次数:695
The Time in Words
1 def main(): 2 time = ["one", "two", "three", "four", "five", "six", "seven", "eight","nine", 3 "ten", "eleven", "twelve", "thirteen", "...
分类:其他好文   时间:2015-05-17 11:56:43    阅读次数:111
Swift数组的迭代访问
1 你可以通过for-in循环来迭代访问整个数组的值。 2 for item in shoppingList { 3 println(item) 4 } 5 // Six eggs 6 // Milk 7 // Flour 8 // Baking Powder 9 // Bananas 1...
分类:编程语言   时间:2015-05-16 16:26:54    阅读次数:151
Swift学习4---枚举和结构体
1.枚举 使用enum创建枚举——注意Swift的枚举可以关联方法:enum Rank: Int { case Ace = 1 case Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten case Jack, Queen, ...
分类:编程语言   时间:2015-05-14 00:37:21    阅读次数:137
How can I get a name of enumerated value
For example, if you have the some enum typeTyourEnumType = (One, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten)and you want in run-time to get ...
分类:编程语言   时间:2015-05-11 00:01:17    阅读次数:332
北大ACM2139——Six Degrees of Cowvin Bacon
题目的意思是:输入牛的头数和电影的个数N和M,接下来M + 1 行为每一部电影涉及到的牛的个数已经哪一头牛。在同一部电影里面的牛与其他的牛(也在这部电影中)的度为1. 求解的是每一头牛到其他的牛的度之和最小。(题目的是求平均的,将度之和除以N - 1 再乘以100就行了) 典型的最短路径问题,只是这里求解的是任意两头牛直接的,用Floyd算法可以搞定,N最大为300,时间复杂度为N^3,完全可...
分类:其他好文   时间:2015-05-05 21:58:34    阅读次数:137
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