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Codeforces Round #324 (Div. 2) D. Dima and Lisa (哥德巴赫猜想 + 暴力)
D. Dima and Lisa D. Dima and Lisa Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three ...
分类:其他好文   时间:2016-07-12 21:29:38    阅读次数:193
opencv-视频处理-实时的前景检测-Vibe算法
vibe算法是一种像素级的前景检测算法,实时性高,内存占有率低,前景检测准确率高。但是会出现“鬼影”。《 ViBe: a powerful random technique to estimate the background in video sequences》《Background Subtraction: Experiments and Improvements for ViBe. 》《V...
分类:编程语言   时间:2016-07-10 18:59:35    阅读次数:958
E - Lovely Palindromes
Description Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321 ...
分类:其他好文   时间:2016-07-09 16:24:37    阅读次数:137
什么是基因间隔区和内含子?
基因间隔区:An Intergenic region (IGR) is a stretch of DNA sequences located between genes. Intergenic regions are a subset of Noncoding DNA. Occasionally s ...
分类:其他好文   时间:2016-07-05 00:54:56    阅读次数:694
hdu 5656 CA Loves GCD dp
CA Loves GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Problem Description CA is a fine comrade who loves the ...
分类:其他好文   时间:2016-07-02 19:00:41    阅读次数:182
NOI2016模拟赛Zbox loves stack
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 #define PI pair<int, ...
分类:其他好文   时间:2016-06-20 22:09:54    阅读次数:228
hdu-5587 Array(回溯)
题目链接: Array Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Problem Description Vicky is a magician who loves math. ...
分类:其他好文   时间:2016-06-17 23:47:43    阅读次数:216
二分套二分 hrbeu.acm.1211Kth Largest
Kth Largest TimeLimit: 1 Second MemoryLimit: 32 Megabyte Description There are two sequences A and B with N (1<=N<=10000) elements each. All of the el ...
分类:其他好文   时间:2016-06-15 23:45:49    阅读次数:276
cppreference
地址 ASCII Escape Sequences C++ Complier Support C++标准现在状态 C语言格式化IO修饰符 C++中的数值类型 C++字符相关判断函数 STL容器列表 变量的storage duration and linkage lvalue xvalue prval ...
分类:其他好文   时间:2016-06-07 16:11:04    阅读次数:284
HDU5709 : Claris Loves Painting
对于每个点维护两棵线段树$T1[x],T2[x]$: $T1[x]$维护$x$子树内,深度在$[l,r]$内的点数,同种颜色有多个的话,保留深度最小的那个。 $T2[x]$维护$x$子树内每种颜色的最小深度。 从底向上合并线段树,先合并$T1$,然后合并$T2$的时候,发现有重复点,那么在$T1$里 ...
分类:其他好文   时间:2016-05-29 06:23:14    阅读次数:293
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