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CodeForces - 794C:Naming Company(博弈&简单贪心)
Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but the ...
分类:其他好文   时间:2018-08-28 20:14:53    阅读次数:212
Naming Company CodeForces - 794C
Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but the ...
分类:其他好文   时间:2018-08-24 02:00:41    阅读次数:195
POJ3255 Roadblocks [Dijkstra,次短路]
题目传送门 Roadblocks Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to ge ...
分类:数据库   时间:2018-08-17 20:02:43    阅读次数:182
187. Repeated DNA Sequences
问题描述: All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes usef ...
分类:其他好文   时间:2018-08-16 10:43:44    阅读次数:160
A1035 Password (20)(20 分)
A1035 Password (20)(20 分) To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always ...
分类:其他好文   时间:2018-08-10 10:51:41    阅读次数:125
[LeetCode] Expressive Words 富于表现力的单词
Sometimes people repeat letters to represent extra feeling, such as "hello" -> "heeellooo", "hi" -> "hiiii". Here, we have groups, of adjacent letters ...
分类:其他好文   时间:2018-08-05 21:36:28    阅读次数:220
PAT 甲级 1035 Password
https://pintia.cn/problem-sets/994805342720868352/problems/994805454989803520 To prepare for PAT, the judge sometimes has to generate random passwords ...
分类:其他好文   时间:2018-08-05 17:27:41    阅读次数:145
题解报告:hdu 2588 GCD
Description The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For ...
分类:其他好文   时间:2018-08-05 13:01:25    阅读次数:130
扩展欧几里德算法
0、欧几里德定理 一切的基础,自然就是欧几里德定理了。它的形式非常简单(sometimes naive) gcd(a,b)=gcd(b,a mod b) 证明: 假设a,b的公约数为g,且$${a}={bx+y}{(x,y\in Z)}$$则显然有$${g \mid a},\qquad {g \mi ...
分类:编程语言   时间:2018-08-05 00:30:19    阅读次数:197
[HDU5405]Sometimes Naive
题意:一棵带点权的树,支持修改点权和查询,查询$(u,v)$时答案为$\sum\limits_{(i,j)\cap(u,v)\ne\varnothing}v_iv_j$ 路径交不为空不好处理,考虑转为全部减去路径交为空 全部就是$\sum\limits_i\sum\limits_jv_iv_j=\l ...
分类:其他好文   时间:2018-08-04 12:37:02    阅读次数:206
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