[HDU2294]Pendant】的更多相关文章

Pendant Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1032    Accepted Submission(s): 535 Problem Description On Saint Valentine's Day, Alex imagined to present a special pendant to his girl f…
题目:Pendant 链接:http://acm.hdu.edu.cn/showproblem.php?pid=2294 分析: 1)f[i][j]表示长度为i,有j种珍珠的吊坠的数目. $f[i][j] = (k - j + 1) * f[i - 1][j - 1] + j * f[i - 1][j] $ 2)用矩阵来转移. 转移矩阵: $\left[ \begin{array}{cccccc} 1 & 0 & 0 & ... & 0 & 0 \\ 0 &…
T<=10组数据问K<=30种珠子每种n<=1e9串成1~n长度的序列共有多少种,mod1234567891. 方程没想到.矩阵不会推.很好. f[i][j]--长度i,j种珠子方案数,f[i][j]=f[i-1][j]*j(放个旧的)+f[i-1][j-1]+(K-(j-1))(放个新的) n太大,推不动.由于f[i-1]->f[i],考虑矩乘优化.设递推用的矩阵为A.F[i]表示f[i][1]~f[i][k]. 方法一:f加多一个数表示ans,初始化{0,k,0,0,……},那…
Pendant Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 104 Accepted Submission(s): 66   Problem Description On Saint Valentine's Day, Alex imagined to present a special pendant to his girl friend…
On Saint Valentine's Day, Alex imagined to present a special pendant to his girl friend made by K kind of pearls. The pendant is actually a string of pearls, and its length is defined as the number of pearls in it. As is known to all, Alex is very ri…
Description On Saint Valentine's Day, Alex imagined to present a special pendant to his girl friend made by K kind of pearls. The pendant is actually a string of pearls, and its length is defined as the number of pearls in it. As is known to all, Ale…
tend/tent: from the Latin tendere, meaning 'to stretch, extend, or spread'. tent: [tent] n. 帐篷 vt.& vi. 住帐篷,宿营 tendon: ['tendən] n. 腱 contentious: [kən'tenʃəs] adj. 好争吵的,爱争论的,有异议的 distend: [dɪ'stend] v. 扩大, 扩张, 吹大 portend: [pɔːr'tend] v. 成为...的前兆,预知,…
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本文作者:大象本文地址:http://www.cnblogs.com/daxiang/p/4653546.html 在分析和切出设计稿,以及部署项目目录文件后,开始写HTML Demo. 首先,弄出HTML布局的主干结构,最原始的基础框架. 一.<head>里加各类<meta>元素,<link>引入样式.</body>前<script>引入脚本: 二.构建HTML主干结构: 1.网站站点(site),可分“页头页尾”部分: 2.网站子站(page…
来自OpenCV2.3.1 sample/c/mushroom.cpp 1.首先读入agaricus-lepiota.data的训练样本. 样本中第一项是e或p代表有毒或无毒的标志位:其他是特征,可以把每个样本看做一个特征向量: cvSeqPush( seq, el_ptr );读入序列seq中,每一项都存储一个样本即特征向量: 之后,把特征向量与标志位分别读入CvMat* data与CvMat* reponses中 还有一个CvMat* missing保留丢失位当前小于0位置: 2.训练样本…