随笔分类 -  数论

ACM-ICPC 2018 南京赛区网络预赛 Sum
摘要:A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 阅读全文

posted @ 2018-09-10 19:21 ZefengYao 阅读(149) 评论(0) 推荐(0)

Uva 11174 Stand in a Line
摘要:Stand in a Line Uva 11174 题意:把n个人排成一列,使得没有人排在他父亲的前面,输出方案数MOD 1000000007 思路:《算法竞赛入门经典》P 111 AC代码: 阅读全文

posted @ 2018-05-24 22:10 ZefengYao 阅读(186) 评论(0) 推荐(0)

2018华南理工大学程序设计竞赛 H-对称与反对称
摘要:H-对称与反对称 题目描述 给出一个N*N的方阵A。构造方阵B,C: 使得A = B + C.其中 B为对称矩阵,C为反对称矩阵。 对于方阵S中的任意元素,若(S)ij = (S)ji,则称S为对称矩阵 对于方阵T中的任意元素,若(T)ij = -(T)ji,则称T为反对称矩阵 注意,所有运算在模M 阅读全文

posted @ 2018-04-11 22:58 ZefengYao 阅读(391) 评论(0) 推荐(0)

pat 甲级 1038. Recover the Smallest Number (30)
摘要:1038. Recover the Smallest Number (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 时间限制 400 ms 时间限制 400 ms 内存限制 65536 kB 内存限制 阅读全文

posted @ 2018-03-15 23:21 ZefengYao 阅读(201) 评论(0) 推荐(0)

pat 甲级 1049. Counting Ones (30)
摘要:1049. Counting Ones (30) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 时间限制 100 ms 时间限制 100 ms 内存限制 65536 kB 内存限制 65536 kB 代码长度限 阅读全文

posted @ 2018-03-13 20:32 ZefengYao 阅读(154) 评论(0) 推荐(0)

poj 3532 Resistance
摘要:恢复内容开始 Resistance Time Limit: 1000MS Memory Limit: 131072K Total Submissions: 1289 Accepted: 418 Description H.L. is preparing a circuit for the next 阅读全文

posted @ 2017-11-23 12:57 ZefengYao 阅读(287) 评论(0) 推荐(0)

poj 2345 Central heating
摘要:Central heating Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 703 Accepted: 321 Description Winter has come, but at the Ural State Univer 阅读全文

posted @ 2017-11-14 20:44 ZefengYao 阅读(309) 评论(0) 推荐(0)

poj 3708 Recurrent Function
摘要:Recurrent Function Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 1233 Accepted: 336 Description Dr. Yao is involved in a secret research 阅读全文

posted @ 2017-11-10 13:39 ZefengYao 阅读(235) 评论(0) 推荐(0)

poj Pseudoprime numbers 3641
摘要:Pseudoprime numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10903 Accepted: 4710 Description Fermat's theorem states that for any p 阅读全文

posted @ 2017-11-10 13:04 ZefengYao 阅读(229) 评论(0) 推荐(0)

poj 2720 Last Digits
摘要:Last Digits Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2275 Accepted: 484 Description Exponentiation of one integer by another often p 阅读全文

posted @ 2017-11-10 12:53 ZefengYao 阅读(380) 评论(0) 推荐(0)

poj 2115 C Looooops
摘要:C Looooops Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 29034 Accepted: 8350 Description A Compiler Mystery: We are given a C-language s 阅读全文

posted @ 2017-11-06 00:45 ZefengYao 阅读(183) 评论(0) 推荐(0)

poj 1150 The Last Non-zero Digit
摘要:The Last Non-zero Digit Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5363 Accepted: 1642 Description In this problem you will be given t 阅读全文

posted @ 2017-11-04 14:26 ZefengYao 阅读(272) 评论(0) 推荐(0)

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