随笔分类 -  数学

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LightOJ - 1197 Help Hanzo筛素数
摘要：Amakusa, the evil spiritual leader has captured the beautiful princess Nakururu. The reason behind this is he had a little problem wit... 阅读全文
posted @ 2018-05-16 09:57 _大美 阅读(207) 评论(0) 推荐(0)
LightOJ - 1213 Fantasy of a Summation
摘要：If you think codes, eat codes then sometimes you may get stressed. In your dreams you may see huge codes, as I have seen once. Here is... 阅读全文
posted @ 2018-05-15 20:19 _大美 阅读(247) 评论(0) 推荐(0)
POJ - 2116 Death to Binary?
摘要：The group of Absurd Calculation Maniacs has discovered a great new way how to count. Instead of using the ordinary decadic numbers, th... 阅读全文
posted @ 2018-05-15 17:42 _大美 阅读(199) 评论(0) 推荐(0)
LightOJ - 1220 Mysterious Bacteria
摘要：Dr. Mob has just discovered a Deathly Bacteria. He named it RC-01. RC-01 has a very strange reproduction system. RC-01 lives exactly x... 阅读全文
posted @ 2018-05-15 16:11 _大美 阅读(214) 评论(0) 推荐(0)
51Nod - 1188 最大公约数之和 V2
摘要：给出一个数N，输出小于等于N的所有数，两两之间的最大公约数之和。相当于计算这段程序（程序中的gcd(i,j)表示i与j的最大公约数）：G=0;for(i=1;i#include#include#include#include#include#include#inclu... 阅读全文
posted @ 2018-05-15 15:23 _大美 阅读(170) 评论(0) 推荐(0)
LightOJ - 1236 Pairs Forming LCM
摘要：Find the result of the following code:long long pairsFormLCM( int n ) { long long res = 0; for( int i = 1; i #include#include#in... 阅读全文
posted @ 2018-05-12 20:51 _大美 阅读(111) 评论(0) 推荐(0)
LightOJ - 1245 Harmonic Number (II)
摘要：I was trying to solve problem '1234 - Harmonic Number', I wrote the following codelong long H( int n ) { long long res = 0; for(... 阅读全文
posted @ 2018-05-12 09:18 _大美 阅读(133) 评论(0) 推荐(0)
LightOJ - 1259 Goldbach`s Conjecture（素数筛法）
摘要：Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:Every even integer,... 阅读全文
posted @ 2018-05-11 16:35 _大美 阅读(148) 评论(0) 推荐(0)
Steps POJ - 2590
摘要：StepsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8219 Accepted: 3747DescriptionOne steps through integer points of the... 阅读全文
posted @ 2018-05-11 10:31 _大美 阅读(128) 评论(0) 推荐(0)
LightOJ - 1336 Sigma Function （因子和）
摘要：Sigma function is an interesting function in Number Theory. It is denoted by the Greek letter Sigma (σ). This function actually denote... 阅读全文
posted @ 2018-05-11 09:42 _大美 阅读(182) 评论(0) 推荐(0)
LightOJ - 1341 Aladdin and the Flying Carpet （筛素数+找素因子+算数基本原理）
摘要：It's said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concer... 阅读全文
posted @ 2018-05-10 20:14 _大美 阅读(179) 评论(0) 推荐(0)
632-掷骰子
摘要：632-掷骰子内存限制:64MB 时间限制:3000ms Special Judge: Noaccepted:9 submit:54题目描述:小明有m个骰子，小红有n个骰子。每一个骰子都是标准六面骰子。他们分别掷出自己的骰子，如果小明的点数和大，则小明胜；否则小红胜。... 阅读全文
posted @ 2018-05-07 20:58 _大美 阅读(209) 评论(0) 推荐(0)
ZOJ - 3336(Friend Number II)
摘要：Given a positive integer x, let S(x) denotes the sum of all x's digits. Two integers x and y are friend numbers if S(x)=S(y). Here com... 阅读全文
posted @ 2018-05-02 21:03 _大美 阅读(191) 评论(0) 推荐(0)
Clock（数学题）
摘要：ClockTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1195 Accepted Submissi... 阅读全文
posted @ 2018-04-30 11:24 _大美 阅读(246) 评论(0) 推荐(0)
The Lucky Week（规律）
摘要：The Lucky WeekEdward, the headmaster of the Marjar University, is very busy every day and always forgets the date.There was one day Ed... 阅读全文
posted @ 2018-04-28 17:10 _大美 阅读(313) 评论(0) 推荐(0)
POJ 1061 青蛙的约会(扩展欧几里德算法)
摘要：青蛙的约会Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 124981 Accepted: 26763Description两只青蛙在网上相识了，它们聊得很开心，于是觉得很有必要见一面。它们很高兴地发... 阅读全文
posted @ 2018-04-26 20:34 _大美 阅读(131) 评论(0) 推荐(0)
51nod 1284 2 3 5 7的倍数(容斥原理)
摘要：1284 2 3 5 7的倍数 基准时间限制：1 秒 空间限制：131072 KB 分值: 5 难度：1级算法题 收藏 关注给出一个数N，求1至N中，有多少个数不是2 3 5 7的倍数。 例如N = 10，只有1不是2 3 5 7的倍数。Input输入1个数N(1... 阅读全文
posted @ 2018-04-17 21:16 _大美 阅读(203) 评论(0) 推荐(0)
对称与反对称
摘要：链接：https://www.nowcoder.com/acm/contest/94/H来源：牛客网对称与反对称时间限制：C/C++ 2秒，其他语言4秒空间限制：C/C++ 262144K，其他语言524288K64bit IO Format: %lld题目描述 给出... 阅读全文
posted @ 2018-04-07 21:40 _大美 阅读(1256) 评论(0) 推荐(0)
Can you find it?
摘要：Problem DescriptionGive you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find... 阅读全文
posted @ 2018-04-06 08:39 _大美 阅读(333) 评论(0) 推荐(0)
C - C HihoCoder - 1701
摘要：给定N个整数A1, A2, ... AN，小Hi希望从中选出M个整数，使得任意两个选出的整数的差都是K的倍数。 请你计算有多少种不同的选法。由于选法可能非常多，你只需要输出对1000000009取模的结果。Input第一行包含三个整数N、M和K。 第二行包含N个整... 阅读全文
posted @ 2018-03-31 19:19 _大美 阅读(214) 评论(0) 推荐(0)

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