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ACM_大数问题
L老师讲解的
大数问题
- 2013.5.20
模板来自吉林大学acm模板及网络。 老师均添加了一些注释及改进。 第一个,普通的大数运算: 1 #include <stdio.h> 2 #include <string.h> 3 /*==================================================*\ 4 | 普通的大数运算 5 \*========
·
2015-11-11 17:14
问题
hdu 1002:A + B Problem II(
大数问题
)
A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 158615 Accepted Submission(s): 3009
·
2015-11-11 17:13
HDU
寻找第K大的数的方法总结
所谓“第(前)k
大数问题
”指的是在长度为n(n>=k
·
2015-11-11 13:16
总结
大数相乘
比赛中经常会遇到
大数问题
,自己写了一个大数相乘的模板,可能会有Bug,先凑合看吧; #include<stdio.h> #include<string.h> const
·
2015-11-11 12:42
把数组排成最小的数
分析:这里一个需要注意的是,组成的数有可能溢出,所以这是一个
大数问题
。对于
大数问题
,我们又想到不能直接这么组成一个数字,可以采用字符串,也就是先转换成字符串,再进行比较,strcmp()。
·
2015-11-11 04:25
数组
打印出从1到最大的n位十进制数
首先这一题会溢出,要考虑的
大数问题
。
·
2015-11-11 04:11
十进制
数值的N次方
不得使用库函数, 同时不需考虑
大数问题
。 思路分析: 要是你秒秒钟想到一个循环搞定估计面试没戏了。要考虑指数为0和负数的情况,如果底数也 为0了?
·
2015-11-09 11:18
值
大数问题
大数问题
主要有以下四种: ================================== 1. 大数相加 2. 大数相乘 3. 大数阶乘 4.
·
2015-11-08 14:41
问题
打印1到最大的N位数
思路分析: 最简单的想法莫过于先算出这个最大的数,然后循环打出,但是没有考虑大溢出和
大数问题
。 下面有两种思路,一个是用数组模拟字符串,一种是用排列组合的方法。
·
2015-11-08 09:04
打印
ACM_
高次同余方程
/*poj3243 *解决高次同余方程的应用,已知X^Y=KmodZ,及X,Z,K的值,求Y的值 */ #include #include #include usingnamespacestd; #definelint__int64 #defineMAXN131071 structHashNode{lintdata,id,next;}; HashNodehash[MAXN=1) { if(b&1
xiaotan1314
·
2015-11-07 10:00
高次同余方程
HNCU专题训练_线段树(1)
大数(模板题)6.just a Hook7.I Hate It8.动态的最长递增子序列(区间更新题)9.图灵树10.覆盖的面积14.买票问题16.村庄问题17.Hotel19.矩形周长问题23.区间第k
大数问题
·
2015-11-07 09:49
线段树
ACM_
扩展欧几里德算法
/* 扩展欧几里德算法 基本算法:对于不完全为0的非负整数a,b,gcd(a,b)表示a,b的最大公约数,必然存在整数对x,y,使得gcd(a,b)=ax+by。 证明:设a>b。 1,显然当b=0,gcd(a,b)=a。此时x=1,y=0; 2,ab!=0时 设ax1+by1=gcd(a,b); bx2+(amodb)y2=gcd(b,amodb); 根据朴素的欧几里德原理有gcd
xiaotan1314
·
2015-11-06 14:00
算法
ACM
gcd
(本来只是一个·简单的地推,只是
大数问题
有点纠结·,本人用数组简单模拟了一下)
简单n! Time Limit: 1000MS Memory limit: 65536K 题目描述 给定一个数n(0 <= n <= 150), 求0到n中所有数的阶乘。 输入 题目有多组数据,处理到文件结尾。输入一个数n。 输出 输出阶乘,形式如:4! = 24.每组数据输出后跟一个空行。 示例输入 1 4 示例输出 0! = 1
·
2015-11-02 19:22
数组
Java 解决一些ACM中
大数问题
大数中算术运算结果的首选标度 运算 结果的首选标度 加 max(addend.scale(), augend.scale()) 减 max(minuend.scale(), subtrahend.scale()) 乘 multiplier.scale() + multiplicand.scale() 除 dividend.scale() - div
·
2015-11-02 17:42
java
[
ACM_
模拟] UVA 12504 Updating a Dictionary [字符串处理 字典增加、减少、改变问题]
Updating a Dictionary In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, and values are non-negative integers. Given
·
2015-11-01 10:30
字符串处理
[
ACM_
水题] UVA 12502 Three Families [2人干3人的活后分钱,水]
Three Families Three families share a garden. They usually clean the garden together at the end of each week, but last week, family C was on holiday, so family A sp
·
2015-11-01 10:29
ACM
[
ACM_
模拟] UVA 12503 Robot Instructions [指令控制坐标轴上机器人移动 水]
Robot Instructions You have a robot standing on the origin of x axis. The robot will be given some instructions. Your task is to predict its position after executin
·
2015-11-01 10:29
struct
第(前)k
大数问题
以前做过这种题,可是又给忘了。脑子还是活动不开…… 在网上搜了搜,发现这种问题有很多种解法,并且衍生出来很多新的问题。贴出来给大家看看。 转自 http://summerbell.javaeye.com/blog/510394 ------------------------------------------------------------------------
·
2015-10-31 17:41
问题
找第k大的数(方法汇总)
所谓“第(前)k
大数问题
”指的是在长度为n(n>=k)的乱序数组中S找出从大到小顺序的第(前)k个数的问题。
·
2015-10-31 17:24
方法
大数问题
(高精度运算)
在处理大数的时候,可以将其作为字符串读入,然后一个数字一个数字的存储到数组中,然后编写相应运算操作的处理函数即可解决
大数问题
。 也就是说在对大数进行运算之前,要先解决对大数进行存储的问题。而这
·
2015-10-31 17:19
问题
POJ 1426 Find The Multiple (BFS基础)
解题思路: 刚开始考虑了
大数问题
,但是仔细想了下,是多虑的,因为就题目的样例来看,给你的这几个答案都很长。。。实际有比这个答案更小的数字。
·
2015-10-31 14:11
find
【剑指offer】的功率值
同一时候不须要考虑
大数问题
。 分析描写叙述: 对于实现一个函数,首先要做的就是全面考虑它的參数的所有可能。对于此
·
2015-10-31 13:47
FF
大数,高精度计算---大数减法
大数是算法语言中的数据类型无法表示的数,其位数超过最大数据类型所能表示的范围,所以,在处理
大数问题
时首先要考虑的是怎样存储大数,然后是在这种存储方式下其处理的实现方法。
·
2015-10-31 12:05
计算
[
ACM_
数据结构] POJ2352 [树状数组稍微变形]
Description Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars tha
·
2015-10-31 11:37
数据结构
[
ACM_
图论] ZOJ 3708 [Density of Power Network 线路密度,a->b=b->a去重]
The vast power system is the most complicated man-made system and the greatest engineering innovation in the 20th century. The following diagram shows a typical 14 bus power system. In
·
2015-10-31 11:36
NetWork
[
ACM_
水题] ZOJ 3712 [Hard to Play 300 100 50 最大最小]
MightyHorse is playing a music game called osu!. After playing for several months, MightyHorse discovered the way of calculating score in osu!: 1. While p
·
2015-10-31 11:36
play
[
ACM_
暴力][
ACM_
几何] ZOJ 1426 Counting Rectangles (水平竖直线段组成的矩形个数,暴力)
Description We are given a figure consisting of only horizontal and vertical line segments. Our goal is to count the number of all different rectangles formed by these segments. As an example, the n
·
2015-10-31 11:36
count
[
ACM_
动态规划] UVA 12511 Virus [最长公共递增子序列 LCIS 动态规划]
Virus We have a log file, which is a sequence of recorded events. Naturally, the timestamps are strictly increasing. However, it is infected by a virus, so random
·
2015-10-31 11:36
动态规划
[
ACM_
动态规划] hdu 1176 免费馅饼 [变形数塔问题]
Problem Description 都说天上不会掉馅饼,但有一天gameboy正走在回家的小径上,忽然天上掉下大把大把的馅饼。说来gameboy的人品实在是太好了,这馅饼别处都不掉,就掉落在他身旁的10米范围内。馅饼如果掉在了地上当然就不能吃了,所以gameboy马上卸下身上的背包去接。但由于小径两侧都不能站人,所以他只能在小径上接。由于gameboy平时老
·
2015-10-31 11:36
动态规划
[
ACM_
模拟] ACM - Draw Something Cheat [n个长12的大写字母串,找出交集,按字母序输出]
Description Have you played Draw Something? It's currently one of the hottest social drawing games on Apple iOS and Android Devices! In this game, you and your friend play in turn. You
·
2015-10-31 11:36
ACM
[
ACM_
数学] LA 3708 Graveyard [墓地雕塑 圈上新加点 找规律]
Description Programming contests became so popular in the year 2397 that the governor of New Earck -- the largest human-inhabited planet of the galaxy -- opened a special Alley of
·
2015-10-31 11:36
ACM
[
ACM_
水题] UVA 11292 Dragon of Loowater [勇士斗恶龙 双数组排序 贪心]
Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for g
·
2015-10-31 11:36
water
[
ACM_
模拟][
ACM_
数学] LA 2995 Image Is Everything [由6个视图计算立方体最大体积]
Description Your new company is building a robot that can hold small lightweight objects. The robot will have the intelligence to determine if an object is light enough to hold. It
·
2015-10-31 11:36
image
[
ACM_
图论] The Perfect Stall 完美的牛栏(匈牙利算法、最大二分匹配)
描述 农夫约翰上个星期刚刚建好了他的新牛棚,他使用了最新的挤奶技术。不幸的是,由于工程问题,每个牛栏都不一样。第一个星期,农夫约翰随便地让奶牛们进入牛栏,但是问题很快地显露出来:每头奶牛都只愿意在她们喜欢的那些牛栏中产奶。上个星期,农夫约翰刚刚收集到了奶牛们的爱好的信息(每头奶牛喜欢在哪些牛栏产奶)。一个牛栏只能容纳一头奶牛,当然,一头奶牛只能在一个牛栏中产奶。 给出奶牛们的爱好的信息,计算最
·
2015-10-31 11:35
ACM
[
ACM_
搜索] ZOJ 1103 || POJ 2415 Hike on a Graph (带条件移动3盘子到同一位置的最少步数 广搜)
Description "Hike on a Graph" is a game that is played on a board on which an undirected graph is drawn. The graph is complete and has all loops, i.e. for any two locations there is exactly
·
2015-10-31 11:35
Graph
[
ACM_
图论] Fire Net (ZOJ 1002 带障碍棋盘布炮,互不攻击最大数量)
Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small cas
·
2015-10-31 11:35
ACM
[
ACM_
图论] Sorting Slides(挑选幻灯片,二分匹配,中等)
Description Professor Clumsey is going to give an important talk this afternoon. Unfortunately, he is not a very tidy person and has put all his transparencies on one big heap. Before giving the talk
·
2015-10-31 11:35
sort
[
ACM_
搜索] Triangles(POJ1471,简单搜索,注意细节)
Description It is always very nice to have little brothers or sisters. You can tease them, lock them in the bathroom or put red hot chili in their sandwiches. But there is also a time when all meanne
·
2015-10-31 11:35
ACM
[
ACM_
动态规划] ZOJ 1425 Crossed Matchings(交叉最大匹配 动态规划)
Description There are two rows of positive integer numbers. We can draw one line segment between any two equal numbers, with values r, if one of them is located in the first row and the other one is
·
2015-10-31 11:35
match
[
ACM_
搜索] POJ 1096 Space Station Shielding (搜索 + 洪泛算法Flood_Fill)
Description Roger Wilco is in charge of the design of a low orbiting space station for the planet Mars. To simplify construction, the station is made up of a series of Airtight Cubical Modules (ACM's
·
2015-10-31 11:35
ACM
[
ACM_
几何] Transmitters (zoj 1041 ,可旋转半圆内的最多点)
Description In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirement that signals don't overlap, or at least that they don't conflict. One way of a
·
2015-10-31 11:35
ACM
[
ACM_
模拟] The Willy Memorial Program (poj 1073 ,联通水管注水模拟)
Description Willy the spider used to live in the chemistry laboratory of Dr. Petro. He used to wander about the lab pipes and sometimes inside empty ones. One night while he was in a pipe, he fell as
·
2015-10-31 11:35
ACM
[
ACM_
其他] Square Ice (poj1099 规律)
Description Square Ice is a two-dimensional arrangement of water molecules H2O, with oxygen at the vertices of a square lattice and one hydrogen atom between each pair of adjacent oxygen atoms. The
·
2015-10-31 11:35
ACM
[
ACM_
图论] Highways (变形说法的最小生成树)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28972#problem/C 题目给出T种情况,每种情况有n个城镇,接下来每一行是第i个城镇到所有城镇的距离(其实就是个可达矩阵)。 求建设一条公路联通所有城镇并且要求最长的一段最小(其实就是最小生成树)!代码如下: #include<
·
2015-10-31 11:34
最小生成树
[
ACM_
数学] Counting Solutions to an Integral Equation (x+2y+2z=n 组合种类)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27938#problem/E 题目大意:Given, n, count the number of solutions to the equation x+2y+2z=n, where x,y,z,n are non negative inte
·
2015-10-31 11:34
count
[
ACM_
几何] The Deadly Olympic Returns!!! (空间相对运动之最短距离)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28235#problem/B 题目大意: 有两个同时再空间中匀速运动的导弹,告诉一个时间以及各自的初始坐标和该时间时的坐标,求运动过程中的最短距离 解题思路: 求出相对初位置、相对速度,则答案就是原点到射线型轨迹的距离,注意是射线!!!
·
2015-10-31 11:34
return
[
ACM_
几何] F. 3D Triangles (三维三角行相交)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28235#problem/A 题目大意:给出三维空间两个三角形三个顶点,判断二者是否有公共点,三角形顶点、边、内部算三角形的一部分。 解题思路:见模板 //**********************************************
·
2015-10-31 11:34
ACM
[
ACM_
动态规划] 轮廓线动态规划——铺放骨牌(状态压缩1)
Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on
·
2015-10-31 11:34
动态规划
[
ACM_
图论] 棋盘问题 (棋盘上放棋子的方案数)
不能同行同列,给定形状和大小的棋盘,求摆放k个棋子的可行方案 Input 2表示是2X2的棋盘,1表示k,#表示可放,点不可放(-1 -1 结束) Output 输出摆放的方案数目C Sample Input 2 1 #. .# 4 4 ...# ..#. .#.. #... -1 -1 Sample Output 2 1
·
2015-10-31 11:34
ACM
[
ACM_
其他] 总和不小于S的连续子序列的长度的最小值——尺缩法
Description: 给定长度为n的整数数列,A[0],A[1],A[2]….A[n-1]以及整数S,求出总和不小于S的连续子序列的长度的最小值。如果解不存在,则输出0。 Input: 输入数据有多组,每组数据第一行输入n,S, (10<n<10^5,S<10^8)第二行输入A[0],A[1],A[2]….A[n-1] ( 0<A[i]≤10000)
·
2015-10-31 11:34
ACM
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