`
水木清华77
  • 浏览: 36900 次
  • 性别: Icon_minigender_1
  • 来自: 杭州
社区版块
存档分类
最新评论

Problem13

阅读更多
package com.yao.Algorithms;

import java.util.Arrays;


public class Problem13 {
	public static void main(String[] args) {
		String str = "37107287533902102798797998220837590246510135740250"
				+ "46376937677490009712648124896970078050417018260538"
				+ "74324986199524741059474233309513058123726617309629"
				+ "91942213363574161572522430563301811072406154908250"
				+ "23067588207539346171171980310421047513778063246676"
				+ "89261670696623633820136378418383684178734361726757"
				+ "28112879812849979408065481931592621691275889832738"
				+ "44274228917432520321923589422876796487670272189318"
				+ "47451445736001306439091167216856844588711603153276"
				+ "70386486105843025439939619828917593665686757934951"
				+ "62176457141856560629502157223196586755079324193331"
				+ "64906352462741904929101432445813822663347944758178"
				+ "92575867718337217661963751590579239728245598838407"
				+ "58203565325359399008402633568948830189458628227828"
				+ "80181199384826282014278194139940567587151170094390"
				+ "35398664372827112653829987240784473053190104293586"
				+ "86515506006295864861532075273371959191420517255829"
				+ "71693888707715466499115593487603532921714970056938"
				+ "54370070576826684624621495650076471787294438377604"
				+ "53282654108756828443191190634694037855217779295145"
				+ "36123272525000296071075082563815656710885258350721"
				+ "45876576172410976447339110607218265236877223636045"
				+ "17423706905851860660448207621209813287860733969412"
				+ "81142660418086830619328460811191061556940512689692"
				+ "51934325451728388641918047049293215058642563049483"
				+ "62467221648435076201727918039944693004732956340691"
				+ "15732444386908125794514089057706229429197107928209"
				+ "55037687525678773091862540744969844508330393682126"
				+ "18336384825330154686196124348767681297534375946515"
				+ "80386287592878490201521685554828717201219257766954"
				+ "78182833757993103614740356856449095527097864797581"
				+ "16726320100436897842553539920931837441497806860984"
				+ "48403098129077791799088218795327364475675590848030"
				+ "87086987551392711854517078544161852424320693150332"
				+ "59959406895756536782107074926966537676326235447210"
				+ "69793950679652694742597709739166693763042633987085"
				+ "41052684708299085211399427365734116182760315001271"
				+ "65378607361501080857009149939512557028198746004375"
				+ "35829035317434717326932123578154982629742552737307"
				+ "94953759765105305946966067683156574377167401875275"
				+ "88902802571733229619176668713819931811048770190271"
				+ "25267680276078003013678680992525463401061632866526"
				+ "36270218540497705585629946580636237993140746255962"
				+ "24074486908231174977792365466257246923322810917141"
				+ "91430288197103288597806669760892938638285025333403"
				+ "34413065578016127815921815005561868836468420090470"
				+ "23053081172816430487623791969842487255036638784583"
				+ "11487696932154902810424020138335124462181441773470"
				+ "63783299490636259666498587618221225225512486764533"
				+ "67720186971698544312419572409913959008952310058822"
				+ "95548255300263520781532296796249481641953868218774"
				+ "76085327132285723110424803456124867697064507995236"
				+ "37774242535411291684276865538926205024910326572967"
				+ "23701913275725675285653248258265463092207058596522"
				+ "29798860272258331913126375147341994889534765745501"
				+ "18495701454879288984856827726077713721403798879715"
				+ "38298203783031473527721580348144513491373226651381"
				+ "34829543829199918180278916522431027392251122869539"
				+ "40957953066405232632538044100059654939159879593635"
				+ "29746152185502371307642255121183693803580388584903"
				+ "41698116222072977186158236678424689157993532961922"
				+ "62467957194401269043877107275048102390895523597457"
				+ "23189706772547915061505504953922979530901129967519"
				+ "86188088225875314529584099251203829009407770775672"
				+ "11306739708304724483816533873502340845647058077308"
				+ "82959174767140363198008187129011875491310547126581"
				+ "97623331044818386269515456334926366572897563400500"
				+ "42846280183517070527831839425882145521227251250327"
				+ "55121603546981200581762165212827652751691296897789"
				+ "32238195734329339946437501907836945765883352399886"
				+ "75506164965184775180738168837861091527357929701337"
				+ "62177842752192623401942399639168044983993173312731"
				+ "32924185707147349566916674687634660915035914677504"
				+ "99518671430235219628894890102423325116913619626622"
				+ "73267460800591547471830798392868535206946944540724"
				+ "76841822524674417161514036427982273348055556214818"
				+ "97142617910342598647204516893989422179826088076852"
				+ "87783646182799346313767754307809363333018982642090"
				+ "10848802521674670883215120185883543223812876952786"
				+ "71329612474782464538636993009049310363619763878039"
				+ "62184073572399794223406235393808339651327408011116"
				+ "66627891981488087797941876876144230030984490851411"
				+ "60661826293682836764744779239180335110989069790714"
				+ "85786944089552990653640447425576083659976645795096"
				+ "66024396409905389607120198219976047599490197230297"
				+ "64913982680032973156037120041377903785566085089252"
				+ "16730939319872750275468906903707539413042652315011"
				+ "94809377245048795150954100921645863754710598436791"
				+ "78639167021187492431995700641917969777599028300699"
				+ "15368713711936614952811305876380278410754449733078"
				+ "40789923115535562561142322423255033685442488917353"
				+ "44889911501440648020369068063960672322193204149535"
				+ "41503128880339536053299340368006977710650566631954"
				+ "81234880673210146739058568557934581403627822703280"
				+ "82616570773948327592232845941706525094512325230608"
				+ "22918802058777319719839450180888072429661980811197"
				+ "77158542502016545090413245809786882778948721859617"
				+ "72107838435069186155435662884062257473692284509516"
				+ "20849603980134001723930671666823555245252804609722"
				+ "53503534226472524250874054075591789781264330331690";
		byte[] sum = new byte[25];
		int breakCount=50;
		for (int i = 0; i < str.length(); i += breakCount) {
			sum = add(myInt(str.substring(i, i + breakCount)), sum);
		}
		int first=10;
		for(int i=0;i<first/2;i++){
			byte b=sum[sum.length-1-i];
			if(b/10==0)
				System.out.print(0);
			System.out.print(b);
	}
	
	
	}

	public static byte[] myInt(String str) {
		/** 一个比特表示2位数 最大99 */
		int length = str.length() % 2 + str.length() / 2;
		byte[] myInt = new byte[length];
		int j = 0;
		if (str.length() % 2 == 1) {
			myInt[length - 1] = (byte) (str.charAt(0) - '0');
			str = str.substring(1);
			j = 1;
		}
		for (int i = 0; i < str.length(); i += 2) {
			int k = Integer.valueOf(str.substring(i, i + 2));
			myInt[length - 1 - j] = (byte) k;
			j++;
		}
		return myInt;
	}

	private static byte[] add(byte[] a, byte[] b) {

		int length = a.length > b.length ? a.length + 1 : b.length + 1;
		byte[] sum = new byte[length];
		int small = a.length;
		int big = b.length;
		if (a.length > b.length) {
			big = a.length;
			small = b.length;
			
		}
		for (int i = 0; i < small; i++) {
			sum[i] += (byte) ((a[i] + b[i]) % 100);
			sum[i + 1] += (byte) ((a[i] + b[i]) / 100);
		}

		for (int i = small; i < big; i++) {
			if (big == a.length)
				sum[i] += a[i];
			else
				sum[i] += b[i];
		}
		
		if(sum[sum.length-1]==0){
			byte[] sum_yao=new byte[sum.length-1];
			System.arraycopy(sum, 0,sum_yao ,0, sum_yao.length);
			return sum_yao;
		}
		
		return sum;
	}

}
分享到:
评论

相关推荐

    problem13p7共4页.pdf.zip

    很抱歉,但根据您给出的信息,"problem13p7共4页.pdf.zip" 和 "赚钱项目" 这些描述并没有提供足够的具体内容来生成一篇超过1000字的IT知识文章。这些信息可能指的是一个压缩文件,包含了可能是问题解答、项目计划或...

    计算机网络第六版答案

    13. a) 2 users can be supported because each user requires half of the link bandwidth. b) Since each user requires 1Mbps when transmitting, if two or fewer users transmit simultaneously, a maximum ...

    MySQL数据库考试试题.docx

    13. 条件语句:WHERE 子句(Problem 13) WHERE 子句用于指定 SELECT 语句的条件,以便仅检索满足条件的数据。 14. 主键的建立方法:三种(Problem 14) 主键可以使用 PRIMARY KEY 约束、UNIQUE 约束或索引来建立...

    Computer-Based.Problem.Solving.Process

    Title: Computer-Based Problem Solving Process Author: Teodor Rus Length: 350 pages Edition: 1 Language: English Publisher: World Scientific Publishing Company Publication Date: 2015-05-30 ISBN-10: ...

    euler project.r.zip_R Euler project_project

    13. **Problem 13: 大数显示** - 知识点:大数处理,字符串操作。 - 解决方法:处理大整数,将其转换为字符串,然后截取指定长度。 14. **Problem 14: 长链回文子串** - 知识点:字符串处理,回文判断,循环结构...

    Problem Solving with C++, 10th Global Edition

    ISBN-13: 9781292222820 For courses in C++ introductory programming. Learn the fundamentals of C++ programming with an emphasis on problem solving Now in its 10th Edition, Problem Solving with C++ is...

    模式识别 中科院刘成林 作业一

    ≤ (c−1)/c (d) Describe a situation for which P(error) = (c−1)/c Question 2 (Pattern Classification, Chapter 2, Problem 13) In many pattern classification problems one has the option either to ...

    Algorithmic Problem Solving

    ISBN-13 书号: 9780470684535 Edition 版本: 1 出版日期: 2011-10-24 pages 页数: 432 An entertaining and captivating way to learn the fundamentals of using algorithms to solve problems The algorithmic ...

    Artificial Intelligence and Problem Solving

    Title: Artificial Intelligence and Problem Solving Author(s): Danny Kopec, Christopher Pileggi, David Ungar, Shweta Shetty Publisher: Mercury Learning & Information Year: 2017 Language: english ...

    Prime Ring Problem 深度探索

    int M, g, a[20] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}, ag[20], c[20]; ``` - 初始化数组`a`,用于存储可能填入环中的数字。 - `ag`数组用于标记数字是否已被使用。 - `c`...

    Problem Solving with C++

    #### 第13章:链表的遍历(p.762) 链表是一种常用的数据结构,在C++中使用指针进行实现。本节介绍了一种遍历链表的方法,并展示了如何通过链表来管理类的对象。 - **遍历**:从链表的头节点开始,依次访问每个...

    C++.Programming.From.Problem.Analysis.to.Program.Design.7th.Edition

    Title: C++ Programming: From Problem Analysis to Program Design, 7th Edition Author: D. S. Malik Length: 1488 pages Edition: 7 Language: English Publisher: Course Technology Publication Date: 2014-02-...

    Problem Solving in Data Structures & Algorithms Using Java

    Title: Problem Solving in Data Structures & Algorithms Using Java: The Ultimate Guide to Programming Author: Hemant Jain Length: 436 pages Edition: First Edition Language: English Publisher: ...

    Problem Solving in Data Structures & Algorithms Using C++

    ISBN-13: 9781542396479 This book is about the usage of data structures and algorithms in computer programming. Designing an efficient algorithm to solve a computer science problem is a skill of ...

    problema_1era convocatoria_12_13_LaPrimera_#phisic_

    【标题】"problema_1era convocatoria_12_13_LaPrimera_#phisic_" 暗示着这是一个关于物理学科的第一次考试问题集,可能是一次学术竞赛或学校考试的初步通知。"La Primera" 指的是“第一次”,而 "#phisic" 是对...

    C++ Programming: From Problem Analysis to Program Design, 8th Edition

    C++ Programming: From Problem Analysis to Program Design By 作者: D. S. Malik ISBN-10 书号: 1337102083 ISBN-13 书号: 9781337102087 Edition 版本: 8 出版日期: 2017-02-13 pages 页数: 1491 Contents ...

    Problem Solving with C++ (7th edition)

    #### Chapter 13: Advanced OOP Features This chapter covers additional advanced OOP features: - **Exception Handling**: Explanation of exception handling mechanisms in C++, including try-catch blocks...

    0-1-knapsack-problem-master (13).zip

    标题 "0-1-knapsack-problem-master (13).zip" 提示我们这是一个关于 0-1 背包问题的项目,而描述 "通讯录排序" 可能是指在解决这个问题时使用的一种数据结构或算法。标签 "c" 指出这个问题的解决方案是用 C 语言...

Global site tag (gtag.js) - Google Analytics