题目要求是:

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?    

看到这题目的那一刻,感觉有点棘手,但是仔细分析起来也还好,并不是很难。

题目给出的序列是一个很规矩的数字组合。我们可以考虑把它存放到等长的二维数组里面,当然也可以直接存放到一位数组。这里我存放到二维数组里面了。

首先,我先把序列进行简化来说明。假设我有以下序列,并存放到二维数组里面。

我从上到下依次给每个数组元素进行编号。

假设我们要找相连的T个数字的乘积最大值,我们可以设置标记从1号位置开始,直到36号位置结束(为什么是36号位置结束呢?这个应该不用解释吧)。当然这里不是每个位置都需要我们处理,稍后我会说我们怎么跳过部分不需要处理的位置。

但是这里还有一个问题就是:我们使用的是二维数组存放,我们需要把我们的位置编号计算成二维数组的行号和列号。因为是数组是规矩的,所以我们计算起来还是挺方便的。

假设我们的标记位置为i,数组的二维长度是n(这里是10),则

行号rowIndex=i/n,而列号columnIndex=i%n;

好了,下一步我们来说如何跳过某些位置。我们知道0乘以任何数都得0,所以在0的位置以及0前面T-1个位置都是可以不用计算的,可以直接跳过。

例如:当前位置为i,数字0的位置为j(j-i<T),则下一次开始计算位置应该为j+1。下面是用java写出来的简单代码。

 int[][] numbers=new int[][]{
{7,3,1,6,7,1,7,6,5,3,1,3,3,0,6,2,4,9,1,9,2,2,5,1,1,9,6,7,4,4,2,6,5,7,4,7,4,2,3,5,5,3,4,9,1,9,4,9,3,4},
{9,6,9,8,3,5,2,0,3,1,2,7,7,4,5,0,6,3,2,6,2,3,9,5,7,8,3,1,8,0,1,6,9,8,4,8,0,1,8,6,9,4,7,8,8,5,1,8,4,3},
{8,5,8,6,1,5,6,0,7,8,9,1,1,2,9,4,9,4,9,5,4,5,9,5,0,1,7,3,7,9,5,8,3,3,1,9,5,2,8,5,3,2,0,8,8,0,5,5,1,1},
{1,2,5,4,0,6,9,8,7,4,7,1,5,8,5,2,3,8,6,3,0,5,0,7,1,5,6,9,3,2,9,0,9,6,3,2,9,5,2,2,7,4,4,3,0,4,3,5,5,7},
{6,6,8,9,6,6,4,8,9,5,0,4,4,5,2,4,4,5,2,3,1,6,1,7,3,1,8,5,6,4,0,3,0,9,8,7,1,1,1,2,1,7,2,2,3,8,3,1,1,3},
{6,2,2,2,9,8,9,3,4,2,3,3,8,0,3,0,8,1,3,5,3,3,6,2,7,6,6,1,4,2,8,2,8,0,6,4,4,4,4,8,6,6,4,5,2,3,8,7,4,9},
{3,0,3,5,8,9,0,7,2,9,6,2,9,0,4,9,1,5,6,0,4,4,0,7,7,2,3,9,0,7,1,3,8,1,0,5,1,5,8,5,9,3,0,7,9,6,0,8,6,6},
{7,0,1,7,2,4,2,7,1,2,1,8,8,3,9,9,8,7,9,7,9,0,8,7,9,2,2,7,4,9,2,1,9,0,1,6,9,9,7,2,0,8,8,8,0,9,3,7,7,6},
{6,5,7,2,7,3,3,3,0,0,1,0,5,3,3,6,7,8,8,1,2,2,0,2,3,5,4,2,1,8,0,9,7,5,1,2,5,4,5,4,0,5,9,4,7,5,2,2,4,3},
{5,2,5,8,4,9,0,7,7,1,1,6,7,0,5,5,6,0,1,3,6,0,4,8,3,9,5,8,6,4,4,6,7,0,6,3,2,4,4,1,5,7,2,2,1,5,5,3,9,7},
{5,3,6,9,7,8,1,7,9,7,7,8,4,6,1,7,4,0,6,4,9,5,5,1,4,9,2,9,0,8,6,2,5,6,9,3,2,1,9,7,8,4,6,8,6,2,2,4,8,2},
{8,3,9,7,2,2,4,1,3,7,5,6,5,7,0,5,6,0,5,7,4,9,0,2,6,1,4,0,7,9,7,2,9,6,8,6,5,2,4,1,4,5,3,5,1,0,0,4,7,4},
{8,2,1,6,6,3,7,0,4,8,4,4,0,3,1,9,9,8,9,0,0,0,8,8,9,5,2,4,3,4,5,0,6,5,8,5,4,1,2,2,7,5,8,8,6,6,6,8,8,1},
{1,6,4,2,7,1,7,1,4,7,9,9,2,4,4,4,2,9,2,8,2,3,0,8,6,3,4,6,5,6,7,4,8,1,3,9,1,9,1,2,3,1,6,2,8,2,4,5,8,6},
{1,7,8,6,6,4,5,8,3,5,9,1,2,4,5,6,6,5,2,9,4,7,6,5,4,5,6,8,2,8,4,8,9,1,2,8,8,3,1,4,2,6,0,7,6,9,0,0,4,2},
{2,4,2,1,9,0,2,2,6,7,1,0,5,5,6,2,6,3,2,1,1,1,1,1,0,9,3,7,0,5,4,4,2,1,7,5,0,6,9,4,1,6,5,8,9,6,0,4,0,8},
{0,7,1,9,8,4,0,3,8,5,0,9,6,2,4,5,5,4,4,4,3,6,2,9,8,1,2,3,0,9,8,7,8,7,9,9,2,7,2,4,4,2,8,4,9,0,9,1,8,8},
{8,4,5,8,0,1,5,6,1,6,6,0,9,7,9,1,9,1,3,3,8,7,5,4,9,9,2,0,0,5,2,4,0,6,3,6,8,9,9,1,2,5,6,0,7,1,7,6,0,6},
{0,5,8,8,6,1,1,6,4,6,7,1,0,9,4,0,5,0,7,7,5,4,1,0,0,2,2,5,6,9,8,3,1,5,5,2,0,0,0,5,5,9,3,5,7,2,9,7,2,5},
{7,1,6,3,6,2,6,9,5,6,1,8,8,2,6,7,0,4,2,8,2,5,2,4,8,3,6,0,0,8,2,3,2,5,7,5,3,0,4,2,0,7,5,2,9,6,3,4,5,0}
};
int number=13;
int total=numbers.length*numbers[0].length-1; //Total numbers subtract 1
int k=0;
long temp=1;
long result=1;
for(int i=0;i<=(total-number+1);){ temp=1;
k=i+number-1; for(int j=i;j<=k;j++){
int oneIndex= j/numbers[0].length;
int twoIndex= j%numbers[0].length; temp*=numbers[oneIndex][twoIndex]; if(numbers[oneIndex][twoIndex]==0)
{
i=j+1;
break;
}
}
if(temp!=0)
{
if(temp>result)
{
//System.out.printf("%3d %d\r\n",i,temp);
result=temp;
}
i++;
}
} System.out.println(result);

Project Euler 8

【Project Euler 8】Largest product in a series的更多相关文章

  1. 【Project Euler 1】Multiples of 3 and 5

    题目要求是: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and ...

  2. 【Project Euler】530 GCD of Divisors 莫比乌斯反演

    [题目]GCD of Divisors [题意]给定f(n)=Σd|n gcd(d,n/d)的前缀和F(n),n=10^15. [算法]莫比乌斯反演 [题解]参考:任之洲数论函数.pdf 这个范围显然 ...

  3. Project Euler 8 Largest product in a series

    题意:寻找这 1000 个数中相邻 13 个数相乘积最大的值 思路:首先想到的是暴力,但是还可以利用之前记录过的数值,什么意思呢?即在计算 2 - 14 后,再计算 3 - 15 时完全可以利用之前计 ...

  4. 【刷题-LeetCode】238. Product of Array Except Self

    Product of Array Except Self Given an array nums of n integers where n > 1, return an array outpu ...

  5. Project Euler:99 Largest exponential C++

    Comparing two numbers written in index form like 211 and 37 is not difficult, as any calculator woul ...

  6. projecteuler Problem 8 Largest product in a series

    The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = ...

  7. Largest product in a series

    这个我开始理解错了,算错了. 我以为是求连续5个数的最大值,结果,是连接5个数相乘的最大值. 英语不好,容易吃亏啊. Find the greatest product of five consecu ...

  8. Problem 8: Largest product in a series

    先粘实现代码,以后需要再慢慢补充思路 s = ''' 73167176531330624919225119674426574742355349194934 9698352031277450632623 ...

  9. Python练习题 036:Project Euler 008:1000位数字中相邻13个数字最大的乘积

    本题来自 Project Euler 第8题:https://projecteuler.net/problem=8 # Project Euler: Problem 8: Largest produc ...

随机推荐

  1. Codeforces Round 548 (Div. 2)

    layout: post title: Codeforces Round 548 (Div. 2) author: "luowentaoaa" catalog: true tags ...

  2. python 多进程操作

    由于python 多线程是无法在多核上发挥优势的,所以才用多进程的方式来折中将这个问题解决. from multiprocessing import Pool import os def f(x): ...

  3. ironic简介

    转:https://doodu.gitbooks.io/openstack-ironic 简介 Bare Metal Servcie 裸机服务 -- 'bear betal' ironic简介 如今O ...

  4. thunk 函数

    function* f() { console.log(1); for (var i = 0; true; i++) { console.log('come in'); var reset = yie ...

  5. [HNOI2006]最短母串问题 --- AC自动机 + 隐式图搜索

    [HNOI2006]最短母串问题 题目描述: 给定n个字符串(S1,S2.....,Sn),要求找到一个最短的字符串T,使得这n个字符串(S1,S2,......,Sn)都是T的子串. 输入格式: 第 ...

  6. POJ 1509 Glass Beads 后缀自动机 模板 字符串的最小表示

    http://poj.org/problem?id=1509 后缀自动机其实就是一个压缩储存空间时间(对节点重复利用)的储存所有一个字符串所有子串的trie树,如果想不起来长什么样子可以百度一下找个图 ...

  7. BZOJ3289[JZYZOJP2018]: Mato的文件管理 莫队+树状数组+离散化

            描述 Description     Mato同学从各路神犇以各种方式(你们懂的)收集了许多资料,这些资料一共有n份,每份有一个大小和一个编号.为了防止他人偷拷,这些资料都是加密过的, ...

  8. 【DP】BZOJ1592-[Usaco2008 Feb]Making the Grade 路面修整

    我活着从期中考试回来了!!!!!!!!!备考NOIP!!!!!!!!! [题目大意] 给出n个整数a1~an,修改一个数的代价为修改前后差的绝对值,问修改成不下降序列或者不上升序列的最小总代价. [思 ...

  9. hdu 1561 树形dp+分组背包

    题意:就是给定n个点,每个地点有value[i]的宝物,而且有的宝物必须是另一个宝物取了才能取,问取m个点可以获得的最多宝物价值. 一个子节点就可以返回m个状态,每个状态表示容量为j(j<=m) ...

  10. AOP流程分析

    1. 注册AnnotationAwareAspectJAutoProxyCreator @EnableAspectJAutoProxy --> @Import(AspectJAutoProxyR ...