项目欧拉：＃8

本文介绍了项目欧拉：＃8的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

尝试回答此问题时：

1000位数字中的四个相邻数字，其中最大产物是9×9×8×9 = 5832。

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 164271714799244429282308634656 74813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450

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

查找13 具有最大产品的1000位数字的相邻数字。这个产品的价值是多少？

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

我得到2091059712然而欧拉说答案是不正确的，有什么我可能做错了吗？

I get 2091059712 however Euler says the answer is incorrect is there anything I may be doing incorrectly?

public class LargestProductThirteen{ public static void main( String[] args ) { final String num = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"; long greatestProduct = 0; for (int i = 0; i < num.length() - 12; i++) { long sum = Character.getNumericValue(num.charAt(i))* Character.getNumericValue(num.charAt(i+1))* Character.getNumericValue(num.charAt(i+2))* Character.getNumericValue(num.charAt(i+3))* Character.getNumericValue(num.charAt(i+4))* Character.getNumericValue(num.charAt(i+5))* Character.getNumericValue(num.charAt(i+6))* Character.getNumericValue(num.charAt(i+7))* Character.getNumericValue(num.charAt(i+8))* Character.getNumericValue(num.charAt(i+9))* Character.getNumericValue(num.charAt(i+10))* Character.getNumericValue(num.charAt(i+11))* Character.getNumericValue(num.charAt(i+12)); if (sum > greatestProduct) greatestProduct = sum; } System.out.println(greatestProduct); } }

推荐答案

你当您将所有这些字符的数值相乘时，它们正在执行整数运算。有了高位数，其中有13个，很可能这样的产品会溢出 int ，其最大值约为20亿（10位）。

You are performing integer arithmetic when you are multiplying all those characters' numeric values together. With high digits, and 13 of them, it is likely that such a product would overflow an int, whose max value is about 2 billion (10 digits).

Character.getNumericValue 方法返回 int 。将第一个返回值转换为 long 以强制 long math。

The Character.getNumericValue method returns an int. Cast the first return value as a long to force long math.

long sum = (long) Character.getNumericValue(num.charAt(i))* Character.getNumericValue(num.charAt(i+1))* ...

顺便提一下，即使你有 greatProduct 已经因某种原因将此变量定义为 sum 。出于语义的考虑，我会将其命名为 product 。

Incidentally, even though you have greatestProduct already, for some reason you defined this variable as sum. Just for semantics' sake, I would name it product.