在1000位数字中找到五个连续数字的最大乘积

小编典典

在1000位数字中找到五个连续数字的最大乘积

java

我想知道如何解决这个问题（这不是学校的东西，我是业余爱好，我正在做欧拉的练习）。我的代码中的问题是：char c = great.charAt（i）;
我想知道如何在great.charAt（i）中看到下一个i。我发布了代码，在此先感谢您的帮助。

// Find the greatest product of five consecutive digits in the 1000-digit number

import java.util.*;
import java.io.*;

public class FIVE
{
    public static void main(String args[]) 
    {
        try{

            String greatest = "73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557668966489504452445231617318564030987111217223831136222989342338030813533627661428280644448664523874930358907296290491560440772390713810515859307960866701724271218839987979087922749219016997208880937766572733300105336788122023542180975125454059475224352584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
            long TheNUMBER = 0; 
            long n = 0;    
            long count = 0;
            long provvisoryBiggest = 0;
            long countZERO = 0;
            long countBLOCK = 0;
            long X = 0;

            for ( int i=0; i < greatest.lenght(); i++)  
            {
                for(int j=0; j<5; j++)
                {
                    if(X == 0)
                    {
                        X = 1;
                    }
                    char c = greatest.charAt(i); 
                    n = Character.getNumericValue(c);
                    count++;
                    System.out.println(count + "th number: " + n);
                    X = (X * n );
                    System.out.println("The product is: " + X);

                    while (provvisoryBiggest < X)
                    {
                        provvisoryBiggest = X;
                    }
                    while(count == 5) 
                    {
                        System.out.println("Now the biggest one is: " + provvisoryBiggest);
                        count = countZERO;
                        X = countZERO;
                        countBLOCK++;
                        System.out.println("Block number: "+ countBLOCK);
                        System.out.println("------------------------------------");
                    }
                }
            } 
            }
           catch(ArrayIndexOutOfBoundsException ex) { System.out.println("ARRAY ERROR"); }
           catch(ArithmeticException ex) { System.out.println("MATH ERROR");}
           catch(NumberFormatException ex) { System.out.println("NUMBER FORMAT EXCEPTION");}
           catch(StringIndexOutOfBoundsException ex) { System.out.println("INDEX ERROR"); }
    }}

阅读 267

2020-11-26

共1个答案

小编典典

查看程序的输出，我看到它具有如下所示的块：

1th number: 8
The product is: 8
2th number: 8
The product is: 64
3th number: 8
The product is: 512
4th number: 8
The product is: 4096
5th number: 8
The product is: 32768
Now the biggest one is: 59049
Block number: 684

这说明，对于五个数字的子序列，您实际上只是将其中一个数字乘以五倍。

您需要确保迭代该子序列才能获得产品。

看来问题出在您的行上greatest.charAt(i)。实际上，您需要添加偏移量j以获取五位数子序列中的每一位数。

2020-11-26