Question

我想知道我怎么能解决这个问题（这不是学校的东西，我作为一个爱好编程，我正在做欧拉的练习）。我的代码中的问题是：char c = great.charAt（i）; 我想知道怎样才能看到最伟大的下一个i.charAt（i）。我发布了我的代码，我提前感谢谁会帮助我。

// 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"); }
    }}

Answer 1

查看程序的输出，我发现它有这样的块：

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，实际上只是将其中一个数字乘以五次。

您需要确保迭代该子序列才能获得该产品。

问题出现在您的行greatest.charAt(i)中。实际上，您需要添加偏移量j以获取五位数子序列中的每个数字。

Answer 2

蛮力：

        String num = "73167176531330624...";

        int prod = 0;

        for (int i=0; i < num.length()-4; i++)
        {
            int tmp = (num.charAt(i) - '0')*
                      (num.charAt(i+1) - '0')*
                      (num.charAt(i+2) - '0')*
                      (num.charAt(i+3) - '0')*
                      (num.charAt(i+4) - '0');
            if (tmp > prod){
                prod = tmp;
            }
        }

        return  prod;

Answer 3

更清洁的代码版本：

    String str = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

    int i=0;
    int max = 0;

    while( i != (str.length() - 4)){
        int val = Character.digit(str.charAt(i), 10) *
                Character.digit(str.charAt(i+1), 10) *
                Character.digit(str.charAt(i+2), 10) *
                Character.digit(str.charAt(i+3), 10) *
                Character.digit(str.charAt(i+4), 10);

        if(val > max){
            max = val;
        }

        i++;
    }

    System.out.println(max);

Answer 4

解决方案是：

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

            String greatest = "731671765313306249192251196744265747423553491949349698352031277450632623957831801698480186947885184385861560789112949495459501737958331952853208805511125406987471585238630507156932909632952274430435576689664895044524452316173185640309871112172238311362229893423380308135336276614282806444486645238749303589072962904915604407723907138105158593079608667017242712188379087922749219016997208880937766572733300105336788122023542180975125454059475224352584907711670556013604839586446706324415722155397 5369781797784617406495514929086256932197846862248283972241375657056057490261407972968652414535100474821663704844031998900088952434506585412275886668811642717147992444292823086346567481391912316282458617866458359124566529476545682848912883142607690042242190226710556263211111093705442175069416589604080719840385096245544436298123098787992724428490918884580156166097919133875499200524063689912560717606058861164671094050775410022569831552000559357297257163626956188267042825248360082325753042075296345099879";
            int n = 0;
            int count = 0;
            long biggest = 0;
            int countBlock = 1;
            int biggestBlock = 1;
            long X = 1;

            for (int i = 0; i < greatest.length()-4; i++) {
                for (int k = i; k < i + 5; k++) {
                    char c = greatest.charAt(k);
                    n = Character.getNumericValue(c);
                    count++;
                    X = (X * n);
                    System.out.println("Product is:" + X);
                }
                if(biggest < X){
                    biggest = X;
                    biggestBlock = countBlock;
                }

                System.out.println("Now the biggest one is: "+ biggest);
                count = 0;
                X = 1;
                countBlock++;
                System.out.println("Block number: " + biggestBlock);
                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");
        }
    }
}

找到1000位数字中连续五位数的最佳乘积

4 个答案: