我正试图遍历20x20网格并计算产品,但我一直得到不正确的结果.你能帮忙发现我的错误吗？

Question

这是Project Euler的第11个问题.它提供了一个20x20的数字网格并询问:

20x20网格中任意方向(上,下,左,右或对角)的四个相邻数字的最大乘积是多少？

我已多次查看我的代码,但我似乎无法找到为什么我的结果不正确.

#include <iostream>
using namespace std;

void find_greatest_product()
{   
    int grid[20][20] =
    {{8, 02, 22, 97, 38, 15, 00, 40, 00, 75, 04, 05, 07, 78, 52, 12, 50, 77, 91, 8},
    {49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 04, 56, 62, 00},
    {81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 03, 49, 13, 36, 65},
    {52, 70, 95, 23, 04, 60, 11, 42, 69, 24, 68, 56, 01, 32, 56, 71, 37, 02, 36, 91},
    {22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
    {24, 47, 32, 60, 99, 03, 45, 02, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
    {32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
    {67, 26, 20, 68, 02, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
    {24, 55, 58, 05, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
    {21, 36, 23, 9, 75, 00, 76, 44, 20, 45, 35, 14, 00, 61, 33, 97, 34, 31, 33, 95},
    {78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 03, 80, 04, 62, 16, 14, 9, 53, 56, 92},
    {16, 39, 05, 42, 96, 35, 31, 47, 55, 58, 88, 24, 00, 17, 54, 24, 36, 29, 85, 57},
    {86, 56, 00, 48, 35, 71, 89, 07, 05, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
    {19, 80, 81, 68, 05, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 04, 89, 55, 40},
    {04, 52, 8, 83, 97, 35, 99, 16, 07, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
    {88, 36, 68, 87, 57, 62, 20, 72, 03, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
    {04, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
    {20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 04, 36, 16},
    {20, 73, 35, 29, 78, 31, 90, 01, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 05, 54},
    {01, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 01, 89, 19, 67, 48}};

    int downRight_diag = 1;
    int upRight_diag = 1;
    int left = 1;
    int right = 1;
    int up = 1;
    int down = 1;
    int firstMax = 0;
    int absoluteMax = 0;

    for(int i = 0; i < 20; i++)
    {
        for(int j = 0; j < 20; j++)
        {
            for(int k = i; k < 17; k++)
            {
                downRight_diag = grid[i][j]*grid[i+1][j+1]*grid[i+2][j+2]*grid[i+3][j+3];
                down = grid[i][j]*grid[i+1][j]*grid[i+2][j]*grid[i+3][j];

                if(downRight_diag > down && downRight_diag > firstMax)
                    firstMax = downRight_diag;
                else if(down > downRight_diag && down > firstMax)
                    firstMax = down;

            }
            for(int l = i; l >= 3 && l < 20; l++)
            {
                upRight_diag = grid[i][j]*grid[i-1][j+1]*grid[i-2][j+2]*grid[i-3][j+3];
                up = grid[i][j]*grid[i-1][j]*grid[i-2][j]*grid[i-3][j];

                if(upRight_diag > firstMax)
                    firstMax = upRight_diag;
                else if(up > firstMax)
                    firstMax = up;
            }
            if(j < 17)
            {
                left = grid[i][j]*grid[i][j+1]*grid[i][j+2]*grid[i][j+3];

                if(left > firstMax)
                    firstMax = left;
            }
            if(j >= 3)
            {
                right = grid[i][j]*grid[i][j-1]*grid[i][j-2]*grid[i][j-3];

                if(right > firstMax)
                    firstMax = right;
            }

            if(firstMax > absoluteMax)
            {
                absoluteMax = firstMax;    
            }

            downRight_diag = 1;
            upRight_diag = 1;
            left = 1;
            right = 1;
            up = 1;
            down = 1;
            firstMax = 0;
        }
    }
    cout << absoluteMax << endl;
}
int main()
{

    find_greatest_product();
    system("pause");
}

澄清:left是从左到右,right从右到左计算产品.

我知道这可能不是最有效的方法,但我仍然感谢一些反馈.谢谢!

Answer 1

正如MRAB所回答的那样,这很可能是由于阵列溢出造成的.

我注意到的另一件事(但由于冗余计算可能不是问题)是这样的:

            if(upRight_diag > firstMax)
                firstMax = upRight_diag;
            else if(up > firstMax)
                firstMax = up;

如果两者都大于firstMaxAND up恰好是全局最大值,那么你会错过它.但这很奇怪,因为它up是多余的(由down前面处理).

我会做一些一般性评论.首先,为什么要多次循环上/下测试？你甚至没有使用循环变量.你只想测试边界(事实上你并没有这样做).此外,left您无需计算何时计算right.同样的,up和down我已经提到.

这是一个简化计算水平,垂直和两个对角线:

for(int i = 0; i < 20; i++)
{
    for(int j = 0; j < 20; j++)
    {
        int horz=0, vert=0, diagDR=0, diagUR=0;

        if( j >= 3 )
            horz = grid[i][j-3]*grid[i][j-2]*grid[i][j-1]*grid[i][j];

        if( i >= 3 )
            vert = grid[i-3][j]*grid[i-2][j]*grid[i-1][j]*grid[i][j];

        if( i >= 3 && j >= 3 ) {
            diagDR = grid[i-3][j-3]*grid[i-2][j-2]*grid[i-1][j-1]*grid[i][j];
            diagUR = grid[i-3][j]*grid[i-2][j-1]*grid[i-1][j-2]*grid[i][j-3];
        }

        if( horz > absoluteMax ) absoluteMax = horz;
        if( vert > absoluteMax ) absoluteMax = vert;
        if( diagDR > absoluteMax ) absoluteMax = diagDR;
        if( diagUR > absoluteMax ) absoluteMax = diagUR;
    }
}