【欧拉计划第 11 题】 网格中的最大乘积 Largest product in a grid

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Problem 11 Largest product in a grid

In the $20×20$ grid below, four numbers along a diagonal line have been marked in red.

The product of these numbers is $26 × 63 × 78 × 14 = 1788696$.
What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the $20×20$ grid?

问题 11 网格中的最大乘积

在下面的 $20×20$ 网格中，沿对角线的四个数字被标记为红色。

这些数字的乘积是 $26 × 63 × 78 × 14 = 1788696$。
在 $20×20$ 的网格中，沿相同方向（上、下、左、右或对角线）的四个相邻数字的最大乘积是多少？

思路分析

将数据表中的数字存入到二维数组中，依次计算相同方向的四个相邻数字乘积并暂存

比较所有乘积 ，找出最大值即所求

代码实现

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/*
 * @Author: coder-jason
 * @Date: 2022-04-18 14:54:44
 * @LastEditTime: 2022-04-18 15:15:33
 */
#include <bits/stdc++.h>
using namespace std;

int main()  
{
    int iMaxNumber = 0;
    // 分别表示四个相邻数字
    int n1, n2, n3, n4;
    // 初始化二维数组,存入数据表
    int array[400] = {8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8,
                      49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 00,
                      81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65,
                      52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 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, 3, 45, 2, 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, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21,
                      24, 55, 58, 5, 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, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92,
                      16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 00, 17, 54, 24, 36, 29, 85, 57,
                      86, 56, 00, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58,
                      19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40,
                      4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66,
                      88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69,
                      4, 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, 4, 36, 16,
                      20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54,
                      1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48};
    
    for (int i = 0; i < 20; i++)
    {
        for (int j = 0; j < 16; j++)
        {
            int a = array[i * 20 + j] * array[i * 20 + j + 1] * array[i * 20 + j + 2] * array[i * 20 + j + 2];
            if (a > iMaxNumber)
            {
                iMaxNumber = a;
                n1 = array[i * 20 + j];
                n2 = array[i * 20 + j + 1];
                n3 = array[i * 20 + j + 2];
                n4 = array[i * 20 + j + 2];
            }
        }
    }

    for (int i = 0; i < 20; i++)
    {
        for (int j = 0; j < 16; j++)
        {
            int a = array[j * 20 + i] * array[(j + 1) * 20 + i] * array[(j + 2) * 20 + i] * array[(j + 3) * 20 + i];
            if (a > iMaxNumber)
            {
                iMaxNumber = a;
                n1 = array[j * 20 + i];
                n2 = array[(j + 1) * 20 + i];
                n3 = array[(j + 2) * 20 + i];
                n4 = array[(j + 3) * 20 + i];
            }
        }
    }

    for (int i = 3; i < 20; i++)
    {
        int j = i;
        int a = 1;

        for (int x = 0; x <= i - 3; x++)
        {
            a = array[i - x + 20 * x] * array[i - x - 1 + 20 * (x + 1)] * array[i - x - 2 + 20 * (x + 2)] * array[i - x - 3 + 20 * (x + 3)];
            if (a > iMaxNumber)
            {
                iMaxNumber = a;
                n1 = array[i - x + 20 * x];
                n2 = array[i - x - 1 + 20 * (x + 1)];
                n3 = array[i - x - 2 + 20 * (x + 2)];
                n4 = array[i - x - 3 + 20 * (x + 3)];
            }
        }
    }

    for (int i = 16; i >= 0; i--)
    {
        int a = 1;

        for (int x = 0; x <= 16 - i; x++)
        {
            a = array[i + x + 20 * x] * array[i + x + 1 + 20 * (x + 1)] * array[i + x + 2 + 20 * (x + 2)] * array[i + x + 3 + 20 * (x + 3)];
            if (a > iMaxNumber)
            {
                iMaxNumber = a;
                n1 = array[i + x + 20 * x];
                n2 = array[i + x + 1 + 20 * (x + 1)];
                n3 = array[i + x + 2 + 20 * (x + 2)];
                n4 = array[i + x + 3 + 20 * (x + 3)];
            }
        }
    }
    printf("ans=%d. a = %d, b = %d, c = %d , d = %d.\n", iMaxNumber, n1, n2, n3, n4);
    return 0;
}

答案：70600674