欧拉计划第十一题--java实现

先做个说明，里面所有的题都是我自己写的解题报告，由于我的能力有限，是个刚学java的小白，有很多不足的地方，还望各位大佬不奢赐教，谢谢！

Largest product in a grid

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

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
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 08 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 09 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 09 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 08 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 08 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

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?

1.问题描述

在这个20×20的矩阵里，4个对角上标红色的数，26，63，78，14的乘积为1788696.

那么在这个矩阵中，4个临近的数（包括竖直，水平，和对角方向）的最大乘积是多少？

答案为：70600674

2.代码java实现

public class pro_11 {
public static void main(String[] args)
{ int max=0;
int i,j,sum1,sum2,sum3,sum4;//i和j是定义的变量，sum1是横向的计算，sum2是纵向的计算

//sum3是右下向的计算，sum4是左下向的计算，为什么我不算左上方和右下方？因为和右下向，左下向重合了
int a[][]=
{ {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, 4,56,62,0},
{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, 0,76,44,20,45,35,14, 0,61,33,97,34,31,33,95},
{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, 0,17,54,24,36,29,85,57},
{86,56, 0,48,35,71,89, 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(i=0;i<16;i++)
for(j=0;j<16;j++)
{sum1=a[i][j]*a[i][j+1]*a[i][j+2]*a[i][j+3];
if(sum1>max)
max=sum1;
}
System.out.print("max1 is:");//横向
System.out.println(max);
for(i=0;i<16;i++)
for(j=0;j<16;j++)
{sum2=a[i][j]*a[i+1][j]*a[i+2][j]*a[i+3][j];
if(sum2>max)
max=sum2;
}
System.out.print("max2 is:");//纵向
System.out.println(max);
for(i=0;i<16;i++)
for(j=0;j<16;j++)
{sum3=a[i][j]*a[i+1][j+1]*a[i+2][j+2]*a[i+3][j+3];
if(sum3>max)
max=sum3;
}
System.out.print("max3 is:");//斜向右下方
System.out.println(max);
for(i=3;i<20;i++)
for(j=0;j<16;j++)
{sum4=a[i][j]*a[i-1][j+1]*a[i-2][j+2]*a[i-3][j+3];
if(sum4>max)
max=sum4;
}
System.out.print("max4 is:");//斜向左下方
System.out.println(max);
}
}

3.输出如下

然后取四者中最大的就行。