In the 2020 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 2020 grid?

http://projecteuler.net/problem=11

需要把矩阵分区，第0~19行&&第0~15列的求横向的，第0~15行&&第0~19列求纵向的，第0~15行&&第0~15列求斜向右下的，第0~15行&&第4~19列求斜向左下的。

#include
     
      #include
      
       using namespace std;int n[20][20];int main(){    freopen("1.txt","r",stdin);    for(int i=0; i<20; i++)        for(int j=0; j<20; j++)            cin>>n[i][j];    int max=1;    for(int i=0; i<20; i++)        for(int j=0; j<16; j++)            if (n[i][j]*n[i][j+1]*n[i][j+2]*n[i][j+3]>max )                max=n[i][j]*n[i][j+1]*n[i][j+2]*n[i][j+3];    for(int i=0; i<16; i++)        for(int j=0; j<20; j++)            if(n[i][j]*n[i+1][j]*n[i+2][j]*n[i+3][j]>max)                max=n[i][j]*n[i+1][j]*n[i+2][j]*n[i+3][j];    for(int i=0; i<16; i++)        for(int j=0; j<16; j++)            if(n[i][j]*n[i+1][j+1]*n[i+2][j+2]*n[i+3][j+3]>max)                max=n[i][j]*n[i+1][j+1]*n[i+2][j+2]*n[i+3][j+3];    for(int i=0; i<16; i++)        for(int j=4; j<20; j++)            if (n[i][j]*n[i+1][j-1]*n[i+2][j-2]*n[i+3][j-3]>max)                max=n[i][j]*n[i+1][j-1]*n[i+2][j-2]*n[i+3][j-3];    cout<
       
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