[Project Euler] add Euler problem 11

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 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 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 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 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 16 92 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 any direction (Up, down, left, right, or diagonally) in the 20 × 20 grid?

 

Calculate the product of the four adjacent numbers in the 20*20 matrix and the maximum value.

Adjacent positions include, up and down, left and right, and diagonal.

 

Ideas:

1. Convert the string to the corresponding two-dimensional integer array.

2. For each vertex in a two-dimensional array, solve the four values associated in these directions.

The four directions starting from this point (note that four directions are used here, instead of eight directions, to avoid repeated computation .)

Upper right corner

Lower right corner

Right side

Below

If there are less than four elements on either side (including the vertex itself), the value is 0.

Compare the maximum values of the four values.

Such a value exists for each vertex.

3. traverse two-dimensional arrays and compare the maximum values of each vertex. That is, the maximum value of the matrix.

 

There are several notes:

1. How to convert the string to a corresponding two-dimensional array.

2. How to Solve the maximum value corresponding to each vertex.

3. It is required to know how to pass parameters of multi-dimensional arrays.

 Question 1 of the Euler Project # Include <stdio. h> # include <stdlib. h> // Maximum value of the product of the Four Directions surrounding each vertex in the analysis matrix  // The four directions are the upper right corner, the right corner, the lower right corner, and the lower right corner.  // Parameters I and j are the corresponding coordinates, and return values are the maximum values in the four directions corresponding to the I and j coordinates.  Unsigned   Long Pointmax ( Int (* ARR) [], Int I, Int J ); Int Main ( Int Argc,Char * Argv []) { Char * P =" 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 28 28 66 33 13 80 \ 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 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 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 85 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 16 16 92 48 61 43 52 01 89 19 67 48";          Int Arr [20] [20]; Int Row = 0, Col = 1; // Convert the string to an integer Array      While (* P! = '\ 0 '){ If (COL % 21! = 0) {arr [row] [col-1] = (* P-48) * 10 + * (p + 1)-48; Col ++; If (ROW = 19) & (COL = 21 )){ Break ;} Else P + = 3 ;} Else {Row ++; Col = 1 ;}} Int Showrow = 0, showcol = 0; // Display a two-dimensional array      For (Showrow = 0; showrow <20; showrow ++) For (Showcol = 1; showcol <21; showcol ++ ){ Printf (" % 3d ", Arr [showrow] [showcol-1]); If (Showcol % 20 = 0) Printf (" \ N ");} Unsigned  Long Max = 0; ROW = 0, Col = 0; For (ROW = 0; row <20; row ++) For (COL = 0; Col <20; Col ++ ){ Unsigned   Long TMP = pointmax (ARR, row, col ); If (TMP> MAX) max = TMP ;} Printf (" % Lu \ n ", Max ); System (" Pause "); Return 0 ;}Unsigned   Long Pointmax ( Int (* ARR) [20], Int I, Int J ){ Unsigned   Long Maxarr [4]; // The temporary array stores values from four directions      Unsigned   Long Max = 0; If (I> = 3 & J <= 16) // Check whether all three elements exist in the upper-right corner. {Maxarr [0] = arr [I-1] [J + 1] * arr [I-2] [J + 2] * arr [I-3] [J + 3] * arr [I] [J];} Else {Maxarr [0] = 0 ;} If (J <= 16) // Determine the three elements on the right {Maxarr [1] = arr [I] [J] * arr [I] [J + 1] * arr [I] [J + 2] * arr [I] [J + 3];} Else {Maxarr [1] = 0 ;} If (I <= 16 & J <= 16) // Determine the three elements in the lower right corner {Maxarr [2] = arr [I + 1] [J + 1] * arr [I + 2] [J + 2] * arr [I + 3] [J + 3] * arr [I] [J];} Else {Maxarr [2] = 0 ;}If (I <= 16) // Determine the three elements below {Maxarr [3] = arr [I + 1] [J] * arr [I + 2] [J] * arr [I + 3] [J] * arr [I] [J];} Else {Maxarr [3] = 0 ;} Int K; For (K = 0; k <4; k ++ ){ If (Maxarr [k]> MAX) max = maxarr [k];} Return Max ;}