Problem：
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 2020 grid?
思路：
考虑一行、一列、左斜、右斜四中情况；
C Code:
#include <stdio.h> #define ROW 20 #define COL 20 #define NUM 4 int GetMaxOfRow(int iArray[][COL],int irow,int icol,int inum) { int i,j; int max = 0; for(i = 0;i < irow;i++) { int tempMax = iArray[i][0] * iArray[i][1] * iArray[i][2] * iArray[i][3]; for(j = 1;j < icol - inum;j++) { if(iArray[i][j + inum -1] > iArray[i][j-1]) { int temp = iArray[i][j] * iArray[i][j + 1] * iArray[i][j + 2] * iArray[i][j + 3]; if(temp > tempMax) { tempMax = temp; } } } if(tempMax > max) { max = tempMax; } } return max; } int GetMaxOfCol(int iArray[][COL],int irow,int icol,int inum) { int i,j; int max = 0; for(j = 0;j < icol;j++) { int tempMax = iArray[0][j] * iArray[1][j] * iArray[2][j] * iArray[3][j]; for(i = 1;i < irow - inum;i++) { if(iArray[i + inum -1][j] > iArray[i-1][j]) { int temp = iArray[i][j] * iArray[i + 1][j] * iArray[i + 2][j] * iArray[i + 3][j]; if(temp > tempMax) { tempMax = temp; } } } if(tempMax > max) { max = tempMax; } } return max; } int GetMaxOfLeftDiagonally(int iArray[][COL],int irow,int icol,int inum) { int i,j; int max = 0; for(i = inum;i < irow;i++) { int tempMax = 0; for(j = 0;j < icol - inum;j++) { int temp = iArray[i][j] * iArray[i - 1][j + 1] * iArray[i - 2][j + 2] * iArray[i - 3][j + 3]; if(temp > tempMax) { tempMax = temp; } } if(tempMax > max) { max = tempMax; } } return max; } int GetMaxOfRightDiagonally(int iArray[][COL],int irow,int icol,int inum) { int i,j; int max = 0; for(i = 0;i < irow - inum;i++) { int tempMax = 0; for(j = 0;j < icol - inum;j++) { int temp = iArray[i][j] * iArray[i + 1][j + 1] * iArray[i + 2][j + 2] * iArray[i + 3][j + 3]; if(temp > tempMax) { tempMax = temp; } } if(tempMax > max) { max = tempMax; } } return max; } void main() { int iArray[ROW][COL]; int i,j; int max1,max2,max3; for(i = 0;i < ROW;i++) { for(j = 0;j < COL;j++) { scanf("%d",&iArray[i][j]); } } max1 = GetMaxOfCol(iArray,ROW,COL,NUM); max2 = GetMaxOfRow(iArray,ROW,COL,NUM); max3 = max1 > max2? max1:max2; max1 = GetMaxOfLeftDiagonally(iArray,ROW,COL,NUM); max2 = GetMaxOfRightDiagonally(iArray,ROW,COL,NUM); max3 = max1 > max3? max1:max3; max3 = max2 > max3? max1:max3; printf("%d\n",max3); }
Result:
70600674
ProjectEuler_P11,布布扣,bubuko.com
原文：http://www.cnblogs.com/zhoueh1991/p/3710380.html