[Project 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 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 any direction (up, down, left, right, or diagonally) in the 20

20 grid?

求四个方向上，连续四个数乘积的最大值

没有太多值得分析的，直接编写程序

#include <iostream>
usingnamespace std;

int grid[20][20] = {
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,
20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57, 5,54,
1,70,54,71,83,51,54,69,16,92,33,48,61,43,52, 1,89,19,67,48
};

int main(){
int product =0;
int tmp1,tmp2;

for(int i=0; i<16; i++){
for(int j=0; j<16; j++){
tmp1 = grid[i][j]*grid[i+1][j+1]*grid[i+2][j+2]*grid[i+3][j+3];
tmp2 = grid[i][j+3]*grid[i+1][j+2]*grid[i+2][j+1]*grid[i+3][j];
if(tmp1 > product)
product = tmp1;
if(tmp2 >product)
product = tmp2;
}
}

for(int i=0; i<20; i++){
for(int j=0; j<16; j++){
tmp1 = grid[i][j]*grid[i][j+1]*grid[i][j+2]*grid[i][j+3];
tmp2 = grid[j][i]*grid[j+1][i]*grid[j+2][i]*grid[j+3][i];
if(tmp1 > product)
product = tmp1;
if(tmp2 >product)
product = tmp2;
}
}

cout << product << endl;
return0;
}



我们发现，向下和向右可以一起遍历，左斜和右斜可以一起遍历

当然，题目还有一点，要把原题中给的字符转换为整型

我直接在编辑器里手动转换了，要编写函数转的话也很方便。