我正在解决Project Euler中的问题11。我已经找到了算法以及我需要做什么。网格保存在文件grid.txt中，其内容为 -
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
问题是 - 20x20网格中任意方向（上，下，左，右或对角）的四个相邻数字的最大乘积是什么？
我知道算法运行正常，因为我尝试使用cout输出nums，它们以正确的顺序显示。它给了我一个不正确的答案，可能是什么错？
void problem11()
{
vector<vector<int>> grid;
ifstream stream("grid.txt");
string line;
char *tok;
if (stream.is_open())
{
while(stream.good())
{
getline(stream, line);
tok = strtok((char *)line.c_str(), " ");
vector<int> row;
while (tok != NULL)
{
int field;
stringstream ss;
ss << tok;
ss >> field;
row.push_back(field);
tok = strtok(NULL, " ");
}
grid.push_back(row);
}
stream.close();
}
unsigned long highest = 0;
/// LEFT TO RIGHT
for (int i=0; i < 20; i++) // i'th row
{
vector<int> row = grid.at(i);
for (int c=0; c < 20-3; c++) // -3 to accomodate for last
{
unsigned long prod = row.at(c) * row.at(c+1) * row.at(c+2) * row.at(c+3); // four consecutive
//cout << row.at(c) << " " << row.at(c+1) << " " << row.at(c+2) << " " << row.at(c+3) << endl;
if (prod > highest)
highest = prod;
}
}
/// TOP TO DOWN
/// This moves from left to right, then top to botom
///
for (int i=0; i < 20-3; i++) // subtract last 3
{
vector<int> row1, row2, row3, row4;
row1 = grid.at(i);
row2 = grid.at(i+1);
row3 = grid.at(i+2);
row4 = grid.at(i+3);
for (int c=0; c < 20; c++)
{
unsigned long prod = row1.at(c) * row2.at(c) * row3.at(c) * row4.at(c);
//cout << row1.at(c) << " " << row2.at(c) << " " << row3.at(c) << " " << row4.at(c) << endl;
if (prod > highest)
highest = prod;
}
}
/// DOWN DIAGONAL
/// This moves diagonally from left to right, top to bottom
for (int i=0; i < 20-3; i++) // subtract last 3
{
vector<int> row1, row2, row3, row4;
row1 = grid.at(i);
row2 = grid.at(i+1);
row3 = grid.at(i+2);
row4 = grid.at(i+3);
for (int c=0; c < 20-3; c++) // omit last 3
{
unsigned long prod = row1.at(c) * row2.at(c+1) * row3.at(c+2) * row4.at(c+3);
//cout << row1.at(c) << " " << row2.at(c+1) << " " << row3.at(c+2) << " " << row4.at(c+3) << endl;
if (prod > highest)
highest = prod;
}
}
/// UP DIAGONAL
/// This moves diagonally from left to right, bottom to top
for (int i=3; i < 20; i++) // start from 3, skipping first four
{
vector<int> row1, row2, row3, row4;
row4 = grid.at(i);
row3 = grid.at(i-1);
row2 = grid.at(i-2);
row1 = grid.at(i-3);
for (int c=0; c < 20-3; c++) // omit last 3
{
unsigned long prod = row4.at(c) * row3.at(c+1) * row3.at(c+2) * row4.at(c+3);
//cout << row4.at(c) << " " << row3.at(c+1) << " " << row2.at(c+2) << " " << row1.at(c+3) << endl;
if (prod > highest)
highest = prod;
}
}
cout << "Required: " << highest;
}
答案 0 :(得分：6)
冒着破坏自己找到答案的乐趣的风险......
打印出对角线。目视检查它们是否符合您的预期。
作为旁注：并且不要创建表行的副本，而是同样访问它们：vector[rowindex][column]。
编辑 -
好的，现在我要破坏它。在NxN矩阵中，您有多少对角线路径？你走过多少条路？ （通过采用2x2矩阵交叉检查，每个方向有3条对角路径）
PS。如果您认真对待编程，遇到错误时，请先验证您的假设。
答案 1 :(得分：4)
尝试以下输入：
0 0 0 1 0 0 ...
0 0 1 0 0 0 ...
0 1 0 0 0 0 ...
1 0 0 0 0 0 ...
0 0 0 0 0 0 ...
.....
并再次执行xtofl推荐的内容。
通常，如果要反转某些操作的逻辑，只需执行一次。 （或者，通常是奇数次。）注意将a<=b替换为b>=a，或left to right and top to bottom替换right to left and bottom to top，或类似的任何内容。
答案 2 :(得分：0)
我用Javascript来解决这个问题。如果您有任何疑问，请告诉我。
<html>
<head>
</head>
<body>
<input type="button" value="Click Me" onClick="onPress()"></input>
</body>
<script>
function onPress()
{
var arr=[
[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]
];
var i, j, product=0, arr2=[], max;
/* Horizontal 4 digit multiplication*/
for(i=0; i<arr.length; i++)
{
for(j=0; j<17;j++)
{
product= arr[i][j]* arr[i][j+1] *arr[i][j+2] * arr[i][j+3]
arr2.push(product);
}
}
/* Vertical 4 digit multiplication*/
for(var j=0; j<arr.length; j++)
{
for(var i=0; i<17; i++)
{
product= arr[i][j] * arr[i+1][j] * arr[i+2][j] * arr[i+3][j];
arr2.push(product);
}
}
/* left to right diagonal*/
for(var j=0 ; j<17; j++)
{
for(var i=0; i<17; i++)
{
product= arr[i][j]* arr[i+1][j+1]*arr[i+2][j+2]*arr[i+3][j+3];
arr2.push(product)
}
}
/* right to left diagonal*/
for(var i=0; i<17; i++)
{
for(var j=19; j>=3; j--)
{
product= arr[i][j]*arr[i+1][j-1]*arr[i+2][j-2]*arr[i+3][j-3]
arr2.push(product);
}
}
max= Math.max.apply(Math, arr2);
console.log(max)
}
</script>
</html>
答案 3 :(得分：-2)
这是我写的代码。给出正确的答案。我希望这有帮助......
#include <iostream>
#include <vector>
using namespace std;
void main()
{
int num_container[20][20] = {
{ 8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91, 8},
{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, 8,40,91,66,49,94,21},
{24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
{21,36,23, 9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
{78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14, 9,53,56,92},
{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, 8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
{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, 8,46,29,32,40,62,76,36},
{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},
{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},
};
int test = num_container[6][8] * num_container[7][9] * num_container[8][10] * num_container[9][11];
cout<<test<<endl;
system("pause");
int start = 0;
int end = 3;
long long mul_result = 1;
vector<long long>final_results;
/////////////////////UP/DOWN/////////////////////
for(int k=0; k<20; k++)
{
for(int i=0; i<=16; i++)
{
for(int j=start; j<=end; j++)
{
mul_result = mul_result * num_container[k][j];
if (j == end)
final_results.push_back(mul_result);
}
mul_result = 1;
start++;
end++;
}
start = 0;
end = 3;
for(int i=0; i<=16; i++)
{
for(int j=start; j<=end; j++)
{
mul_result = mul_result * num_container[j][k];
if (j == end)
final_results.push_back(mul_result);
}
mul_result = 1;
start++;
end++;
}
start = 0;
end = 3;
}
/////////////////////UP/DOWN Ends here//////////////////////
///////////////////Both Ways Diagonal Starts here//////////////////////
int current_row = 0;
for(int i=0; i<=16; i++)
{
for(int j=0; j<=16; j++)
{
current_row = i;
for(int k=start; k<=end; k++)
{
mul_result = mul_result * num_container[current_row][k];
current_row++;
if (k==end)
final_results.push_back(mul_result);
}
mul_result = 1;
start++;
end++;
}
start = 0;
end = 3;
for(int j=0; j<=16; j++)
{
current_row = i+3;
for(int k=start; k<=end; k++)
{
mul_result = mul_result * num_container[current_row][k];
current_row--;
if (k==end)
final_results.push_back(mul_result);
}
mul_result = 1;
start++;
end++;
}
start = 0;
end = 3;
}
/////////////////////Both Ways diagonal ends here///////////////////
////////////////////Compare Thning Starts here//////////////////////
long long the_big_one = 0;
for(int i=0; i<final_results.size(); i++)
{
if (final_results[i] > the_big_one)
the_big_one = final_results[i];
}
cout<<endl<<endl<<"The big one is: "<<the_big_one<<endl;
system("pause");
}