用于 ProjectEuler 问题 11 的程序优化
[英]Optimization of program used for ProjectEuler problem 11
问题描述
I've been learning to program for quite a bit and it seems that one of the greatest competitions between programmers is how few lines one can do a procedure in. Noticing this trend, I'd like to learn to make my programs a bit tighter, cleaner, and preferring functionality without excess.
我一直在学习编程，似乎程序员之间最大的竞争之一是一个程序可以用多少行。注意到这种趋势，我想学习让我的程序更紧凑，更干净，并且更喜欢没有多余的功能。 Here's the code I used to solve ProjectEuler problem 11. It's quite large which kinda worries me when I see code a fourth of its size doing the same thing, hehe.这是我用来解决 ProjectEuler 问题 11 的代码。它非常大，当我看到它的四分之一大小的代码做同样的事情时，我有点担心，呵呵。
#include <iostream>
using namespace std;
int array[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},
{20,73,35,29,78,31,90,1,74,31,49,71,48,86,81,16,23,57,5,54},
{1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48},
};
int s = 0;
int right()
{
int a = 1;
int i = 0;
int n = 0;
int r = 0;
int c = 0;
for(n = 0;n <= 359;n++)
{
if(c <= 16)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[r][(c + i)] << " ";
a *= array[r][(c + i)];
};
//cout << a << " ";
i = 0; c++;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 0; r++;};
};
return s;
};
int left()
{
int a = 1;
int i = 0;
int n = 0;
int r = 0;
int c = 19;
for(n = 0;n <= 359;n++)
{
if(c >= 3)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[r][(c - i)] << " ";
a *= array[r][(c - i)];
};
//cout << a << " ";
i = 0; c--;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 19; r++;};
};
return s;
};
int down()
{
int n = 0;
int i = 0;
int r = 0;
int c = 0;
int a = 1;
for(n = 0;n <= 356;n++)
{
if(c <= 19)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[(r + i)][c] << " ";
a *= array[(r + i)][c];
};
//cout << a << " ";
i = 0; c++;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 0;
if(r <= 16){
r++;
}else{break;};
};
};
return s;
};
int up()
{
int n = 0;
int i = 0;
int r = 19;
int c = 0;
int a = 1;
for(n = 0;n <= 356;n++)
{
if(c <= 19)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[(r - i)][c] << " ";
a *= array[(r - i)][c];
};
//cout << a << " ";
i = 0; c++;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 0;
if(r >= 3){
r--;
}else{break;};
};
};
return s;
};
int diag_left_up()
{
int n = 0;
int i = 0;
int r = 19;
int c = 19;
int a = 1;
for(n = 0;n <= 304;n++)
{
if(c >= 3 && r >= 3)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[(r - i)][(c - i)] << " ";
a *= array[(r - i)][(c - i)];
};
//cout << a << " ";
i = 0; c--;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 19;
if(r >= 3){
r--;
}else{break;};
};
};
return s;
};
int diag_left_down()
{
int n = 0;
int i = 0;
int r = 0;
int c = 19;
int a = 1;
for(n = 0;n <= 304;n++)
{
if(c >= 3 && r <= 16)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[(r + i)][(c - i)] << " ";
a *= array[(r + i)][(c - i)];
};
//cout << a << " ";
i = 0; c--;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 19;
if(r <= 16){
r++;
}else{break;};
};
};
return s;
};
int diag_right_up()
{
int n = 0;
int i = 0;
int r = 19;
int c = 0;
int a = 1;
for(n = 0;n <= 304;n++)
{
if(c <= 16 && r >= 3)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[(r - i)][(c + i)] << " ";
a *= array[(r - i)][(c + i)];
};
//cout << a << " ";
i = 0; c++;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 0;
if(r >= 3){
r--;
}else{break;};
};
};
return s;
};
int diag_right_down()
{
int n = 0;
int i = 0;
int r = 0;
int c = 0;
int a = 1;
for(n = 0;n <= 304;n++)
{
if(c <= 16 && r <= 16)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[(r + i)][(c + i)] << " ";
a *= array[(r + i)][(c + i)];
};
//cout << a << " ";
i = 0; c++;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 0;
if(r <= 16){
r++;
}else{break;};
};
};
return s;
};
int main()
{
cout << "Result from right():" << '\t' << right();
cout << endl;
cout << "Result from left():" << '\t' << left();
cout << endl;
cout << "Result from down():" << '\t' << down();
cout << endl;
cout << "Result from up():" << '\t' << up();
cout << endl;
cout << "Result from diag_right_up(): " << '\t' << diag_right_up();
cout << endl;
cout << "Result from diag_right_down(): " << '\t' << diag_right_down();
cout << endl;
cout << "Result from diag_left_up(): " << '\t' << diag_left_up();
cout << endl;
cout << "Result from diag_left_down(): " << '\t' << diag_left_down();
cout << endl << endl << "Greatest result: " << s;
return 0;
}
3 个解决方案
解决方案1
1 2011-01-13 01:28:36
The first thing I notice is that you've got a lot of functions that do basically the same thing (with some numbers different).我注意到的第一件事是你有很多功能基本上做相同的事情（有些数字不同）。 I would investigate adding a couple of parameters to that function, so that you can describe the direction you're going.我会研究向该函数添加几个参数，以便您可以描述您要前进的方向。 So for example, instead of calling right() you might call traverse(1, 0) and traverse(0, -1) instead of up() .因此，例如，您可以调用traverse(1, 0)和traverse(0, -1)而不是up() ，而不是调用right() up() 。
Your traverse() function declaration might look like:您的traverse()函数声明可能如下所示：
int traverse(int dx, int dy)
with the appropriate changes inside to adapt its behaviour for different values of dx and dy .在内部进行适当的更改以使其行为适应dx和dy不同值。
解决方案2
0 2011-01-13 01:39:11
Well, for starters, you only need four of those directions: right/left, up/down, right-up/down-left and right-down/up-left.好吧，对于初学者来说，您只需要四个方向：右/左、上/下、右上/左下和右下/左上。 Multiplication is commutative, so it doesn't matter which direction you go from a given pair (if you find "abcd" in one direction, you'll find "dcba" in the opposite direction, and you get the same result when multiplying those numbers together).乘法是可交换的，所以从给定的对往哪个方向走并不重要（如果你在一个方向找到“abcd”，你会在相反的方向找到“dcba”，并且在将它们相乘时得到相同的结果数在一起）。
Secondly, use more descriptive variable names.其次，使用更具描述性的变量名称。 A variable name like s is meaningless;像s这样s变量名是没有意义的； something like maximum is better, because that tells you what that variable is for.像maximum这样的东西更好，因为它告诉你那个变量是做什么的。 That doesn't mean you should never ever use a single-character variable name - eg, using i as a for-loop counter is perfectly fine, and if you're dealing with coordinates, x and y can also be just fine, but when at all possible, you should use a descriptive name to make the code more self-documenting.这并不意味着你永远不应该使用单字符变量名——例如，使用i作为 for 循环计数器是非常好的，如果你正在处理坐标， x和y也可以，但是在可能的情况下，您应该使用描述性名称来使代码更具自文档性。
Thirdly, you can look into what Greg suggests and refactor your methods to take a direction instead.第三，您可以查看 Greg 的建议并重构您的方法以取而代之。 This would allow you to ditch all of the similar methods (and just call that one method 4 times with different parameters to cover all of the necessary directions).这将允许您放弃所有类似的方法（只需使用不同的参数调用该方法 4 次即可涵盖所有必要的方向）。
And finally, you may want to be more consistent about your formatting - I know that can be hard to start doing, but it helps you in the long run.最后，您可能希望在格式方面更加一致 - 我知道这很难开始，但从长远来看它对您有帮助。 To understand what I mean here, take a good look at this snippet, taken from your down() method:要理解我在这里的意思，请仔细查看取自down()方法的这段代码：
if(c <= 19)
{
for(i = 0;i <= 3;i++)
{
//cout << " " << array[(r + i)][c] << " ";
a *= array[(r + i)][c];
};
//cout << a << " ";
i = 0; c++;
if(a > s)
{
s = a;
a = 1;
};
//cout << s << " " << endl;
a = 1;
}else{c = 0;
if(r <= 16){
r++;
}else{break;};
};
Notice how you place a line break before the opening curly bracket of your ifs, but write }else{ on a single line.请注意如何在 ifs 的左大括号之前放置换行符，但将}else{写在一行中。 Additionally, inside the first else , you have another if-block where you don't place a line break before the curly bracket, and the closing bracket has the same indent level as the block content ( r++; ).此外，在第一个else ，您还有另一个 if 块，其中您不在大括号之前放置换行符，并且右括号与块内容 ( r++; ) 具有相同的缩进级别。 This is quite inconsistent, and makes it harder to read.这是非常不一致的，并且使阅读变得更加困难。
解决方案3
0
static void largestProduct11() {
int[][] 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}
};
/*
* |A11 A12 A13 * * A1N|
* |A21 A22 A23 * * A2N|
* |A31 A32 A33 A3N|
* * *
* * *
* |An1 A2N A3N * * ANN|
*
* */
String line;
int[] temp = new int[4];
int compre = 1;
int result = 1;
for (int i = 0; i < arr.length - 3; i++) { // optimize:- the condition is true
// if the remaining index are at least four therefore the length should be reduced by
// three that means only the first 16 members can can satisfy the condition
for (int j = 0; j < arr[0].length - 3; j++) {
result = arr[i][j] * arr[i + 1][j + 1] * arr[i + 2][j + 2] * arr[i + 3][j + 3];
if (compre < result) {
compre = result;
}
}
}
//
System.out.println(compre + " Right sie test ");
for (int i = 0; i < arr.length - 3; i++) {
line = "{";
for (int j = arr[0].length - 1; j > 3; j--) {
result = arr[i][j] * arr[i + 1][j - 1] * arr[i + 2][j - 2] * arr[i + 3][j
- 3];
if (compre < result) {
compre = result;
}
}
}
System.out.println(compre + " final result"); // solution= 70600674
}
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