Optimization of program used for ProjectEuler problem 11 - c++

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. 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.
#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;
}

The first thing I notice is that you've got a lot of functions that do basically the same thing (with some numbers different). I would investigate adding a couple of parameters to that function, so that you can describe the direction you're going. So for example, instead of calling right() you might call traverse(1, 0) and traverse(0, -1) instead of up().
Your traverse() function declaration might look like:
int traverse(int dx, int dy)
with the appropriate changes inside to adapt its behaviour for different values of dx and dy.

Well, for starters, you only need four of those directions: right/left, up/down, right-up/down-left and right-down/up-left. Multiplication is commutative, so it doesn't matter which direction you go from a given pair (if you find "a b c d" in one direction, you'll find "d c b a" in the opposite direction, and you get the same result when multiplying those numbers together).
Secondly, use more descriptive variable names. A variable name like s is meaningless; something like maximum is better, because that tells you what that variable is for. That doesn't mean you should never ever use a single-character variable name - e.g., 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.
Thirdly, you can look into what Greg suggests and refactor your methods to take a direction instead. 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).
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. To understand what I mean here, take a good look at this snippet, taken from your down() method:
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. 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++;). This is quite inconsistent, and makes it harder to read.

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
}

Related

Finding the most divisible number in a 100,000 range

I'm a student in the 10th grade and our teacher assigned some exercises. I'm pretty advanced in my class but one exercise is just isn't coming together as I want. The exercise is as follows:
Given the numbers between 100,000 and 200,000, find the number with the most divisors (as in which number can be divided with the most numbers, any number).
I wrote something but it executes in 32 seconds(!) and I just can't figure out the logic behind it (I can't even check if the answer is correct).
This is the code that I wrote:
void f3() {
int mostcount = 0, most, divisors = 0;
for (int i = 100000; i <= 200000; i++) {
for (int j = 1; j<=i/2; j++) {
if (i % j == 0) {
divisors++;
}
}
if (divisors > mostcount) {
mostcount = divisors;
most = i;
}
divisors = 0;
}
cout << most << endl;
return;
}
The output:
166320
Edit: My question is how can I reduce the runtime?
I'd be really thankful if you could explain your answer if you do decide to help, instead of just saying that I could do this with an XYZ type binary tree.
Since you tagged this with performance...
You can find all the prime factors of your input value (and the number of times they occur), and calculate the number of ALL divisors by multiplying those counts+1. This example cut the time by almost 5 times!
void f() {
int mostcount = 0, most = 0;
for (int i = 100000; i <= 200000; i++) {
int divisors = 1;
int x = i;
int limit = sqrt(i);
int count = 0;
for (int j = 2; j <= limit; j++) {
int count = 0;
while (x % j == 0) {
count++;
x /= j;
limit = sqrt(x);
}
divisors *= count + 1;
if (j > 2)
++j;
}
if (x > 1)
divisors *= 2;
if (divisors > mostcount) {
mostcount = divisors;
most = i;
}
}
cout << most << ", " << mostcount << endl;
return;
}
If you find a list of prime numbers, you can cut that time in half:
static vector<int> primes = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79,
83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193,
197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317,
331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449 };
void f() {
int mostcount = 0, most;
for (int i = 100000; i <= 200000; i++) {
int divisors = 1;
int x = i;
int limit = sqrt(i);
for (int j: primes) {
if (j > limit) break;
int count = 0;
while (x % j == 0) {
count++;
x /= j;
}
limit = sqrt(x);
divisors *= count + 1;
}
if(x > 1)
divisors *= 2;
if (divisors > mostcount) {
mostcount = divisors;
most = i;
}
}
cout << most << ", " << mostcount << endl;
return;
}
Explanation: let's take 166320. It's prime factors are:
2 (4 times)
3 (3 times)
5 (once)
7 (once)
11 (once)
Here, 2 produces 5 combination (it may go into the divisor 0 to 4 times), 3 - 3 combinations, and so on.
So you get 5 * 4 * 2 * 2 * 2 = 160 divisors. Strictly speaking, you should add one more - the input number itself.
Using #AhmedAEK 's response:
replace j<=i/2 with j<=sqrt(i), you only need to loop up to that, also #include <math.h> at the top, you also need to multiply the total divisors by 2, since there is a number above the sqrt that reflects the number below the sqrt. ie: 1000 is 10 x 100.
void f3() {
int mostcount = 0, most, divisors = 0;
for (int i = 100000; i <= 200000; i++) {
for (int j = 2; j<=sqrt(i); j++) {
if (i % j == 0) {
divisors++;
}
}
if (divisors > mostcount) {
mostcount = divisors;
most = i;
}
divisors = 0;
}
cout << most << endl;
return;
}

C++ iteration on first and last elements of array

Let's say I have this array:
int oldv[10] = {16, 12, 24, 96, 45, 22, 18, 63, 47, 56};
and another one like
int newv[8];
and I want to fill new from alternating ends of old until a certain condition is met such that I'd have:
newv = [16, 56, 12, 47, 24, 63 ...]
Let's say I want to put in new only 3 numbers taken from old (that is: 16, 56, 12).
I've tried with the following for loop, but of course is not enough...
for(int i = 0; i < 3; i++)
newv[i] = oldv[i*(sizeof(oldv)-1)];
Any help?
int _old[10] = {16, 12, 24, 96, 45, 22, 18, 63, 47, 56};
int _new[8];
const int old_size = sizeof(_old)/sizeof(int);
const int new_size = sizeof(_new)/sizeof(int);
for (int i = 0; i < new_size; ++i)
{
if (i % 2)
_new[i] = _old[old_size - i / 2 - 1];
else
_new[i] = _old[i / 2];
std::cout << _new[i] << " ";
}
std::cout << std::endl;
Returns 16 56 12 47 24 63 96 18
See it live
Enjoy.
#include <algorithm>
#include <iostream>
#include <iterator>
using namespace std;
int main(int, char**)
{
int oldv[] = { 16, 12, 24, 96, 45, 22, 18, 63, 47, 56 };
int newv[8];
size_t numbers_i_want = 3;
size_t oldv_b = 0;
size_t oldv_e = sizeof oldv / sizeof *oldv;
size_t newv_e = sizeof newv / sizeof *newv;
for(size_t i = 0;
i != min(numbers_i_want, newv_e) && oldv_b != oldv_e;
++i)
{
newv[i] = (i % 2) ? oldv[--oldv_e] : oldv[oldv_b++];
}
copy(newv, newv + min(numbers_i_want, newv_e),
ostream_iterator<decltype(*newv)>(cout, " "));
return 0;
}

Allow User To Input Data For A Sudoku Solving Program?

I am currently trying to allow the user to input values into a Sudoku solver however I keep getting an error. Here is the code:
#include <iostream> //
#include <fstream>
using namespace std;
class SudokuBoard;
void printB(SudokuBoard sb);
typedef unsigned int uint;
const uint MAXVAL = 9;
const uint L = 9;
const uint C = 9;
const uint S = L * C;
const uint ZONEL = 3;
const uint ZONEC = 3;
const uint ZONES = ZONEL * ZONEC;
const uint lineElements[L][C] = {
{ 0, 1, 2, 3, 4, 5, 6, 7, 8},
{ 9, 10, 11, 12, 13, 14, 15, 16, 17},
{18, 19, 20, 21, 22, 23, 24, 25, 26},
{27, 28, 29, 30, 31, 32, 33, 34, 35},
{36, 37, 38, 39, 40, 41, 42, 43, 44},
{45, 46, 47, 48, 49, 50, 51, 52, 53},
{54, 55, 56, 57, 58, 59, 60, 61, 62},
{63, 64, 65, 66, 67, 68, 69, 70, 71},
{72, 73, 74, 75, 76, 77, 78, 79, 80}
};
const uint columnElements[C][L] = {
{ 0, 9, 18, 27, 36, 45, 54, 63, 72},
{ 1, 10, 19, 28, 37, 46, 55, 64, 73},
{ 2, 11, 20, 29, 38, 47, 56, 65, 74},
{ 3, 12, 21, 30, 39, 48, 57, 66, 75},
{ 4, 13, 22, 31, 40, 49, 58, 67, 76},
{ 5, 14, 23, 32, 41, 50, 59, 68, 77},
{ 6, 15, 24, 33, 42, 51, 60, 69, 78},
{ 7, 16, 25, 34, 43, 52, 61, 70, 79},
{ 8, 17, 26, 35, 44, 53, 62, 71, 80}
};
const uint zoneElements[S / ZONES][ZONES] = {
{ 0, 1, 2, 9, 10, 11, 18, 19, 20},
{ 3, 4, 5, 12, 13, 14, 21, 22, 23},
{ 6, 7, 8, 15, 16, 17, 24, 25, 26},
{27, 28, 29, 36, 37, 38, 45, 46, 47},
{30, 31, 32, 39, 40, 41, 48, 49, 50},
{33, 34, 35, 42, 43, 44, 51, 52, 53},
{54, 55, 56, 63, 64, 65, 72, 73, 74},
{57, 58, 59, 66, 67, 68, 75, 76, 77},
{60, 61, 62, 69, 70, 71, 78, 79, 80}
};
class SudokuBoard {
public:
SudokuBoard() :
filledIn(0)
{
for (uint i(0); i < S; ++i)
table[i] = usedDigits[i] = 0;
}
virtual ~SudokuBoard() {
}
int const at(uint l, uint c) { // Returns the value at line l and row c
if (isValidPos(l, c))
return table[l * L + c];
else
return -1;
}
void set(uint l, uint c, uint val) { // Sets the cell at line l and row c to hold the value val
if (isValidPos(l, c) && ((0 < val) && (val <= MAXVAL))) {
if (table[l * C + c] == 0)
++filledIn;
table[l * C + c] = val;
for (uint i = 0; i < C; ++i) // Update lines
usedDigits[lineElements[l][i]] |= 1<<val;
for (uint i = 0; i < L; ++i) // Update columns
usedDigits[columnElements[c][i]] |= 1<<val;
int z = findZone(l * C + c);
for (uint i = 0; i < ZONES; ++i) // Update columns
usedDigits[zoneElements[z][i]] |= 1<<val;
}
}
void solve() {
try { // This is just a speed boost
scanAndSet(); // Logic approach
goBruteForce(); // Brute force approach
} catch (int e) { // This is just a speed boost
}
}
void scanAndSet() {
int b;
bool changed(true);
while (changed) {
changed = false;
for (uint i(0); i < S; ++i)
if (0 == table[i]) // Is there a digit already written?
if ((b = bitcount(usedDigits[i])) == MAXVAL - 1) { // If there's only one digit I can place in this cell, do
int d(1); // Find the digit
while ((usedDigits[i] & 1<<d) > 0)
++d;
set(i / C, i % C, d); // Fill it in
changed = true; // The board has been changed so this step must be rerun
} else if (bitcount(usedDigits[i]) == MAXVAL)
throw 666; // Speed boost
}
}
void goBruteForce() {
int max(-1); // Find the cell with the _minimum_ number of posibilities (i.e. the one with the largest number of /used/ digits)
for (uint i(0); i < S; ++i)
if (table[i] == 0) // Is there a digit already written?
if ((max == -1) || (bitcount(usedDigits[i]) > bitcount(usedDigits[max])))
max = i;
if (max != -1) {
for (uint i(1); i <= MAXVAL; ++i) // Go through each possible digit
if ((usedDigits[max] & 1<<i) == 0) { // If it can be placed in this cell, do
SudokuBoard temp(*this); // Create a new board
temp.set(max / C, max % C, i); // Complete the attempt
temp.solve(); // Solve it
if (temp.getFilledIn() == S) { // If the board was completely solved (i.e. the number of filled in cells is S)
for (uint j(0); j < S; ++j) // Copy the board into this one
set(j / C, j % C, temp.at(j / C, j % C));
return; // Break the recursive cascade
}
}
}
}
uint getFilledIn() {
return filledIn;
}
private:
uint table[S];
uint usedDigits[S];
uint filledIn;
bool const inline isValidPos(int l, int c) {
return ((0 <= l) && (l < (int)L) && (0 <= c) && (c < (int)C));
}
uint const inline findZone(uint off) {
return ((off / C / ZONEL) * (C / ZONEC) + (off % C / ZONEC));
}
uint const inline bitcount(uint x) {
uint count(0);
for (; x; ++count, x &= (x - 1));
return count;
}
};
void printB(SudokuBoard sb) {
cout << " ***** Sudoku Solver By Diego Freitas ***** " << endl;
cout << " | ------------------------------- |" << endl;
for (uint i(0); i < S; ++i) {
if (i % 3 == 0)
cout << " |";
cout << " " << sb.at(i / L, i % L);
if (i % C == C - 1) {
if (i / C % 3 == 2)
cout << " |" << endl << " | -------------------------------";
cout << " |" << endl;
}
}
cout << endl;
}
int main(int argc, char *argv[]) {
SudokuBoard sb;
ifstream fin("Sudoku.in");
int aux;
for (uint i(0); i < S; ++i) {
fin >> aux;
sb.set(i / L, i % L, aux);
}
fin.close();
printB(sb);
sb.solve();
printB(sb);
system ("PAUSE");
return 0;
}
If anyone has any alterations to the code then it will be appreciated. I am currently making this during my own time and I roughly want an idea how this works.

Simulate Matrix background

I am traying to simulate the the background of Matrix film and I wrote the code below:
#include <iostream>
#include<windows.h>
#include <conio.h>
using namespace std;
void CursorPosition(short x, short y) {
COORD coordScreen = {x, y};
SetConsoleCursorPosition(GetStdHandle(STD_OUTPUT_HANDLE), coordScreen);
}
void color(int nb) {
SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), nb);
}
int main(int argc, char *argv[]) {
unsigned int y[79];
unsigned int ymin[79];
unsigned int a, b;
char i[36] = {48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 65, 66, 67, 68, 69,
70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87,
88, 89, 90};
system("title The Matrix !");
color(10);
for (a = 0; a < 79; a++) {
y[a] = 0;
ymin[a] = 1;
}
do {
for (unsigned int x = 0; x < 79; x++) {
for (unsigned int j = 0; j < 5; j++) {
a = rand() % 79;
if (y[a] != 0) {
b = rand() % y[a] + ymin[a];
if (b < 26) {
if (b < ymin[a] + 2 & b > 2) color(2);
CursorPosition(a + 1, b);
cout << i[rand() % 36];
color(10);
}
}
}
if (y[x] != 0 || (rand() % 50) == 0) {
if ((rand() % (y[x] + 1)) < 20 & ymin[x] == 1) {
y[x]++;
} else {
CursorPosition(x + 1, ymin[x]);
cout << " ";
ymin[x]++;
if (ymin[x] < 25) {
color(2);
CursorPosition(x + 1, ymin[x] + 1);
cout << i[rand() % 36];
color(10);
}
}
if ((y[x] + ymin[x]) < 26) {
CursorPosition(x + 1, y[x] + ymin[x]);
color(15);
cout << i[rand() % 36];
color(10);
}
if ((y[x] + ymin[x]) < 27) {
CursorPosition(x + 1, y[x] + ymin[x] - 1);
cout << i[rand() % 36];
}
}
if (ymin[x] > 25) {
ymin[x] = 1;
y[x] = 0;
}j
}
Sleep(35);
} while (!kbhit());
return 0;
}
Do you have any recommendation to improve the code ?
My purpose is like this:
Your code is not writing those chars. It is writing simple alphabet chars numbers 0-9.
What you need is half-width Kana Characters:
。 「 」 、 ・ ヲ ァ ィ ゥ ェ ォ ャ ュ ョ ッ
ー ア イ ウ エ オ カ キ ク ケ コ サ シ ス セ ソ
タ チ ツ テ ト ナ ニ ヌ ネ ノ ハ ヒ フ ヘ ホ マ
ミ ム メ モ ヤ ユ ヨ ラ リ ル レ ロ ワ ン ゙ ゚
You need boost libraries' boundary part to get these i think.
Some example i found:
boost::locale::generator gen;
using namespace boost::locale::boundary;
std::string text="生きるか死ぬか、それが問題だ。";
ssegment_index map(word,text.begin(),text.end(),gen("ja_JP.UTF-8"));
for(ssegment_index::iterator it=map.begin(),e=map.end();it!=e;++it) {
std::cout << "Segment " << *it << " contains: ";
if(it->rule() & word_none)
std::cout << "white space or punctuation marks ";
if(it->rule() & word_kana)
std::cout << "kana characters ";
if(it->rule() & word_ideo)
std::cout << "ideographic characters";
std::cout<< std::endl;
}
output:
Segment 生 contains: ideographic characters
Segment きるか contains: kana characters
Segment 死 contains: ideographic characters
Segment ぬか contains: kana characters
Segment 、 contains: white space or punctuation marks
Segment それが contains: kana characters
Segment 問題 contains: ideographic characters
Segment だ contains: kana characters
Segment 。 contains: white space or punctuation marks
Source

C++ ProjectEuler #11

I'm trying to solve the 11th problem on ProjectEuler.net. The goal for this problem is to find the largest product in a 20x20 grid of 4 adjacent numbers in any direction (up, down, right, left, diagonal).
I'm using the BigInteger library because I don't know how large will the numbers be and I don't want an overflow - but I think this might be the problem. Every time I restart the program I get a different answer. :/ I also tried to use an unsigned long long int just to see what will happen - the answer remained the same.
This is the code (Nothing complicated; I just test every element in the grid to see if it has 3 adjacent numbers in any direction, compute the product and set it as the new largest if it is larger than the previous one. On the end I print the largest product.):
//NO.11
#include <iostream>
#include <BigIntegerLibrary.hh>
#include <windows.h>
int main()
{
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, 01, 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, 03, 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, 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, 4, 89, 55, 40},
{ 4, 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, 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, 04, 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}
};
BigInteger biggestProduct = 0;
//unsigned long long int biggestProduct = 0;
for (int row = 0; row < 20; row++)
{
for (int col = 0; col < 20; col++)
{
BigInteger product;
//unsigned long long int product;
//std::cout << grid[row][col] << ":" << std::endl; system("pause>nul");
//UP
if ((row-1 >= 0) && (row-2 >= 0) && (row-3 >= 0))
{
product = grid[row][col] * grid[row-1][col] * grid[row-2][col] * grid[row-3][col];
//std::cout << " U: " << grid[row][col] << "*" << grid[row-1][col] << "*" << grid[row-2][col] << "*" << grid[row-3][col] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
//DOWN
if ((row+1 >= 0) && (row+2 >= 0) && (row+3 >= 0))
{
product = grid[row][col] * grid[row+1][col] * grid[row+2][col] * grid[row+3][col];
//std::cout << " D: " << grid[row][col] << "*" << grid[row+1][col] << "*" << grid[row+2][col] << "*" << grid[row+3][col] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
//RIGHT
if ((col+1 >= 0) && (col+2 >= 0) && (col+3 >= 0))
{
product = grid[row][col] * grid[row][col+1] * grid[row][col+2] * grid[row][col+3];
//std::cout << " R: " << grid[row][col] << "*" << grid[row][col+1] << "*" << grid[row][col+2] << "*" << grid[row][col+3] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
//LEFT
if ((col-1 >= 0) && (col-2 >= 0) && (col-3 >= 0))
{
product = grid[row][col] * grid[row][col-1] * grid[row][col-2] * grid[row][col-3];
//std::cout << " L: " << grid[row][col] << "*" << grid[row][col-1] << "*" << grid[row][col-2] << "*" << grid[row][col-3] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
//UP-RIGHT
if ((row-1 >= 0) && (row-2 >= 0) && (row-3 >= 0) && (col+1 >= 0) && (col+2 >= 0) && (col+3 >= 0))
{
product = grid[row][col] * grid[row-1][col+1] * grid[row-2][col+2] * grid[row-3][col+3];
//std::cout << " U-R: " << grid[row][col] << "*" << grid[row-1][col+1] << "*" << grid[row-2][col+2] << "*" << grid[row-3][col+3] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
//DOWN-RIGHT
if ((row+1 >= 0) && (row+2 >= 0) && (row+3 >= 0) && (col+1 >= 0) && (col+2 >= 0) && (col+3 >= 0))
{
product = grid[row][col] * grid[row+1][col+1] * grid[row+2][col+2] * grid[row+3][col+3];
//std::cout << " D-R: " << grid[row][col] << "*" << grid[row+1][col+1] << "*" << grid[row+2][col+2] << "*" << grid[row+3][col+3] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
//DOWN-LEFT
if ((row+1 >= 0) && (row+2 >= 0) && (row+3 >= 0) && (col-1 >= 0) && (col-2 >= 0) && (col-3 >= 0))
{
product = grid[row][col] * grid[row+1][col-1] * grid[row+2][col-2] * grid[row+3][col-3];
//std::cout << " D-L: " << grid[row][col] << "*" << grid[row+1][col-1] << "*" << grid[row+2][col-2] << "*" << grid[row+3][col-3] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
//UP-LEFT
if ((row-1 >= 0) && (row-2 >= 0) && (row-3 >= 0) && (col-1 >= 0) && (col-2 >= 0) && (col-3 >= 0))
{
product = grid[row][col] * grid[row-1][col-1] * grid[row-2][col-2] * grid[row-3][col-3];
//std::cout << " U-L: " << grid[row][col] << "*" << grid[row-1][col-1] << "*" << grid[row-2][col-2] << "*" << grid[row-3][col-3] << "= \t" << product << std::endl; system("pause>nul");
if (product > biggestProduct)
biggestProduct = product;
}
}
}
std::cout << biggestProduct;
return 0;
}
Does anyone know what's wrong?
your problem are your range checks:
example:
// DOWN
if ((row+1 >= 0) && (row+2 >= 0) && (row+3 >= 0))
should be:
// DOWN
//if ((row+1 < 20 ) && (row+2 < 20 ) && (row+3 < 20 ))
//which still contains redundant comparisons (as pointed out by Blastfurnace),
//and thus can be can be simplified to :
if( row + 3 < 20 )
If you have random results on each run for a program, and you dont use concurrency, random number generators or something like that, most likely that randomness i caused by undefined behaviour like reading unallocated memory or reading using unintialised variables.
Looks like you're reading off the bottom and right edges:
if ((row+1 >= 0) && (row+2 >= 0) && (row+3 >= 0))
if ((col+1 >= 0) && (col+2 >= 0) && (col+3 >= 0))
and the same test with the diagonals.
You should be checking that the index doesn't reach 20.
visit this link for code
http://codingloverlavi.blogspot.in/2013/06/project-euler-solution-11.html
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
long find_max(int **);
int main()
{
FILE *f;
int **arr,i=0,j;
arr=(int **)malloc(sizeof(int*)*20);
for(i=0;i<20;i++)
arr[i]=(int *)malloc(sizeof(int)*20);
f=fopen("D:/b.txt","r");
if(f)
for(i=0;i<20;i++)
for(j=0;j<20;j++)
fscanf(f,"%d",&arr[i][j]);
else
printf("file operation not successful\n");
printf("Answer = %ld\n",find_max(arr));
system("pause");
}
long find_max(int **arr)
{
int i,j,k;
long product,max=0;
for(i=0;i<20;i++)
for(j=0;j<20;j++)
{
/*checking for horizontal sequence*/
product=1;
for(k=0;k<4;k++)
{
if(j+k<20)
{
product*=arr[i][j+k];
}
}
if(product>max)
max=product;
/*checking for vertical sequence*/
product=1;
for(k=0;k<4;k++)
{
if(i+k<20)
{
product*=arr[i+k][j];
}
}
if(product>max)
max=product;
product=1;
for(k=0;k<4;k++)
{
if( (i+k<20) && (j+k<20) )
{
product*=arr[i+k][j+k];
}
}
if(product>max)
max=product;
product=1;
for(k=0;k<4;k++)
{
if( (i+k<20) && (j-k>0) )
{
product*=arr[i+k][j-k];
}
}
if(product>max)
max=product;
}
return max;
}