3*3 matrix, the value is limited to 1-9 is not repeated, known as the number of and, and an oblique value, to find the matrix, what is the fast way?



x x x 11
x x x 15
x x x 19

16 14 15 15

Discussion stickers:

http://bbs.csdn.net/topics/391816265

The exact matching method of three rows is first obtained:

Puzzl.cpp: Defines the entry point of the console application. #include "stdafx.h"//Puzzle.cpp:Defines the entry point for the console application.//#include "stdafx.h" #include &lt ;vector> #include <set> #include <iostream>using namespace Std;int result[3][3] = {0};bool check (int a, int b, int c, int sum) {if ((A + B + c) = = sum) return true; else return false;} int lastvalue (int a, int b,int sum) {return (sum-a-B);} Min maxvector<vector<int> > Zuhe (int min,int max,int sum) {vector<vector<int> > Quanji; for ( int i = Min;i < Max;   ++i) {for (int j = 1;j < Max; ++j) {if (i + J > sum) continue;      else {for (int k = 1; k <max; ++k) {if (check (i,j,k,sum)) {vector<int> Ziji;      Ziji.push_back (i);      Ziji.push_back (j);      Ziji.push_back (k);      if (i!=j&&j!=k&&i!=k) {quanji.push_back (Ziji);     }} else {continue; }}}}} return Quanji;} Vector<vector<int> > Filter_zuhe (int key,vector<vector<int> > Temp) {vector<vector<int> >::iterator ITER; for (iter = Temp.begin (); Iter!=temp.end ();)  {if (*iter) [0]! = key) iter = Temp.erase (ITER); else ITER + +; } return temp;} BOOL IsOk () {set<int> myset;for (int i = 0;i <3; ++i) {for (int j = 0;j<3;++j) {if (result[i][j]>0&& RESULT[I][J]&LT;10) {Myset.insert (result[i][j]);}} if (Myset.size ()!=9) {return false;} Else{if (Myset.size () ==9) {for (int i = 0;i <3; ++i) {cout <<endl;for (int j = 0;j<3;++j) {cout<<result[i][ j]<< "";}} return true;}}} void Dayin () {for (int i = 0;i <3; ++i) {cout <<endl;for (int j = 0;j<3;++j) {cout<<result[i][j]<< ""; }}cout<<endl;} void Qingling () {for (int i = 0;i <3; ++i) {for (int j = 0;j<3;++j) {(result[i][j]=0);}}} int _tmain (int argc, _tchar* argv[]) {vector<vector<int> > Quanjiheng = Zuhe (1,9,11); vector<vector< int> > Quanjishu = Zuhe (1,9,16); Vector<vector<int> > Quanjixie = Zuhe (1,9,15); int Sizeheng = Quanjiheng.size (); int Sizeshu = Quanjishu.size (); int sizexie = Quanjixie.size (); int key = 0,last_key = 0; vector<vector<int> > Quanjishufilter; vector<vector<int> > Quanjixiefilter;   for (int i = 0;i < Sizeheng; ++i) {key = Quanjiheng[i][0];   if (key!=last_key) {quanjishufilter = Filter_zuhe (Key,quanjishu); quanjixiefilter = Filter_zuhe (Key,quanjixie);   } Last_key = key;    Assign a value for (int j = 0;j< 3;++j) {result[0][j] = quanjiheng[i][j];  } int sizeshu = Quanjishufilter.size ();  for (int k = 0; k< sizeshu;++k) {//Assign value to vertical lines for (int j = 0;j<3;++j) {result[j][0] = quanjishufilter[k][j];  } int sizexie = Quanjixiefilter.size ();  for (int x = 0, x < Sizexie; ++x) {for (int j = 0; j < 3; ++j) {result[j][j] = quanjixiefilter[x][j]; }RESULT[2][1] = 14-result[0][1]-result[1][1];result[1][2] = 15-result[1][0]-result[1][1];if (IsOk ()) {//dayin (); GetChar ();return 0;}   else{}}} qingling (); } getchar (); System ("pause"); return 0;}

Methods for arranging combinations:

Template <typename t>void swap (t* array, unsigned int i, unsigned int j) {T t = array[i];array[i] = Array[j];array[j] = t;} void FullArray (int* array, size_t array_size, unsigned int index) {if (index >= array_size) {if ((Array[0] + array[1] + a RRAY[2] = = one) && (Array[3] + array[4] + array[5] = =) && (array[6] + array[7] + array[8] = =) &&amp ; (Array[0] + array[3] + array[6] = = && (array[1] + array[4] + array[7] = =) && (array[2] + array[5] + a RRAY[8] = = && (array[0] + array[4] + array[8] = =)) {printf ("%d,%d,%d\n", Array[0], array[1], array[2]);p rintf ("%d,%d,%d\n", Array[3], array[4], array[5]);p rintf ("%d,%d,%d\n\n", Array[6], array[7], array[8]);} return;} for (unsigned int i = index; i < array_size; ++i) {swap (array, I, index); FullArray (array, array_size, index + 1); swap (array, I, index);}} int main (int argc, char* argv[]) {int value[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; FullArray (value, 9, 0);}

Imitate the person to do the method, arbitrarily fill in 3 empty space, guarantee not to walk different column, or not a slash. We can figure out the rest.

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3*3 matrix, the value is limited to 1-9 is not repeated, known as the number of and, and an oblique value, to find the matrix, what is the fast way?