形成矩阵线圈 · GeeksForGeeks 中文系列教程

# 形成矩阵线圈 > 原文： [https://www.geeksforgeeks.org/form-coils-matrix/](https://www.geeksforgeeks.org/form-coils-matrix/) 给定一个正整数 n，该整数表示 4n x 4n 矩阵的维，值从 1 到 n 从左到右，从上到下填充。从矩阵形成两个线圈，然后打印线圈。例子： ``` Input : n = 1; Output : Coil 1 : 10 6 2 3 4 8 12 16 Coil 2 : 7 11 15 14 13 9 5 1 Explanation : Matrix is 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Input : n = 2; Output : Coil 1 : 36 28 20 21 22 30 38 46 54 53 52 51 50 42 34 26 18 10 2 3 4 5 6 7 8 16 24 32 40 48 56 64 Coil 2 : 29 37 45 44 43 35 27 19 11 12 13 14 15 23 31 39 47 55 63 62 61 60 59 58 57 49 41 33 25 17 9 1 ``` 矩阵中的元素总数为 16n <sup>2</sup> 。所有元件均分为两个线圈。每个线圈具有 8n <sup>2</sup> 个元素。我们制作两个这样大小的数组。我们首先以给定的顺序遍历来填充线圈 1 中的元素。一旦我们在线圈 1 中填充了元素，就可以使用公式 coil2 [i] = 16 * n * n +1 -coil1 [i]获得其他线圈 2 的元素。 ## C++ ```cpp // C++ program to print 2 coils of a // 4n x 4n matrix. #include<iostream> using namespace std; // Print coils in a matrix of size 4n x 4n void printCoils(int n) { // Number of elements in each coil int m = 8*n*n; // Let us fill elements in coil 1\. int coil1[m]; // First element of coil1 // 4*n*2*n + 2*n; coil1[0] = 8*n*n + 2*n; int curr = coil1[0]; int nflg = 1, step = 2; // Fill remaining m-1 elements in coil1[] int index = 1; while (index < m) { // Fill elements of current step from // down to up for (int i=0; i<step; i++) { // Next element from current element curr = coil1[index++] = (curr - 4*n*nflg); if (index >= m) break; } if (index >= m) break; // Fill elements of current step from // up to down. for (int i=0; i<step; i++) { curr = coil1[index++] = curr + nflg; if (index >= m) break; } nflg = nflg*(-1); step += 2; } /* get coil2 from coil1 */ int coil2[m]; for (int i=0; i<8*n*n; i++) coil2[i] = 16*n*n + 1 -coil1[i]; // Print both coils cout << "Coil 1 : "; for(int i=0; i<8*n*n; i++) cout << coil1[i] << " "; cout << "\nCoil 2 : "; for (int i=0; i<8*n*n; i++) cout << coil2[i] << " "; } // Driver code int main() { int n = 1; printCoils(n); return 0; } ``` ## Java ```java // Java program to print 2 coils // of a 4n x 4n matrix. class GFG { // Print coils in a matrix of size 4n x 4n static void printCoils(int n) { // Number of elements in each coil int m = 8 * n * n; // Let us fill elements in coil 1\. int coil1[] = new int[m]; // First element of coil1 // 4*n*2*n + 2*n; coil1[0] = 8 * n * n + 2 * n; int curr = coil1[0]; int nflg = 1, step = 2; // Fill remaining m-1 elements in coil1[] int index = 1; while (index < m) { // Fill elements of current step from // down to up for (int i = 0; i < step; i++) { // Next element from current element curr = coil1[index++] = (curr - 4 * n * nflg); if (index >= m) break; } if (index >= m) break; // Fill elements of current step from // up to down. for (int i = 0; i < step; i++) { curr = coil1[index++] = curr + nflg; if (index >= m) break; } nflg = nflg * (-1); step += 2; } /* get coil2 from coil1 */ int coil2[] = new int[m]; for (int i = 0; i < 8 * n * n; i++) coil2[i] = 16 * n * n + 1 - coil1[i]; // Print both coils System.out.print("Coil 1 : "); for (int i = 0; i < 8 * n * n; i++) System.out.print(coil1[i] + " "); System.out.print("\nCoil 2 : "); for (int i = 0; i < 8 * n * n; i++) System.out.print(coil2[i] + " "); } // Driver code public static void main(String[] args) { int n = 1; printCoils(n); } } // This code is contributed by Anant Agarwal. ``` ## Python3 ```py # Python3 program to pr2 coils of a # 4n x 4n matrix. # Prcoils in a matrix of size 4n x 4n def printCoils(n): # Number of elements in each coil m = 8*n*n # Let us fill elements in coil 1\. coil1 = [0]*m # First element of coil1 # 4*n*2*n + 2*n coil1[0] = 8*n*n + 2*n curr = coil1[0] nflg = 1 step = 2 # Fill remaining m-1 elements in coil1[] index = 1 while (index < m): # Fill elements of current step from # down to up for i in range(step): # Next element from current element curr = coil1[index] = (curr - 4*n*nflg) index += 1 if (index >= m): break if (index >= m): break # Fill elements of current step from # up to down. for i in range(step): curr = coil1[index] = curr + nflg index += 1 if (index >= m): break nflg = nflg*(-1) step += 2 #get coil2 from coil1 */ coil2 = [0]*m i = 0 while(i < 8*n*n): coil2[i] = 16*n*n + 1 -coil1[i] i += 1 # Prboth coils print("Coil 1 :", end = " ") i = 0 while(i < 8*n*n): print(coil1[i], end = " ") i += 1 print("\nCoil 2 :", end = " ") i = 0 while(i < 8*n*n): print(coil2[i], end = " ") i += 1 # Driver code n = 1 printCoils(n) # This code is contributed by shubhamsingh10 ``` ## C# ```cs // C# program to print 2 coils // of a 4n x 4n matrix. using System; class GFG { // Print coils in a matrix of // size 4n x 4n static void printCoils(int n) { // Number of elements in // each coil int m = 8 * n * n; // Let us fill elements in // coil 1\. int [] coil1 = new int[m]; // First element of coil1 // 4*n*2*n + 2*n; coil1[0] = 8 * n * n + 2 * n; int curr = coil1[0]; int nflg = 1, step = 2; // Fill remaining m-1 elements // in coil1[] int index = 1; while (index < m) { // Fill elements of current // step from down to up for (int i = 0; i < step; i++) { // Next element from // current element curr = coil1[index++] = (curr - 4 * n * nflg); if (index >= m) break; } if (index >= m) break; // Fill elements of current step // from up to down. for (int i = 0; i < step; i++) { curr = coil1[index++] = curr + nflg; if (index >= m) break; } nflg = nflg * (-1); step += 2; } /* get coil2 from coil1 */ int [] coil2 = new int[m]; for (int i = 0; i < 8 * n * n; i++) coil2[i] = 16 * n * n + 1 - coil1[i]; // Print both coils Console.Write("Coil 1 : "); for (int i = 0; i < 8 * n * n; i++) Console.Write(coil1[i] + " "); Console.Write("\nCoil 2 : "); for (int i = 0; i < 8 * n * n; i++) Console.Write(coil2[i] + " "); } // Driver code public static void Main() { int n = 1; printCoils(n); } } // This code is contributed by KRV. ``` ## PHP ```php <?php // PHP program to print 2 coils of a // 4n x 4n matrix. // Print coils in a matrix of size 4n x 4n function printCoils( $n) { // Number of elements in each coil $m = 8 * $n * $n; // Let us fill elements in coil 1\. $coil1 = array(); // First element of coil1 // 4*n*2*n + 2*n; $coil1[0] = 8 * $n * $n + 2 * $n; $curr = $coil1[0]; $nflg = 1; $step = 2; // Fill remaining m-1 elements in coil1[] $index = 1; while ($index < $m) { // Fill elements of current step from // down to up for ( $i = 0; $i < $step; $i++) { // Next element from current element $curr = $coil1[$index++] = ($curr - 4*$n*$nflg); if ($index >= $m) break; } if ($index >= $m) break; // Fill elements of current step from // up to down. for ( $i=0; $i<$step; $i++) { $curr = $coil1[$index++] = $curr + $nflg; if ($index >= $m) break; } $nflg = $nflg * (-1); $step += 2; } /* get coil2 from coil1 */ $coil2 = array(); for ( $i = 0; $i < 8 * $n * $n; $i++) $coil2[$i] = 16 * $n * $n + 1 -$coil1[$i]; // Print both coils echo "Coil 1 : "; for( $i = 0; $i < 8 * $n * $n; $i++) echo $coil1[$i] , " "; echo "\nCoil 2 : "; for ( $i = 0; $i < 8 * $n * $n; $i++) echo $coil2[$i] , " "; } // Driver code $n = 1; printCoils($n); // This code is contributed by anuj_67\. ?> ``` Output: ``` Coil 1 : 10 6 2 3 4 8 12 16 Coil 2 : 7 11 15 14 13 9 5 1 ``` 在**中询问 Yahoo**