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//經(jīng)典矩陣問題是利用數(shù)字生成一個矩陣,而該矩陣剛好是一個正方形,該矩陣內(nèi)的數(shù)字是有 //規(guī)律的排序而形成矩陣。比較常見有以下形式 //1. // 1 2 9 10 // 4 3 8 11 // 5 6 7 12 // 16 15 14 13 //2. // 1 2 6 7 // 3 5 8 13 // 4 9 12 14 // 10 11 15 16 //3. // 1 2 3 4 // 12 13 14 5 // 11 16 15 6 // 10 9 8 7 //4. // 7 6 5 16 // 8 1 4 15 // 9 2 3 14 // 10 11 12 13 #include<iostream> using namespace std; const int N = 5;//可修改 int a[N][N]; void Print(int n) { for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) printf("%-2d ",a[i][j]); cout << endl; } cout << endl; } void Rectangle1() { int i = 0, j = 0; int lap = 1, m = 1, n; //cout<< "please input n行數(shù)"; //cin >> n; n = 4; a[i][j] = m++; lap++; while (lap <= n) { if (lap % 2 == 0) { for (j++; i < lap; i++) a[i][j] = m++; i--; for (j--; j >= 0; j--) a[i][j] = m++; j++; } else { for (i++; j < lap; j++) a[i][j] = m++; j--; for (i--; i >= 0; i--) a[i][j] = m++; i++; } lap++; } Print(n); } void Rectangle2() { int i = 0, j = 0; int s = 1, nNum = 1;//s標(biāo)記升降方向,斜向上為升s=1,斜向下為降s=-1 int n = 4; while (1) { if (s == 1) { a[i][j] = nNum; if (i - 1 < 0) { if (j + 1 == n) i++; else j++; s = -1; } else if (j + 1 == n) { i++; s = -1; } else { i--; j++; } } else { a[i][j] = nNum; if (j - 1 < 0) { if (j + 1 == n) j++; else i++; s = 1; } else if (i + 1 == n) { j++; s = 1; } else { i++; j--; } } nNum++; if (nNum > n*n) break; } Print(n); } void Rectangle3() { int i = 0, j = 0; int n = 4; int x1=0, x2=n, y1=0, y2=n;//分別表示上下左右 int m = 1, s = 1;//s=1表示升序,s=-1表示降序 while (1) { if (s == 1) { for (j; j < y2; ++j) a[i][j] = m++; j--; i++; y2--; for (i; i < x2; ++i) a[i][j] = m++; i--; j--; x2--; s = -1; } else { for (j; j >=y1; --j) a[i][j] = m++; j++; i--; y1++; for (i; i>=x1+1; --i) a[i][j] = m++; i++; j++; x1++; s = 1; } if (m > n*n) break; } Print(n); } void Rectangle4() { int i = 0, j = 0; int n = 4; int m = n*n; int x1 = 0, y1 = 0, x2 = n, y2 = n; int s = 1; if (n % 2 == 0) { j = n - 1; y2 = n - 1; s = 1; } else { i = n - 1; y1 = 1; s = -1; } while (1) { if (s == 1) { for (i; i < x2; ++i) a[i][j] = m--; i--; j--; x2--; for (j; j >= y1; --j) a[i][j] = m--; j++; i--; y1++; s = -1; } else { for (i; i >=x1; --i) a[i][j] = m--; i++; j++; x1++; for (j; j<y2; ++j) a[i][j] = m--; j--; i++; y2--; s = 1; } if (m < 1) break; } Print(n); } int main() { Rectangle1(); Rectangle2(); Rectangle3(); Rectangle4(); system("pause"); return 0; }
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