示例
Input: 5Output: X X O X O X X O X O X X O X O X X
整个问题包含3个不同的分区 −
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对于奇数n，打印上半部分的n-1行，对于偶数n，打印上半部分的n-2行。
打印中间行，对于奇数n，打印1行，对于偶数n，打印3行。
打印下半部分，对于奇数n，打印n-1行，对于偶数n，打印n-2行。
算法
START STEP 1: IF (n % 2 == 0) then x = x - 1; Define p as n – 1 for spaces Define s = 1 for characters STEP 2: LOOP FOR i= 1 AND i <= (x - 1) / 2 AND i++ LOOP FOR j = 1 AND j <= p AND j++ PRINT SPACE END LOOP IF i % 2 != 0 then GOTO STEP 11 Printx(s) ELSE GOTO STEP 12 Printo(s) END PRINT NEW LINE INCREMENT p BY 1 LOOP FOR j = 1 AND j <= p AND j++ PRINT SPACE END LOOP IF i % 2 != 0 then, GOTO STEP 11 Printx(s) ELSE GOTO STEP 12 Printo(s) END IF PRINT NEWLINE DECREMENT p BY 1 INCREMENT s BY 1 END LOOP STEP 3: IF n % 2 == 0 { LOOP FOR i = 1 AND i <= p AND i++ PRINT SPACE IF n % 4 != 0 then, GOTO STEP 11 Printx(n/2) ELSE GOTO STEP 12 Printo(n/2) END IF PRINT SPACE END IF STEP 4: IF n % 2 != 0 then, GOTO STEP ELSE IF n % 4 != 0 then, DIVIDE n BY 2 AND GOTO STEP 11 Printx(n/2) and Printx(n/2) ELSE DIVIDE n BY 2 AND GOTO STEP 11 Printx(n/2) and Printo(n/2) DIVIDE n BY 2 AND GOTO STEP END IF END IF PRINT NEW LINE STEP 5: IF n % 2 == 0 then, { PRINT SPACE DIVIDE n BY 2 AND GOTO STEP 11 Printx(n/2) PRINT NEWLINE END IF STEP 6: SET p = 1 STEP 7: IF n % 2 == 0 DECREMENT x BY 1 SET p = 2 END IF STEP 8: SET q = x / 2 STEP 9: LOOP FOR i = 1 AND i <= x AND i++) { LOOP FOR j = 1 AND j <= p AND j++ PRINT SPACE END FOR PASS q TO STEP 11 Printx(q) IF i % 2 == 0 THEN DECREMENT q BY 1 END IF PRINT NEWLINE INCREMENT p BY 1 END FOR STEP 10: PRINT NEWLINE STEP 11: Printx(n) LOOP FOR i = 1 AND i< = n AND i++ IF i % 2 != 0 then, PRINT x ELSE PRINT o END IF END FOR STEP 12: Printo(n) LOOP FOR i = 1 AND i< = n AND i++ IF i % 2 != 0 then, PRINT o ELSE PRINT xSTOP
示例
#include // printing alternate x o starting with xint printx(int n) { int i; for ( i = 1; i < = n; i++) { if (i % 2! = 0) printf("x "); else printf("o "); } return 0;}// printing alternate x o starting with oint printo(int n) { int i; for ( i = 1; i < = n; i++) { if (i % 2! = 0) printf("o "); else printf("x "); } return 0;}// To print the pattern for nint printpattern(int n) { // upper half // n-1 lines for odd, n-2 lines for even int x = n; int i, j; if (n % 2 == 0) x = x - 1; // number of spaces to leave in each line int p = n - 1; // number of characters in each line int s = 1; // prints double lines in each iteration for ( i = 1; i < = (x - 1) / 2; i++) { for ( j = 1; j < = p; j++) { printf(" "); } if (i % 2! = 0) printx(s); else printo(s); printf(""); p++; for ( j = 1; j < = p; j++) printf(" "); if (i % 2! = 0) printx(s); else printo(s); printf("
"); p--; s++; } // extra upper middle for even if (n % 2 == 0) { for ( i = 1; i < = p; i++) printf(" "); if (n % 4! = 0) printx(n / 2); else printo(n / 2); printf("
"); } // middle line if (n % 2! = 0) printx(n); else { if (n % 4! = 0) { printx(n / 2); printx(n / 2); } else { printx(n / 2); printo(n / 2); } } printf("
"); // extra lower middle for even if (n % 2 == 0) { printf(" "); printx(n / 2); printf("
"); } // lower half p = 1; if (n % 2 == 0) { x--; p = 2; } int q = x / 2; // one line for each iteration for ( i = 1; i < = x; i++) { for (int j = 1; j < = p; j++) printf(" "); printx(q); if (i % 2 == 0) q--; printf("
"); p++; } printf("
"); return 1;}int main() { int n = 5; printpattern(n); return 0;}
输出
如果我们运行上面的程序，它将生成以下输出。
X X O X O X X O X O X X O X O X X
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