acm--Xiang da oj 1086--diamond--water

Xiang da OJ Address: http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1086

Diamond Time limit: +Ms | Memory Limit:65536Kb

Description

Xiao Ming is now a student of the 10-degree computer Department of Xiangtan University. His C language is still available, but today the teacher assigned a class after the homework problem but he was baffled. The title means to print a diamond and the diamond diagonal in a n*n (N-odd) diamond Square with the character ' * ', which is filled with the character ' _ ' In addition to the ' * ' character in the diamond. Figure I and figure two are the patterns of n=7 and n=9 respectively. Xiao Ming is a competitive person, he does not want to ask other students in the class, so he secretly run to help you, I hope you can help him.

                        &NBS P                                                                

Figure A figure II

Input

The input contains multiple samples, each of which is an odd n (5 <= N < 50). When n is 0, the input ends, and this 0 does not need to be processed.

Output

for each sample, the diamond pattern is printed with the character ' * ' in the square of the n*n. (Note: The remainder of the diamond is filled with the ' _ ' character except for the character ' * ')

Sample Input

5

7

0

Sample Output

__*__

_***_

*****

_***_

__*__

___*___

__***__

_*_*_*_

*******

_*_*_*_

__***__

___*___

#include <stdio.h> #include <iostream>using namespace std;/* used to print diamond */void ha (int n) {int i,j,b;    int index=1;    int temp;    int flag=1;        Header for (i=1;i<=n;i++) {if (i!=n/2+1) {printf ("_");        }else{printf ("*");        }} printf ("\ n");        For (i=1;i<= (n-3)/2;i++) {temp= (n-3-(i-1) *)/2;        for (j=1;j<=temp;j++) {printf ("_");        } printf ("*");        for (b=1;b<i;b++) {printf ("_");            } printf ("*");        for (b=1;b<i;b++) {printf ("_");                } printf ("*");        for (j=1;j<=temp;j++) {printf ("_");    } printf ("\ n");    } for (i=1;i<=n;i++) {printf ("*");        } printf ("\ n"); for (i= (n-3)/2;i>=1;i--) {temp= (n-3-(i-1) *)/2;        for (j=1;j<=temp;j++) {printf ("_");        } printf ("*");        for (b=1;b<i;b++) {printf ("_");       }     printf ("*");        for (b=1;b<i;b++) {printf ("_");                } printf ("*");        for (j=1;j<=temp;j++) {printf ("_");    } printf ("\ n");        }//tail for (i=1;i<=n;i++) {if (i!=n/2+1) {printf ("_");        }else{printf ("*"); }} printf ("\ n");}    int main () {int n;            while (scanf ("%d", &n) &&n!=0) {ha (n); }}

acm--Xiang da oj 1086--diamond--water