Hduoj2137【水题】

来源：互联网发布：写真机蒙泰软件编辑：程序博客网时间：2024/05/29 04:16

/*circumgyrate the string Time Limit : 10000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 2   Accepted Submission(s) : 1Font: Times New Roman | Verdana | Georgia Font Size: ← →Problem Description  Give you a string, just circumgyrate. The number N means you just   circumgyrate the string N times, and each time you circumgyrate the   string for 45 degree anticlockwise.Input  In each case there is string and a integer N. And the length of the string is always odd, so the center of the string will not be changed,   and the string is always horizontal at the beginning. The length of the string will not exceed 80, so we can see the complete result on the   screen.Output  For each case, print the circumgrated string.Sample Inputasdfass 7Sample Outputa s  d   f    a     s      sAuthorwangye SourceHDU 2007-11 Programming Contest_WarmUp */#include<stdio.h>#include<string.h>int main(){char s[84];int i, j, k, n, l;while(scanf("%s %d", s, &n) != EOF){n %= 8;if(n < 0){n += 8;}l = strlen(s);if(n == 1){for(i = 0; i < l; ++i){for(j = 0; j < l-1-i; ++j)printf(" ");printf("%c\n", s[l-1-i]);}}else if(n == 2){for(i = 0; i < l; ++i){for(j = 0; j < l/2; ++ j)printf(" ");printf("%c\n",s[l-1-i]);}}else if(n == 3){for(i = 0; i < l; ++i){for(j = 0; j < i; ++j)printf(" ");printf("%c\n", s[l-1-i]);}}else if(n == 4){for(i = 0; i < l; ++i)printf("%c", s[l-1-i]);printf("\n");}else if(n == 5){for(i = 0; i < l; ++i){for(j = 0; j < l-1-i; ++j)printf(" ");printf("%c\n", s[i]);}}else if(n == 6){for(i = 0; i < l; ++i){for(j = 0; j < l/2; ++j)printf(" ");printf("%c\n",s[i]);}}else if(n == 7){for(i = 0; i < l; ++i){for(j = 0; j < i; ++j)printf(" ");printf("%c\n", s[i]);}}else{printf("%s\n",s);}}return 0;}

题意：给定一个字符串，长度为奇数，并给出一个n，以字符串中心为旋转原点旋转，每次旋转45°，输出n次旋转后的字符串。

注意：

1.N可以是负数2.当横向输出时，不用输出上下的N个空格和换行，当竖直输出的时候，应该输出N行左边的空格3.所有输出字符的后面没有空格

0 0