51nod 1347 旋转字符串

S[0...n-1]是一个长度为n的字符串，定义旋转函数Left(S)=S[1…n-1]+S[0].比如S=”abcd”,Left(S)=”bcda”.一个串是对串当且仅当这个串长度为偶数，前半段和后半段一样。比如”abcabc”是对串,”aabbcc”则不是。现在问题是给定一个字符串，判断他是否可以由一个对串旋转任意次得到。Input第1行：给出一个字符串（字符串非空串，只包含小写字母，长度不超过1000000）Output对于每个测试用例，输出结果占一行，如果能，输出YES，否则输出NO。Input示例aaabOutput示例YESNO虽然是道很简单的题，，然而搞得我好忧伤。。自己的解题方法是挺笨的吧#include "stdio.h"#include "string.h"char str[20000012];bool is_judge(char *ptr, int num){for (int i = 0; i < num; ++i){if ((ptr[i] >= 65 && ptr[i] <= 90) || ptr[i] == 32){return false;}}return true;}bool is_RotateString(char *p, int num, int length){int count = 0, i;for (i = 0; i < (num - length) / 2; ++i){if (p[i] == p[(num - length) / 2 + length + i]){count++;}else{break;}}if (count == (num - length) / 2){return true;}else{return false;}}int main(){int num, i;bool flag = false;memset(str, 0, sizeof(str));scanf("%s", str);num = strlen(str);if (0 == num % 2 && is_judge(str, num)){if (is_RotateString(str, num, 0)){flag = 1;}else{for (i = 0; i < num - 1; i++){str[num + i] = str[i];if (is_RotateString(str + 1 + i, num + i, i)){flag = true;break;}}}}if (flag){printf("YES\n");}else{printf("NO\n");}return 0;}