https://vjudge.net/problem/UESTC-1703
题意:略
思路:
枚举+字符串hash。
ans从1到len开始枚举字符串的长度,然后就依次比较各段长度为ans的字符串的hash值是否和hash(0,ans)的hash值相等。对于剩余的长度为tlen小于长度为ans的字符串,比较它的hash和hash(0,tlen)。一旦遇到一个满足条件的ans,直接跳出,就可以保证这个是最小的了。
代码:
1 #include <stdio.h> 2 #include <string.h> 3 4 char s[1000005]; 5 6 unsigned int mha(int st,int len) 7 { 8 unsigned int sed = 131; 9 unsigned int h = 0; 10 11 for (int i = 0;i < len;i++) 12 { 13 h = h * sed + s[st+i]; 14 } 15 16 return (h & 0x7FFFFFFF); 17 } 18 19 int main() 20 { 21 scanf("%s",s); 22 23 int len = strlen(s); 24 25 int ans; 26 27 for (int i = 1;i <= len;i++) 28 { 29 bool f = 0; 30 31 int code = mha(0,i); 32 33 for (int j = 0;j + i < len;j += i) 34 { 35 int tmp = mha(j,i); 36 37 if (tmp != code) 38 { 39 f = 1; 40 break; 41 } 42 } 43 44 int re = len % i; 45 46 code = mha(0,re); 47 48 int tmp = mha(len-re,re); 49 50 if (tmp != code) f = 1; 51 52 if (!f) 53 { 54 ans = i; 55 break; 56 } 57 } 58 59 printf("%d\n",ans); 60 61 return 0; 62 }
原文:http://www.cnblogs.com/kickit/p/7237968.html