poj 2406 KMP

题意:给定一个串,求此串可由最多几个子串连接而成.

思路:套用模板next函数,若该串可由某子串连接而成,len(s)-next[len] 为子串长度(未能证明,自己写了几个例子,感觉是对的,相当不严谨)

代码:

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAX = 1000009;int next[MAX];char s[MAX];int len;void next1(){int i=0, j=-1;next[0] = -1;while(i<len){if(j==-1 || s[i]==s[j]){++i;++j;next[i] = j;}elsej = next[j];}}int main(void){while(scanf("%s",s)!=EOF){if(strcmp(s,".")==0)break;len = strlen(s);next1();int ans = 1;if(len%(len-next[len])==0)ans = len/(len-next[len]);printf("%d/n",ans);}return 0;}