POJ

题目链接：http://poj.org/problem?id=2406

题意：给出一个字符串 问它最多由多少相同的字串组成。

思路：用next数组，如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]，如果不存在那么就只能使循环节为自己。

代码：

#include <cstdio>#include <cmath>#include <iostream>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <map>#include <numeric>#include <set>#include <string>#include <cctype>#include <sstream>#define INF 0x3f3f3f3f#define INF 0x3f3f3f3f#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1using namespace std;typedef long long LL;typedef pair<LL, LL> P;const int maxn = 1e6 + 5;const int mod = 1e8 + 7;int Next[maxn];char T[maxn];int n;void getNext() {    int j, k;    j = 0;    k = -1;    Next[0] = -1;    while(j < n)        if(k == -1 || T[j] == T[k])            Next[++j] = ++k;        else            k = Next[k];}int main() {    //freopen ("in.txt", "r", stdin);    while (~scanf ("%s", T)) {        if (T[0]=='.') break;        n = strlen(T);        getNext();        if (n % (n - Next[n]) == 0)            printf ("%d\n",n / (n - Next[n]));        else printf ("1\n");    }    return 0;}