如果 k 是字符串的循环节,
/**********************jibancanyang************************** *Author* :jibancanyang *Created Time* : 六 5/ 7 21:17:06 2016 *File Name* : poj2406.cpp **Problem**:扩展kmp **Analyse**: **Code**: ***********************[email protected]**********************/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
vector<int> vi;
#define pr(x) cout << #x << ": " << x << " "
#define pl(x) cout << #x << ": " << x << endl;
#define pri(a) printf("%d\n",(a));
#define xx first
#define yy second
#define sa(n) scanf("%d", &(n))
#define sal(n) scanf("%lld", &(n))
#define sai(n) scanf("%I64d", &(n))
#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++)
const int mod = int(1e9) + 7, INF = 0x3fffffff;
const int maxn = 1e6 + 13;
char t[maxn], s[maxn];
int nexts[maxn], extend[maxn], slen, tlen;
//计算next数组,保存到next参数中。
void getextendnexts(void)
{
tlen = slen = strlen(s);
for (int i = 1, j = -1, a, p; i < tlen; i++, j--)
if (j < 0 || i + nexts[i - a] >= p) {
if (j < 0) j = 0, p = i;
while (p < tlen && t[j] == t[p]) j++, p++;
nexts[i] = j, a = i;
}
else nexts[i] = nexts[i - a];
}
void getextend (void)
{
getextendnexts(); //计算next数组。mx是表示最大长度的常数。
for (int i = 0, j = -1, a, p; i < slen; i++, j--)
if (j < 0 || i + nexts[i - a] >= p) {
if (j < 0) j = 0, p = i;
while (p < slen && j < tlen && s[p] == t[j]) j++, p++;
extend[i] = j, a = i;
}
else extend[i] = nexts[i - a];
}
int main(void)
{
#ifdef LOCAL
// freopen("/Users/zhaoyang/in.txt", "r", stdin);
//freopen("/Users/zhaoyang/out.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
while (true) {
scanf("%s", s);
if (s[0] == '.' && s[1] == '\0') break;
memcpy(t, s, sizeof(s));
getextend();
int ans = slen;
for (int i = 1; i < slen; i++) {
if (extend[i] == slen - i && slen % i == 0) {
ans = i;
break;
}
}
pri(slen / ans);
}
return 0;
}