题目链接
题意
对于任意的字符串，定义它的 重复次数 为：它最多可被划分成的完全相同的子串个数。例如：ababab 的重复次数为3，ababa 的重复次数为1.
现给定一字符串，求它的一个子串，其重复次数取到最大值，且字典序取到最小值。
思路
参考 hzwer.
首先，重复次数显然至少为\(1\)，所以下面只考虑重复次数\(\geq 2\)的情况。
首先枚举子串长度\(L\)。对于长度为\(L\)的子串，其重复次数至少为\(2\)，意味着它的其中两个重复部分必为\(s[0],s[L],s[2L],...\)中相邻的两个。
所以只需枚举\(i\)，看\(s[i*L]\)和\(s[(i+1)*L]\)向左\((le)\)向右\((ri)\)最多能匹配到多远，记总长度为\(k\)，则重复次数\(=k/L+1\).
至于开头位置，则落在\(i*L-le\)和\(i*L-le+(le+ri)\%k\)之间，取字典序最小者。
对于所有重复次数相同的子串，要使其字典序取到最小值，即取\(rank\)值最小者。
上述步骤中，看最长匹配多远（即看最长公共前缀）与取rank最小值均借助\(ST\)表实现。
Code
#include <stdio.h>
#include <iostream>
#include <string.h>
#define maxn 100010
using namespace std;
typedef long long LL;
int wa[maxn], wb[maxn], wv[maxn], wt[maxn], r1[maxn], r2[maxn],
h1[maxn], h2[maxn], rk1[maxn], rk2[maxn], sa[maxn], n, kas, bin[20], Log[maxn];
struct rmqNode { int val, p; } mn[maxn][20];
struct node { int val, pos, len; };
int mn1[maxn][20], mn2[maxn][20];
char s[maxn];
void rmqInit(int n) {
Log[0] = -1; bin[0] = 1;
for (int i = 1; i < 20; ++i) bin[i] = bin[i-1] << 1;
for (int i = 1; i <= n; ++i) Log[i] = Log[i>>1] + 1;
}
bool cmp(int* r, int a, int b, int l) { return r[a] == r[b] && r[a+l] == r[b+l]; }
void init(int* r, int* rk, int* h, int n, int m) {
int* x=wa, *y=wb, *t, i, j, p;
for (i = 0; i < m; ++i) wt[i] = 0;
for (i = 0; i < n; ++i) ++wt[x[i] = r[i]];
for (i = 1; i < m; ++i) wt[i] += wt[i - 1];
for (i = n-1; i >= 0; --i) sa[--wt[x[i]]] = i;
for (j = 1, p = 1; p < n; j <<= 1, m = p) {
for (p = 0, i = n-j; i < n; ++i) y[p++] = i;
for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
for (i = 0; i < n; ++i) wv[i] = x[y[i]];
for (i = 0; i < m; ++i) wt[i] = 0;
for (i = 0; i < n; ++i) ++wt[wv[i]];
for (i = 1; i < m; ++i) wt[i] += wt[i - 1];
for (i = n-1; i >= 0; --i) sa[--wt[wv[i]]] = y[i];
t = x, x = y, y = t, x[sa[0]] = 0;
for (p = 1, i = 1; i < n; ++i) x[sa[i]] = cmp(y, sa[i], sa[i-1], j) ? p - 1 : p++;
}
for (i = 0; i < n; ++i) rk[sa[i]] = i;
int k = 0;
for (i = 0; i < n - 1; h[rk[i++]] = k) {
for (k = k ? --k : 0, j = sa[rk[i] - 1]; r[i+k] == r[j+k]; ++k);
}
}
void rmq1(int* a, int (*mn)[20], int n) {
for (int i = 1; i <= n; ++i) mn[i][0] = a[i];
for (int j = 1; bin[j] <= n; ++j) {
for (int i = 1; i+bin[j-1]-1 <= n; ++i) {
mn[i][j] = min(mn[i][j-1], mn[i+bin[j-1]][j-1]);
}
}
}
void rmq2(int* a, rmqNode (*mn)[20], int n) {
for (int i = 1; i <= n; ++i) mn[i][0] = {a[i-1], i};
for (int j = 1; bin[j] <= n; ++j) {
for (int i = 1; i+bin[j-1]-1 <= n; ++i) {
if (mn[i][j-1].val <= mn[i+bin[j-1]][j-1].val) mn[i][j] = mn[i][j-1];
else mn[i][j] = mn[i+bin[j-1]][j-1];
}
}
}
int query1(int (*mn)[20], int l, int r) {
int k = Log[r-l+1];
return min(mn[l][k], mn[r-bin[k]+1][k]);
}
int query2(rmqNode (*mn)[20], int l, int r) {
int k = Log[r-l+1];
return mn[l][k].val < mn[r-bin[k]+1][k].val ? mn[l][k].p : mn[r-bin[k]+1][k].p;
}
int match(int* rk, int (*mn)[20], int p, int len) {
int rk1 = rk[p], rk2 = rk[p+len];
if (rk1 > rk2) swap(rk1, rk2);
return query1(mn, rk1+1, rk2);
}
void work() {
int tot1=0, tot2=0, m=0, len=strlen(s);
for (int i = 0; i < len; ++i) m = max(r1[tot1++] = s[i], m); r1[tot1++] = 0;
for (int i = len-1; i >= 0; --i) r2[tot2++] = s[i]; r2[tot2++] = 0;
rmqInit(len);
init(r1, rk1, h1, tot1, ++m);
init(r2, rk2, h2, tot2, m);
rmq1(h1, mn1, len);
rmq1(h2, mn2, len);
rmq2(rk1, mn, len);
node ans = {1,0,0};
for (int l = 1; l <= len; ++l) {
int lim = len / l, upp;
if (ans.val >= lim+1) break;
if (len % l) upp = lim; else upp = lim-1;
for (int i = 0; i < upp; ++i) {
int ri = match(rk1, mn1, i*l, l),
le = i ? match(rk2, mn2, len-(i+1)*l, l) : 0,
k = le + ri;
int cnt = k/l + 1;
if (cnt >= ans.val) {
int l1 = i*l-le, l2 = l1+k%l,
p = query2(mn, l1+1, l2+1)-1;
if (cnt > ans.val || rk1[p]<rk1[ans.pos]) ans = {cnt, p, cnt*l};
}
}
}
printf("Case %d: ", ++kas);
if (ans.val == 1) {
char ch='z'+1; for (int i = 0; s[i]; ++i) ch = min(ch, s[i]);
putchar(ch); puts("");
}
else {
s[ans.pos+ans.len] = '\0';
puts(s+ans.pos);
}
}
int main() {
while (scanf("%s", s) != EOF && s[0]!='#') work();
return 0;
}