题面
Sol
题目的提示说的也非常好
我对求(LCP(P - L + len \% l, P + len \% L))做补充
(len=LCP(P, P + L))
为什么只要求(LCP(P - L + len \% l, P + len \% L))呢？
考虑在(P - L + len \% l)右边到(P)之间，它不比这里的重复次数大
考虑在(P - L + len \% l)左边到(P-1)之间，一样的它也是不增的
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(100010);
int n, a[_], sa[_], rk[_], y[_], height[_], t[_], vis[_];
int st[20][_], lg[_];
char s[_];
IL bool Cmp(RG int i, RG int j, RG int k){ return y[i] == y[j] && y[i + k] == y[j + k]; }
IL void Sort(){
RG int m = 30;
for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];
for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
for(RG int k = 1; k <= n; k <<= 1){
RG int l = 0;
for(RG int i = n - k + 1; i <= n; ++i) y[++l] = i;
for(RG int i = 1; i <= n; ++i) if(sa[i] > k) y[++l] = sa[i] - k;
for(RG int i = 0; i <= m; ++i) t[i] = 0;
for(RG int i = 1; i <= n; ++i) ++t[rk[y[i]]];
for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
for(RG int i = n; i; --i) sa[t[rk[y[i]]]--] = y[i];
swap(rk, y); rk[sa[1]] = l = 1;
for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
if(l >= n) break;
m = l;
}
for(RG int i = 1, j = 0; i <= n; ++i){
j = max(0, j - 1);
while(a[j + i] == a[sa[rk[i] - 1] + j]) ++j;
height[rk[i]] = j;
}
}
IL int LCP(RG int xx, RG int yy){
xx = rk[xx]; yy = rk[yy];
if(xx > yy) swap(xx, yy);
++xx;
RG int l = lg[yy - xx + 1];
return min(st[l][xx], st[l][yy - (1 << l) + 1]);
}
IL int Calc(){
RG int ans = 0;
for(RG int l = 1; l <= n; ++l)
for(RG int i = 1; i + l <= n; i += l){
RG int len = LCP(i, i + l);
ans = max(ans, len / l + 1);
if(i >= l - len % l) ans = max(ans, LCP(i - l + len % l, i + len % l) / l + 1);
}
return ans;
}
int main(RG int argc, RG char* argv[]){
scanf(" %s", s + 1);
n = strlen(s + 1);
for(RG int i = 1; i <= n; ++i) a[i] = s[i] - 'a' + 1;
for(RG int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
Sort();
for(RG int i = 1; i <= n; ++i) st[0][i] = height[i];
for(RG int i = 1; i <= lg[n]; ++i)
for(RG int j = 1; j + (1 << i) - 1 <= n; ++j)
st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
printf("%d
", Calc());
return 0;
}