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# 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\n", Calc()); return 0;}