嘟嘟嘟
这题不得不吐槽一下,(O(n ^ 2))哈希能得95分,那么考场上有人写正解吗?或许只有队爷儿吧
正解思路特别妙,这篇题解也很妙:题解:[NOI2016]优秀的拆分(洛谷第一篇题解,带图的那个)。
我这绝对不是在水博客,因为那篇题解讲的太清楚了,我都不知道补充啥好。
然后因为(n)不是很大,所以找lcp和lcs的时候可以(O(logn))用哈希二分,总复杂度(O(n log ^ 2n))。(毕竟我不会SA也不想写SAM)。
还有,第13个点最后一组数据卡自然溢出哈希,得换成取模版哈希或双哈希(就是写一个自然溢出和一个取模,判等的话必须这两者都相等才算相等)
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<ctime>
#include<cassert>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e6 + 5;
const ll BAS = 31;
const ll mod = 998244353;
In ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
In void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
char s[maxn];
int n;
ll has[maxn], p[maxn];
ll Beg[maxn], End[maxn];
In ll H(int L, int R) {return (has[R] - has[L - 1] * p[R - L + 1] % mod + mod) % mod;}
In int lcs(int x, int y)
{
int L = 0, R = x;
while(L < R)
{
int mid = (L + R + 1) >> 1;
if(H(x - mid + 1, x) == H(y - mid + 1, y)) L = mid;
else R = mid - 1;
}
return L;
}
In int lcp(int x, int y)
{
int L = 0, R = n - y + 1;
while(L < R)
{
int mid = (L + R + 1) >> 1;
if(H(x, x + mid - 1) == H(y, y + mid - 1)) L = mid;
else R = mid - 1;
}
return L;
}
In void solve(int len)
{
for(int i = len; i <= n; i += len)
{
int j = i + len;
if(j > n) break;
int l1 = min(len, lcs(i, j)), l2 = min(len, lcp(i, j));
if(l1 + l2 >= len + 1)
{
int tp = l1 + l2 - len - 1;
++Beg[i - l1 + 1], --Beg[i - l1 + 1 + tp + 1];
++End[j + l2 - 1 - tp], --End[j + l2];
}
}
}
In void init()
{
Mem(Beg, 0), Mem(End, 0);
for(int i = 1; i <= n; ++i) has[i] = (has[i - 1] * BAS + s[i] - 'a') % mod;
}
int main()
{
//freopen("ha.in", "r", stdin);
int T = read();
p[0] = 1;
for(int i = 1; i < maxn; ++i) p[i] = p[i - 1] * BAS % mod;
while(T--)
{
scanf("%s", s + 1);
n = strlen(s + 1); init();
for(int i = 1; i <= (n >> 1); ++i) solve(i);
ll ans = 0;
for(int i = 1; i <= n; ++i)
{
Beg[i] = Beg[i - 1] + Beg[i];
End[i] = End[i - 1] + End[i];
ans += End[i - 1] * Beg[i];
}
write(ans), enter;
}
return 0;
}