Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF17E Palisection

LuoGu: CF17E Palisection

CF: E. Palisection

相交是不好做的,考虑求不相交。记 $f _ i$ 表示在 $i$ 结尾的回文串的个数,$g _ i$ 表示在 $i$ 开头的回文串的个数,这可以由 Manacher 得到,因为要区间加,所以先差分修改,最后前缀和。统计答案时将 $f$ 再做前缀和,对于每一个 $g _ i$ 统计答案。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 4e6 + 10, mod = 51123987;
char s[N];
int n, d[N], f[N], g[N];
void init()
{
s[n++] = '$', s[n++] = '#';
char c = getchar();
while (c < 'a' || c > 'z') c = getchar();
while (c >= 'a' && c <= 'z')
s[n++] = c, s[n++] = '#', c = getchar();
s[n++] = '^';
}
int main()
{
scanf("%*d");
init();
for (int i = 0, l = 0, r = -1; i < n; i++)
{
int k = i <= r ? min(d[l + r - i], r - i + 1) : 1;
while (i - k >= 0 && i + k < n && s[i - k] == s[i + k]) ++k;
d[i] = k;
if (i + k - 1 > r) l = i - k + 1, r = i + k - 1;
}
for (int i = 0; i < n; ++i) if (d[i] > 1)
++f[i + 1 >> 1], --f[i + d[i] + 1 >> 1], ++g[i >> 1], --g[i - d[i] >> 1];
n = n - 3 >> 1;
for (int i = 2; i <= n; ++i) f[i] += f[i - 1];
for (int i = n - 1; i; --i) g[i] += g[i + 1];
for (int i = 2; i <= n; ++i) (f[i] += f[i - 1]) %= mod;
int res = f[n] * (f[n] - 1ll) / 2 % mod;
for (int i = 1; i < n; ++i)
res = (res - (LL)f[i] * g[i + 1]) % mod;
printf("%d", (res + mod) % mod);
return 0;
}