Blog of RuSun

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

P1659 [国家集训队]拉拉队排练

P1659 [国家集训队]拉拉队排练

求出所有回文串长度个数后统计答案即可。

查看代码
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#include <cstdio>
#include <algorithm>
#include <string>
using namespace std;
typedef long long LL;
const int N = 2e6 + 10, mod = 19930726;
char s[N];
int n, d[N], f[N];
LL m;
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 binpow (int b, int k)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
int main()
{
int _n;
scanf("%d %lld\n", &_n, &m);
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++) ++f[d[i] - 1];
n = _n;
for (int i = n - 2; i; --i) f[i] += f[i + 2];
int res = 1;
for (int i = n - !(n & 1); i > 0; i -= 2)
{
if (m <= f[i]) res = (LL)res * binpow(i, m) % mod, m = 0;
else res = (LL)res * binpow(i, f[i]) % mod, m -= f[i];
}
printf("%d", m ? -1 : res);
return 0;
}