给定字符串S, 求有多少长为$n$的字符串T, 使得S为T的子序列.
可以得到转移矩阵为
$egin{equation}
A
=egin{bmatrix}
25 & 0 & 0 &cdots &0 &0\
1 & 25 & 0 &cdots &0 &0\
0 & 1 & 25 & cdots & 0 & 0\
vdots & vdots & vdots & ddots & vdots &vdots\
0 & 0 & 0 &cdots &1& 26\
end{bmatrix}
end{equation}$
设$|S|=m$, 答案就为$A^n$的$(m,0)$项, 也就是说答案只与$S$的长度有关, 但是用矩阵幂的话复杂度是$O(m^3logn)$显然过不去.
实际上我们可以直接设$f(n)$为长为$n$的字符串的答案, 不去维护匹配的状态.
可以得到$f(n) = egin{cases} 0, & n< m \26f(n-1)+25^{n-m}inom{n-1}{m-1}, & nge m end{cases}$
前一部分表示在前$n-1$位已经有子序列等于$S$的情形, 那么第$n$位可以任取值.
后一部分表示在第$n$位时第一次出现子序列等于$S$的情形, 那么枚举前$m-1$位在$T$中的第一次出现位置, 其余位置可以任取其余的$25$个值.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 1e5+10;
int t, clk;
char s[N];
vector<pii> g[N];
int ans[N], f[N];
ll p25[N], fac[N], ifac[N];
ll C(int n, int m) {
return fac[n]*ifac[n-m]%P*ifac[m]%P;
}
int main() {
p25[0] = fac[0] = ifac[0] = 1;
REP(i,1,N-1) {
p25[i] = p25[i-1]*25%P;
fac[i] = fac[i-1]*i%P;
ifac[i] = inv(fac[i]);
}
scanf("%d%s", &t, s);
int now = strlen(s);
REP(i,1,t) {
int op, x;
scanf("%d", &op);
if (op==1) scanf("%s", s), now = strlen(s);
else scanf("%d", &x), g[now].pb(pii(x,++clk));
}
REP(i,1,N-1) if (g[i].size()) {
sort(g[i].begin(),g[i].end());
int sz = g[i].size(), now = 0;
while (now<sz&&g[i][now].x<i) ++now;
f[i-1] = 0;
REP(j,i,N-1) {
f[j] = (f[j-1]*26ll+C(j-1,i-1)*p25[j-i])%P;
while (now<sz&&g[i][now].x==j) ans[g[i][now++].y] = f[j];
if (now>=sz) break;
}
}
REP(i,1,clk) printf("%d
",ans[i]);
}