9 months hate big heart finally cut off! ! ! !
A very good question, I do not know why uoj is on a 70-point difference in assessment.
Topic links: http://uoj.ac/problem/214
Topic effect: Please read on their own.
answer:
The official made a very clear solution to a problem, here add some of their own understanding.
First look \ (O (2 ^ {nm } \ times poly (n, m)) \) approach.
A Way of Understanding is the official explanations.
Set \ (S \) is the total number of courses ( \ (n-\) total length of the string), \ (P (S) \) represents the end position set \ (S \) string matching all need a total of how many different courses completed. Set \ (f (t) \) represents \ (T \) probability point during the process is not terminated, \ (Prob (S, T) \) represents \ (S \) sets of strings at the end of the (T \) \ time or probability before the match has been completed. \ (ans \) is the answer.
\(ans=\sum^{+\inf}_{t=0} f(t)=\sum^{+\inf}_{t=0} \sum_{S, |S|\ge 0}(-1)^{|S|} prob(S,t)=\sum^{+\inf}_{t=0} \sum_{S,|S|\ge 0}(-1)^{|S|} \sum^{f(S)}_{i=0} (-1)^i{f(S)\choose i}(\frac{s-i}{s})^t\\ =\sum^{+\inf}_{t=0} \sum_{S,|S|\ge 0}(-1)^{|S|} \sum^{f(S)}_{i=1} (-1)^i{f(S)\choose i}(\frac{s-i}{s})^t\\ =\sum_{S,|S|\ge 0}(-1)^{|S|} \sum^{f(S)}_{i=0} (-1)^i{f(s)\choose i}\sum^{\inf}_{t=0}(\frac{s-i}{s})^t=\sum_{S,|S|\ge 0}(-1)^{|S|} \sum^{f(S)}_{i=0} (-1)^i{f(S)\choose i} \frac{s}{i}\)
Can be calculated.
In fact, there is another way of understanding. We require that all the time string matching the minimum desired, according to the inclusion and exclusion Min-Max, we can find the pair of each subset of the subset matching string conversion time by enumerating a subset of the maximum desired, and then converted to per a moment of probability did not end.
Then look harder \ (O (2 ^ m \ times poly (n, m)) \) approach.
For all \ (p (S) \) the same \ (S \) , its contribution to the answers no difference.
Thus for all \ (K \) , is calculated considering \ (p (S) = k \) a \ (S \) number.
Dp can then directly.
Error Code record: Note solution1 If \ (nm = 16 \) , then the array to open to \ (2 ^ {17} \) .
Code
#include<cstdio>
#include<cstdlib>
#include<cstring>
#define llong long long
using namespace std;
const int N = 30;
const int S = 26;
const int P = 998244353;
bool ok[N+3];
int f[N+3][S+3];
char str[N+3];
char a[N+3];
llong fact[2000003],finv[2000003],inv[2000003];
int cnt[(1<<17)+3];
int n,m,s;
llong quickpow(llong x,llong y)
{
llong cur = x,ret = 1ll;
for(int i=0; y; i++)
{
if(y&(1ll<<i)) {y-=(1ll<<i); ret = ret*cur%P;}
cur = cur*cur%P;
}
return ret;
}
llong comb(llong x,llong y) {return x<0 || y<0 || x<y ? 0ll : fact[x]*finv[y]%P*finv[x-y]%P;}
namespace Solution1
{
int ff[N+3][S+3];
llong ans;
void solve()
{
ans = 0ll;
for(int i=1; i<(1<<(n-m+1)); i++)
{
llong cur = 0ll; bool gg = false;
for(int j=0; j<=n-m; j++)
{
if(i&(1<<j))
{
for(int k=0; k<m; k++)
{
ff[j+k][a[k]-96] = true;
if(f[j+k][a[k]-96]==false) {gg = true; break;}
}
if(gg==true) break;
}
}
if(gg)
{
for(int j=0; j<n; j++)
{
for(int k=1; k<=S; k++)
{
ff[j][k] = false;
}
}
continue;
}
int num = 0;
for(int j=0; j<n; j++)
{
for(int k=1; k<=S; k++)
{
if(ff[j][k]==true) {num++;}
}
}
for(int j=1; j<=num; j++)
{
llong tmp = (llong)s*inv[j]%P*comb(num,j)%P;
if(!(j&1)) {cur = (cur+tmp)%P;}
else {cur = (cur-tmp+P)%P;}
}
for(int j=0; j<n; j++)
{
for(int k=1; k<=S; k++)
{
ff[j][k] = false;
}
}
if(cnt[i]&1) {ans = (ans-cur+P)%P;}
else {ans = (ans+cur+P)%P;}
}
printf("%lld\n",ans);
}
}
namespace Solution2
{
#define U ((1<<m)-1)
const int M = 14;
llong dp[N+3][(1<<M)+3][N*M+3];
bool ff[N+3][M+3];
void update(llong &x,llong y) {x = (x+y)%P;}
void solve()
{
llong ans = 0ll;
dp[m-1][0][0] = 1ll;
for(int i=m-1; i<n; i++)
{
for(int j=0; j<(1<<m); j++)
{
bool okk = true;
for(int k=0; k<m; k++) if((j&(1<<k)) && ok[i-k-1]==false) {okk = false; break;}
if(!okk) continue;
int p = 0;
for(int k=0; k<m; k++)
{
if(j&(1<<k))
{
for(int l=0; l<m; l++)
{
ff[i-k-1-m+1+l][a[l]-96] = true;
}
}
}
for(int l=0; l<m; l++)
{
if(ff[i-m+1+l][a[l]-96]==false)
{
p++;
}
}
for(int k=0; k<m; k++)
{
if(j&(1<<k))
{
for(int l=0; l<m; l++)
{
ff[i-k-1-m+1+l][a[l]-96] = false;
}
}
}
for(int k=0; k<=n*m; k++)
{
if(dp[i][j][k]==0) continue;
update(dp[i+1][(j<<1)&U][k],dp[i][j][k]);
update(dp[i+1][(j<<1|1)&U][k+p],P-dp[i][j][k]);
dp[i][j][k] = 0ll;
}
}
}
for(int i=1; i<=n*m; i++)
{
llong coe = 0ll;
for(int j=1; j<=i; j++)
{
llong tmp = s*inv[j]%P*comb(i,j)%P;
if(j&1) {coe = (coe-tmp+P)%P;}
else {coe = (coe+tmp)%P;}
}
llong sum = 0ll;
for(int j=0; j<(1<<m); j++)
{
sum = (sum+dp[n][j][i])%P;
dp[n][j][i] = 0ll;
}
ans = (ans+sum*coe)%P;
}
printf("%lld\n",ans);
}
}
int main()
{
fact[0] = 1ll; for(int i=1; i<=2000000; i++) fact[i] = fact[i-1]*i%P;
finv[2000000] = quickpow(fact[2000000],P-2); for(int i=1999999; i>=0; i--) finv[i] = finv[i+1]*(i+1)%P;
for(int i=1; i<=2000000; i++) inv[i] = finv[i]*fact[i-1]%P;
cnt[0] = 0; for(int i=1; i<(1<<17); i++) cnt[i] = cnt[i>>1]+(i&1);
int T; scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m); s = 0;
for(int i=0; i<n; i++)
{
scanf("%s",str); int l = strlen(str);
for(int j=0; j<l; j++)
{
f[i][str[j]-96] = true; s++;
}
}
scanf("%s",a); bool exist = false;
for(int i=m-1; i<n; i++)
{
ok[i] = true;
for(int j=0; j<m; j++)
{
if(f[i-m+1+j][a[j]-96]==false) {ok[i] = false; break;}
}
if(ok[i]) {exist = true;}
}
if(exist==false) {printf("-1\n");}
else
{
if(n-m<=16)
{
Solution1::solve();
}
else
{
Solution2::solve();
}
}
for(int i=0; i<n; i++)
{
for(int j=1; j<=S; j++) f[i][j] = false;
ok[i] = false;
}
}
return 0;
}