Title effect: the definition \ (m \) operator in hexadecimal \ (\ Oplus \) , which satisfy the commutative, associative law, law cycle, given that the calculation rules in any position \ (A_ {i, } J \) , represents \ (I \ A_ Oplus J = I {,} J \) , from ) \) \ ([0, m ^ n to select a certain probability \ (k + 1 \) number \ (S_0, S_2 \ DOTS, S_k \) , find \ ([0, m ^ n ) \) each number is exactly equal to \ (s_0 \ oplus s_1 \ oplus \ dots \ oplus s_k \) probability. For \ (232792561 = 2 ^ 4 \ times3 ^ 2 \ times5 \ times7 \ times11 \ times13 \ times17 \ times19 \) modulo.
solution:
God, that ah, I saw only two days to read explanations
First of all,According to the label,It is contemplated that a subject is to let you transform and its inverse transform configuration, such that the probability of the original sequence \ (p_0, p_1, \ dots , p_ {m ^ n-1} \) after this conversion, for every request \ ( k \) power after the inverse transform can get the answer sequence.
Can then be set transformation matrix \ (T \) , there are \ (p \ times T \) and \ (q \ times T \) corresponding to the position of the multiplied \ ((p * q) \ times T \) is equal to , i.e. \ (T_ {K, I} \ Times T_ {K, T_ {J} = K, I \ J} Oplus \) .
Since the operation to meet cycle law, it is provided to meet the \ (i ^ j = i ( j> 1) \) minimum \ (J \) of \ (len_i \) , analog \ (\ text {DFT} \ ) was transferred matrix, can guess \ (T_ {i, j} \) is \ (0 \) or \ (len_j-1 \) subunit root \ ([0, len_j-1 ) \) power.
Then the foregoing rules can \ (O (m ^ m) \) was added to prune this seizure burst matrix inversion to obtain the inverse transform matrix, and then the \ (P \) is converted, can be obtained the target array.
code:
#include<stdio.h>
#include<algorithm>
#define inf 232792561
int a[102][202],ts[102],st[102],l=0;
int p[102][102],q[102][102],w[102][102],t[102][102];
int n,m;long long k;
int A[1048576],B[1048576];
inline int ksm(long long a,long long b){int ans=1;while(b)(b&1)&&(ans=a*ans%inf),a=a*a%inf,b>>=1;return ans;}
inline void gs(int n,int m){
for(int i=0;i<n;i++){
int pos=i;
while(pos<n&&!a[pos][i])pos++;
if(pos>=n)exit(-1);
if(pos!=i)
for(int j=i;j<m;j++)
a[i][j]^=a[pos][j]^=a[i][j]^=a[pos][j];
int val=ksm(a[i][i],inf-2);
for(int j=i;j<m;j++)a[i][j]=1ull*val*a[i][j]%inf;
for(int j=0;j<n;j++)
if(i!=j&&a[j][i])
for(int k=m-1;k>=i;k--)
a[j][k]=(1ull*(inf-a[j][i])*a[i][k]+a[j][k])%inf;
}
}
void dfs(int pos){
if(l==m)return;
if(pos==m){
for(int i=0;i<m;i++)p[l][i]=st[i];
l++;return;
}bool ff=st[pos]=0;
for(int i=0;i<pos;i++){
for(int j=i;j<pos;j++){
if(t[i][j]==pos){
if(!ff)st[pos]=1ll*st[i]*st[j]%inf,ff=1;
else if(st[pos]!=1ll*st[i]*st[j]%inf)return;
}
}
}if(ff){
for(int j=0;j<=pos;j++){
if(t[j][pos]<=pos&&1ll*st[j]*st[pos]%inf!=st[t[j][pos]])return;
}dfs(pos+1);
}
else{
for(int i=0;i<=ts[pos];i++){
st[pos]=w[ts[pos]][i];
bool f=1;
for(int j=0;j<=pos;j++){
if(t[j][pos]<=pos&&1ll*st[j]*st[pos]%inf!=st[t[j][pos]])f=0;
}if(f)dfs(pos+1);
}
}
}
void ntt(int a[],int b[],int n,bool typ){
int dt=n/m,p1,p2;
if(dt!=1)for(int i=0;i<n;i+=dt)ntt(a+i,b,dt,typ);
for(int i=0;i<n;i++)b[i]=0;
for(int i=0;i<m;i++)
for(int j=0;j<m;j++)
for(int k=0;k<dt;k++){
int p1=i*dt+k,p2=j*dt+k;
b[p1]=(1ll*a[p2]*(typ?q[i][j]:p[i][j])+b[p1])%inf;
}
for(int i=0;i<n;i++)a[i]=b[i];
}
int main(){
scanf("%d%d%lld",&n,&m,&k);
n=ksm(m,n);
for(int i=0;i<m;i++)
for(int j=0;j<m;j++)
scanf("%d",&t[i][j]);
for(int i=1;i<=m;i++){
w[i][0]=1;
if(i!=1)w[i][1]=ksm(71,(inf-1)/i);
for(int j=2;j<i;j++)
w[i][j]=1ll*w[i][j-1]*w[i][1]%inf;
}for(int i=0;i<m;i++){
int P=i;ts[i]=0;
while(1){
ts[i]++;
P=t[P][i];
if(P==i)break;
}
}dfs(0);
for(int i=0;i<m;i++)
for(int j=0;j<m;j++)
a[i][j]=p[i][j];
for(int i=0;i<m;i++)
a[i][i+m]=1;
gs(m,m+m);
for(int i=0;i<m;i++)
for(int j=0;j<m;j++)
q[i][j]=a[i][j+m];
for(int i=0;i<n;i++)scanf("%d",&A[i]);
ntt(A,B,n,0);
for(int i=0;i<n;i++)A[i]=ksm(A[i],k+1);
ntt(A,B,n,1);
for(int i=0;i<n;i++)printf("%d\n",A[i]);
}