标签：roo getchar line struct sum mat put utc mod
题意：白兔初始在\((0,x)\)的位置，每次跳跃要求第一维单增，第二维从\(u\)到\(v\)共\(w_{u,v}\)条边。第一维不能跳超过\(L\)。第二维取值在\(1\)到\(n\)之间。
设步数为\(m\)，询问\(m\;mod\;k=t\)时的跳跃方案对\(p\)取模的结果。
题解：
单位根反演的式子为：\(\frac{1}{n}\sum\limits_{i=0}^{n-1}\omega_{n}^{ik}=[n|k]\)
当\(n=1\)时，设\(W\)为\(w_{1,1}\)的结果，则答案为：
\(\sum\limits_{i=0}^L [k|(i-t)]W^i\binom{L}{i}\)
\(=\sum\limits_{i=0}^L\frac{1}{k}\sum\limits_{j=0}^{k-1}\omega_k^{j(i-t)}W^i\binom{L}{i}\)
\(=\frac{1}{k}\sum\limits_{i=0}^LW^i\binom{L}{i}\sum\limits_{j=0}^{k-1}\omega_k^{-jt+ij}\)
\(=\frac{1}{k}\sum\limits_{i=0}^{k-1}\omega_k^{-it}\sum\limits_{j=0}^LW^j\binom{L}{j}\omega_k^{ij}\)
\(=\frac{1}{k}\sum\limits_{i=0}^{k-1}\omega^{-it}_k(W\omega _k^i+1)^L\)
设\(C_i=(W\omega_k^i+1)^L\)
则原式\(=\frac{1}{k}\sum\limits_{i=0}^{k-1}\omega_k^{\binom{i}{2}+\binom{t}{2}-\binom{i+t}{2}}C_i\)
\(=\frac{1}{k}\omega_k^{\binom{t}{2}}\sum\limits_{i=0}^{k-1}\omega_k^{\binom{i}{2}}C_i\omega_k^{-\binom{i+t}{2}}\)
发现为卷积的形式。可以用任意模数NTT解决。
当\(n>1\)时，设\(Begin\)为初始矩阵，（只有\((1,x)\)位置为1），\(W\)为转移矩阵，
则\(C_i\)为\(Begin\times (W\omega_k^i+1)^L\)这个矩阵的\((1,y)\)位置的值。
代码：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#define mod1 998244353
#define mod2 1004535809
#define mod3 469762049
#define M 1000005
#define g 3
using namespace std;
int max_L,max_len,rev[M],C[M],W_k[M],n,k,L,x,y,p;
int read(){
char c=getchar();int ans=0;bool flag=1;
while (c<'0'||c>'9') flag&=(c!='-'),c=getchar();
while (c>='0'&&c<='9') ans=ans*10+c-'0',c=getchar();
return flag?ans:-ans;
}
void Write(long long x){
if (x<10) putchar(x^48);
else Write(x/10),putchar((x%10)^48);
return;
}
int power(int x,int y,int mod){
long long ans=1,now=x;
for (int i=y;i;i>>=1,now=now*now%mod)
if (i&1) ans=ans*now%mod;
return ans;
}
int add(int u,int v,int mod){u+=v-mod;return u+((u>>31)&mod);}
int sub(int u,int v,int mod){u-=v;return u+((u>>31)&mod);}
int invs(int x,int mod){return x==1?1:(mod-mod/x+0ll)*invs(mod%x,mod)%mod;}
const long long inv1=invs(mod1%mod2,mod2);
const long long mod1_2=mod1*(mod2+0ll);
const long long inv2=invs(mod1_2%mod3,mod3);
int get_root(int p){
static int S[105];int tmp=p-1;S[0]=0;
for (register int i=2;i*i<=tmp;i++){
if (tmp%i) continue;S[++S[0]]=i;
while (tmp%i==0) tmp/=i;
}
if (tmp!=1) S[++S[0]]=tmp;
for (register int i=2;;i++){
bool flag=1;
for (register int j=1;j<=S[0];j++)
if (power(i,(p-1)/S[j],p)==1){flag=0;break;}
if (flag) return i;
}
return 0;
}
struct number{
int num1,num2,num3;
number operator+ (const number b) const{
return (number){add(num1,b.num1,mod1),add(num2,b.num2,mod2),add(num3,b.num3,mod3)};
}
number operator+= (const number &b){return *this=*this+b;}
number operator* (const number b) const{
return (number){(int)((num1+0ll)*b.num1%mod1),(int)((num2+0ll)*b.num2%mod2),(int)((num3+0ll)*b.num3%mod3)};
}
number operator*= (const number &b){return *this=*this*b;}
number operator- (const number b) const{
return (number){sub(num1,b.num1,mod1),sub(num2,b.num2,mod2),sub(num3,b.num3,mod3)};
}
number operator-= (const number &b){return *this=*this-b;}
number operator^ (const int b) const{
number ans=(number){1,1,1},now=*this;
for (register int i=b;i;i>>=1,now*=now)
if (i&1) ans*=now;
return ans;
}
}W[M],inv[M],A[M],B[M];
number make_number (int x){return (number){x%mod1,x%mod2,x%mod3};}
long long get_ans(number now){
long long x4=(now.num2-now.num1+mod2)*inv1%mod2*mod1+now.num1;
return (x4+mod1_2%p*((now.num3+mod3-x4%mod3)*inv2%mod3))%p;
}
void prepare(){
max_len=1<<19,max_L=19;
for (register int i=0;i<max_len;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<max_L-1);
number wn=(number){power(g,(mod1-1)/max_len,mod1),power(g,(mod2-1)/max_len,mod2),power(g,(mod3-1)/max_len,mod3)};
W[max_len>>1]=(number){1,1,1};inv[1]=(number){1,1,1};
for (register int i=2;i<=max_len;i++){
int X=(mod1-mod1/i+0ll)*inv[mod1%i].num1%mod1;
int Y=(mod2-mod2/i+0ll)*inv[mod2%i].num2%mod2;
int Z=(mod3-mod3/i+0ll)*inv[mod3%i].num3%mod3;
inv[i]=(number){X,Y,Z};
}
for (register int i=(max_len>>1)+1;i<max_len;i++) W[i]=W[i-1]*wn;
for (register int i=(max_len>>1)-1;i;i--) W[i]=W[i<<1];
return;
}
void NTT(number *A,int len,int L,int flag){
static number F[M];
for (register int i=0,s=max_L-L;i<len;i++) F[rev[i]>>s]=A[i];
for (register int l=1;l<len;l<<=1)
for (register int i=0;i<len;i+=(l<<1))
for (register int j=0;j<l;j++){
number u=F[i|j],v=F[i|j|l]*W[j|l];
F[i|j]=u+v,F[i|j|l]=u-v;
}
if (flag){
A[0]=F[0]*inv[len];
for (register int i=1;i<len;i++) A[i]=F[len-i]*inv[len];
}
else for (register int i=0;i<len;i++) A[i]=F[i];
return;
}
struct Matrix{
int a[3][3];
Matrix operator* (const Matrix b) const{
Matrix c;
for (register int i=0;i<3;i++)
for (register int j=0;j<3;j++)
c.a[i][j]=(a[i][0]*(b.a[0][j]+0ll)+a[i][1]*(b.a[1][j]+0ll)+a[i][2]*(b.a[2][j]+0ll))%p;
return c;
}
Matrix operator*= (const Matrix &b){return *this=*this*b;}
Matrix operator* (const int b) const{
Matrix c=*this;
for (register int i=0;i<3;i++)
for (register int j=0;j<3;j++)
c.a[i][j]=c.a[i][j]*(b+0ll)%p;
return c;
}
Matrix operator*= (const int &b){return *this=*this*b;}
Matrix operator+ (const Matrix b) const{
Matrix c;
for (register int i=0;i<3;i++)
for (register int j=0;j<3;j++) c.a[i][j]=add(a[i][j],b.a[i][j],p);
return c;
}
Matrix operator+= (const Matrix &b){return *this=*this+b;}
Matrix operator^ (const int b) const{
Matrix ans=*this,now=*this;
if (!b){
ans.a[0][0]=ans.a[1][1]=ans.a[2][2]=1;
ans.a[0][1]=ans.a[1][2]=ans.a[2][0]=0;
ans.a[0][2]=ans.a[1][0]=ans.a[2][1]=0;
return ans;
}
for (register int i=b-1;i;i>>=1,now*=now)
if (i&1) ans*=now;
return ans;
}
}G,Begin,I;
void mul(number *A,number *B,int len){
int rlen=1,L=0;while (rlen<=(len<<1)) rlen<<=1,++L;
NTT(A,rlen,L,0),NTT(B,rlen,L,0);
for (register int i=0;i<rlen;i++) A[i]*=B[i];
NTT(A,rlen,L,1);
for (register int i=len;i<rlen;i++) A[i]=make_number(0);
return;
}
int main(){
n=read(),k=read(),L=read(),x=read()-1,y=read()-1,p=read();
for (register int i=0;i<n;i++)
for (register int j=0;j<n;j++) G.a[i][j]=read();
I.a[0][0]=I.a[1][1]=I.a[2][2]=1;
I.a[0][1]=I.a[1][2]=I.a[2][0]=0;
I.a[0][2]=I.a[1][0]=I.a[2][1]=0;
Begin.a[0][x]=1;W_k[0]=1;W_k[1]=power(get_root(p),(p-1)/k,p);prepare();
for (register int i=2;i<=k;i++) W_k[i]=W_k[i-1]*(W_k[1]+0ll)%p;
for (register int i=0;i<k;i++) C[i]=(Begin*((G*W_k[i]+I)^L)).a[0][y];
for (register int i=0;i<k;i++) A[i]=make_number(W_k[i*(i-1ll)/2%k]*(C[i]+0ll)%p);
for (register int i=0;i<k+k;i++) B[k+k-1-i]=make_number(W_k[k-i*(i-1ll)/2%k]%p);
mul(A,B,k<<1);int invk=invs(k,p);
for (register int i=0;i<k;i++)
Write(get_ans(A[k+k-i-1])*invk%p*W_k[i*(i-1ll)/2%k]%p),putchar('\n');
return 0;
}标签：roo getchar line struct sum mat put utc mod
原文地址：https://www.cnblogs.com/bestlxm/p/12329278.html
