首先只有质数个$4$且个数不超过$10$的限制条件才有用,
也就是长度不能为$44,444,44444,4444444$的倍数。
考虑容斥,计算长度必须是它们$lcm$的倍数,且没有连续$4$个$4$的方案数。
将DP转移方程用矩阵表示,则长度为$L$的方案数为$G^{L-1}\times V$。
所以对于一个$lcm$,答案为$(\sum_{k=0}^{\lfloor\frac{n}{lcm}\rfloor-1} (G^{lcm})^k)\times G^{lcm-1}\times V$,分治求解即可。
#include<cstdio> #define rep(i) for(int i=0;i<4;i++) typedef long long ll; const int P=1000000007; struct mat{ int v[4][4]; mat(){rep(i)rep(j)v[i][j]=0;} mat operator+(const mat&b){ mat c; rep(i)rep(j)c.v[i][j]=(v[i][j]+b.v[i][j])%P; return c; } mat operator*(const mat&b){ mat c; rep(i)rep(j)rep(k)c.v[i][j]=(1LL*v[i][k]*b.v[k][j]+c.v[i][j])%P; return c; } }G,A,B,I; ll a[4]={44,444,44444,4444444},n,t;int S,ans,ret; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} mat pow(mat a,ll b){mat t=I;for(;b;b>>=1,a=a*a)if(b&1)t=t*a;return t;} struct PI{mat x,y;PI(){}PI(mat _x,mat _y){x=_x,y=_y;}}; PI cal(const mat&a,ll b){ if(b==1)return PI(I,a); if(b&1){ PI c=cal(a,b-1); return PI(c.x*a+I,c.y*a); } PI c=cal(a,b>>1); return PI(c.x*(c.y+I),c.y*c.y); } int main(){ rep(i)G.v[0][i]=9,I.v[i][i]=1; G.v[1][0]=G.v[2][1]=G.v[3][2]=1; A.v[0][0]=8,A.v[1][0]=1; scanf("%lld",&n); for(S=0;S<16;S++){ t=1; rep(i)if(S>>i&1)t=lcm(t,a[i]); if(t>n)continue; B=cal(pow(G,t),n/t).x*pow(G,t-1)*A; ret=0; rep(i)ret=(ret+B.v[i][0])%P; if(__builtin_popcount(S)&1)ans=(ans+P-ret)%P;else ans=(ans+ret)%P; } return printf("%d",ans),0; }
时间: 2024-11-02 13:56:08