[loj138]类欧几里得算法

模板题

定义$lfloor x floor$表示小于等于$x$的最大整数，$lceil x ceil$表示大于等于$x$的最小整数

不难发现，若$ain Z^{+}$且$b,xin Z$，则$axle biff xle lfloorfrac{b}{a} floor$、$axge biff xge lfloorfrac{b-1}{a} floor+1$

通过拉格朗日插值法，预处理出$S_{k}(n)=sum_{i=0}^{n}i^{k}$（定义$0^{0}=1$）关于$n$的$k+1$次多项式，并记其中的$i$次项系数为$S_{k}[i]$，即有$S_{k}(n)=sum_{i=0}^{k+1}S_{k}[i]cdot n^{i}$

记$F_{k_{1},k_{2}}(n,a,b,c)=sum_{x=0}^{n}x^{k_{1}}lfloorfrac{ax+b}{c} floor^{k_{2}}$，问题即求该式$F_{k_{1},k_{2}}(n,a,b,c)$

对其分类讨论：

1.若$n<0$，那么原式即为0

2.若$a otin [0,c)$，不妨假设$a=ic+a'$（其中$iin Z$且$a'in [0,c)$​），那么原式即
$$
egin{array}{ll}F_{k_{1},k_{2}}(n,a,b,c)
&=sum_{x=0}^{n}x^{k_{1}}(lfloorfrac{a'x+b}{c} floor+ix)^{k_{2}}\
&=sum_{x=0}^{n}x^{k_{1}}sum_{k=0}^{k_{2}}{k_{2}choose k}lfloorfrac{a'x+b}{c} floor^{k}(ix)^{k2-k}\
&=sum_{k=0}^{k_{2}}{k2choose k}i^{k_{2}-k}F_{k_{1}+k_{2}-k,k}(n,a',b,c)end{array}
$$
3.若$b otin [0,c)$，不妨假设$b=ic+b'$（其中$iin Z$且$b'in [0,c)$），那么原式即
$$
egin{array}{ll}F_{k_{1},k_{2}}(n,a,b,c)
&=sum_{x=0}^{n}x^{k_{1}}(lfloorfrac{ax+b'}{c} floor+i)^{k_{2}}\
&=sum_{x=0}^{n}x^{k_{1}}sum_{k=0}^{k_{2}}{k_{2}choose k}lfloorfrac{ax+b'}{c} floor^{k}i^{k2-k}\
&=sum_{k=0}^{k_{2}}{k2choose k}i^{k_{2}-k}F_{k_{1},k}(n,a,b',c)end{array}
$$
4.不为以上几种情况，先处理$k_{2}=0$的部分，即$F_{k_{1},0}(n,a,b,c)=sum_{x=0}^{n}x^{k_{1}}=S_{k_{1}}(n)$

接下来，若$a=0$则其余部分均为0，否则原式即
$$
egin{array}{ll}F_{k_{1},k_{2}}(n,a,b,c)
&=sum_{x=0}^{n}x^{k_{1}}sum_{i=0}^{lfloorfrac{ax+b}{c} floor-1}left((i+1)^{k_{2}}-i^{k_{2}} ight)\
&=sum_{x=0}^{n}x^{k_{1}}sum_{i=0}^{lfloorfrac{ax+b}{c} floor-1}sum_{k=0}^{k_{2}-1}{k_{2}choose k}i^{k}\
&=sum_{k=0}^{k_{2}-1}{k_{2}choose k}sum_{i=0}^{lfloorfrac{an+b}{c} floor-1}i^{k}sum_{x=lfloorfrac{ic+c-b-1}{a} floor+1}^{n}x^{k_{1}}\
&=sum_{k=0}^{k_{2}-1}{k_{2}choose k}sum_{i=0}^{lfloorfrac{an+b}{c} floor-1}i^{k}left(S_{k_{1}}(n)-S_{k_{1}}(lfloorfrac{ic+c-b-1}{a} floor) ight)\
&=C-sum_{k=0}^{k_{2}-1}{k_{2}choose k}sum_{i=0}^{lfloorfrac{an+b}{c} floor-1}i^{k}sum_{k'=0}^{k_{1}+1}S_{k_{1}}[k']cdot lfloorfrac{ic+c-b-1}{a} floor^{k'}\
&=C-sum_{k=0}^{k_{2}-1}{k_{2}choose k}sum_{k'=0}^{k_{1}+1}S_{k_{1}}[k']cdot F_{k,k'}(lfloorfrac{an+b}{c} floor-1,c,c-b-1,a)end{array}
$$
其中$k_{2}>0$，$C$为$S_{k_{1}}(n)$对应的部分，即
$$
C=sum_{k=0}^{k_{2}-1}{k_{2}choose k}S_{k}(lfloorfrac{an+b}{c} floor-1)S_{k_{1}}(n)
$$
考虑递归计算，即对于状态$(n,a,b,c)$求出所有$k_{1},k_{2}in N^{*}$的$F_{k_{1},k_{2}}(n,a,b,c)$，注意到转移的$k_{1}+k_{2}$单调不增，因此只需要考虑$k_{1}+k_{2}le K=10$的状态即可

由此，递归状态内的计算复杂度为$o(K^{4})$，而状态中的$a$和$c$即为辗转相除，即递归层数为$o(log n)$

综上，时间复杂度为$o(tK^{4}log n)$，可以通过

#include<bits/stdc++.h>
using namespace std;
#define N 15
#define K 10
#define mod 1000000007
#define ll long long
struct Data{
    int a[N][N];
    Data(){
        memset(a,0,sizeof(a));
    }
};
int t,n,a,b,c,k1,k2,inv[N],C[N][N],mi[N],g[N],S[N][N];
void init(){
    inv[0]=inv[1]=1;
    for(int i=2;i<N;i++)inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod;
    for(int i=0;i<N;i++){
        C[i][0]=C[i][i]=1;
        for(int j=1;j<i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
    }
    for(int i=0;i<N;i++)mi[i]=1;
    for(int i=0;i<=K;i++){
        int s=0;
        for(int j=0;j<=i+1;j++){
            s=(s+mi[j])%mod;
            int sum=s;
            memset(g,0,sizeof(g));
            g[0]=1;
            for(int k=0;k<=i+1;k++)
                if (j!=k){
                    sum=(ll)sum*inv[abs(j-k)]%mod;
                    if (j<k)sum=mod-sum;
                    for(int l=i+1;l;l--)g[l]=(g[l-1]-(ll)g[l]*k%mod+mod)%mod;
                    g[0]=mod-(ll)g[0]*k%mod;
                }
            for(int k=0;k<=i+1;k++)S[i][k]=(S[i][k]+(ll)sum*g[k])%mod;
        }
        for(int j=0;j<N;j++)mi[j]=(ll)mi[j]*j%mod;
    }
}
int get_sum(int k,int n){
    int s=1,ans=0;
    for(int i=0;i<=k+1;i++){
        ans=(ans+(ll)S[k][i]*s)%mod;
        s=(ll)s*n%mod;
    }
    return ans;
}
Data dfs(int n,int a,int b,int c){
    Data ans;
    if (n<0)return ans;
    if (a>=c){
        Data s=dfs(n,a%c,b,c);
        mi[0]=1;
        for(int i=1;i<N;i++)mi[i]=(ll)mi[i-1]*(a/c)%mod;
        for(int k1=0;k1<=K;k1++)
            for(int k2=0;k1+k2<=K;k2++)
                for(int k=0;k<=k2;k++)ans.a[k1][k2]=(ans.a[k1][k2]+(ll)C[k2][k]*mi[k2-k]%mod*s.a[k1+k2-k][k])%mod;
        return ans;
    }
    if (b>=c){
        Data s=dfs(n,a,b%c,c);
        mi[0]=1;
        for(int i=1;i<N;i++)mi[i]=(ll)mi[i-1]*(b/c)%mod;
        for(int k1=0;k1<=K;k1++)
            for(int k2=0;k1+k2<=K;k2++)
                for(int k=0;k<=k2;k++)ans.a[k1][k2]=(ans.a[k1][k2]+(ll)C[k2][k]*mi[k2-k]%mod*s.a[k1][k])%mod;
        return ans;
    }
    for(int k1=0;k1<=K;k1++)ans.a[k1][0]=get_sum(k1,n);
    if (!a)return ans;
    int nn=((ll)a*n+b)/c;
    Data s=dfs(nn-1,c,c-b-1,a);
    for(int k1=0;k1<=K;k1++)
        for(int k2=1;k1+k2<=K;k2++)
            for(int k=0;k<k2;k++){
                int sum=0;
                for(int kk=0;kk<=k1+1;kk++)sum=(sum+(ll)S[k1][kk]*s.a[k][kk])%mod;
                sum=((ll)get_sum(k,nn-1)*ans.a[k1][0]-sum+mod)%mod;
                sum=(ll)sum*C[k2][k]%mod;
                ans.a[k1][k2]=(ans.a[k1][k2]+sum)%mod;
            }
    return ans;
}
int main(){
    init();
    scanf("%d",&t);
    while (t--){
        scanf("%d%d%d%d%d%d",&n,&a,&b,&c,&k1,&k2);
        printf("%d
",dfs(n,a,b,c).a[k1][k2]);
    }
    return 0;
}