我又對著跑出正解的程式調了好久

怕不是眼瞎了

這就是個分段矩陣，我們很容易就得到了遞推式

$\(f[i]=f[i-1]*10^k+i\)

其中\(k=log_{10}i\)

於是就是分段矩陣

之後就是程式碼了，沒有加快速乘WA了好久

#include<iostream>
#include<cstring>
#include<cstdio>
#define re register
#define LL long long
LL n,m;
LL ans[4][4],a[4][4];
LL Ans=0;
inline LL mul(LL a,LL b)
{
    LL s=0;
    while(b)
    {
        if(b&1ll) s=s+a%m;
        b>>=1ll;
        a=a+a%m;
    }
    return s;
}
inline void did_a()
{
    LL mid[4][4];
    for(re int i=1;i<=3;i++)
        for(re int j=1;j<=3;j++)
            mid[i][j]=a[i][j],a[i][j]=0;
    for(re int i=1;i<=3;i++)
        for(re int j=1;j<=3;j++)
            for(re int p=1;p<=3;p++)
                a[i][j]=(a[i][j]+mul(mid[i][p],mid[p][j])%m)%m;
}
inline void did_ans()
{
    LL mid[4][4];
    for(re int i=1;i<=3;i++)
        for(re int j=1;j<=3;j++)
            mid[i][j]=ans[i][j],ans[i][j]=0;
    for(re int i=1;i<=3;i++)
        for(re int j=1;j<=3;j++)
            for(re int p=1;p<=3;p++)
                ans[i][j]=(ans[i][j]+mul(mid[i][p],a[p][j])%m)%m;
}
inline void Rebuild(LL t)
{
    memset(a,0,sizeof(a)),memset(ans,0,sizeof(ans));
    ans[1][1]=a[1][1]=1ll;
    ans[2][1]=a[2][1]=1ll;
    ans[2][2]=a[2][2]=1ll;
    ans[3][2]=a[3][2]=1ll;
    ans[3][3]=a[3][3]=t;
}
inline void Quick(LL b)
{
    while(b)
    {
        if(b&1ll) did_ans();
        b>>=1ll;
        did_a();
    }
}
inline void work()
{
    LL now=1;
    LL t=10;
    while(now<=n)
    {
        if(t<0) return;
        Rebuild(t);
        if(n>=t-1) Quick(t-1-now);
        else Quick(n-now);
        Ans=(ans[3][1]%m+mul(now,ans[3][2])%m+mul(Ans,ans[3][3])%m)%m;
        now=t;
        t*=10;
    }
}
int main()
{
    scanf("%lld%lld",&n,&m);
    work();
    printf("%lld\n",Ans);
    return 0;
}