我又对着跑出正解的程序调了好久
怕不是眼瞎了
这就是个分段矩阵，我们很容易就得到了递推式
[f[i]=f[i-1]*10^k+i$
其中$k=log_{10}i$
于是就是分段矩阵
之后就是代码了，没有加快速乘WA了好久
```cpp
#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
",Ans);
return 0;
}
```]