2326: [HNOI2011]数学作业

#include<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=(n);i++)
using namespace std;
typedef unsigned long long ll;
ll mod;ll n;
struct ju{
ll a[4][4];ju(){memset(a,0,sizeof(a));}
}ans,tmp;
ju operator *(ju a,ju b){ju c;
rep(k,1,3)rep(i,1,3)rep(j,1,3)c.a[i][j]=(c.a[i][j]+(1ll*a.a[i][k]*b.a[k][j])%mod)%mod;
return c;
}
void solve(ll k,ll t){
rep(i,1,3)rep(j,1,3)tmp.a[i][j]=0;
tmp.a[1][1]=k%mod;
tmp.a[2][1]=tmp.a[2][2]=tmp.a[3][1]=tmp.a[3][2]=tmp.a[3][3]=1ll;
ll p=t-k/10+1;
for(;p;p>>=1){if(p&1)ans=ans*tmp;
tmp=tmp*tmp;
}

}
int main(){//freopen("in.in","r",stdin);
scanf("%lld%lld",&n,&mod);rep(i,1,3)ans.a[i][i]=1ll;
ll t=10;
while(t<=n){
solve(t,t-1);
t=t*10;
}
solve(t,n);
rep(i,1,3)rep(j,1,3)tmp.a[i][j]=0;tmp.a[1][1]=0,tmp.a[1][2]=0,tmp.a[1][3]=1ll; ans=tmp*ans;
printf("%lld\n",ans.a[1][1]);
}
/*f
=f[n-1]*矩阵 矩阵由小到大
f
=矩阵*f[n-1] 矩阵由大到小
*/