BZOJ 2326: [HNOI2011]数学作业

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int m;
struct matrix{
ll a[3][3];
void print(){
for(int i=0;i<3;i++){
for(int j=0;j<3;j++)
printf("%lld ",a[i][j]);
putchar('\n');
}
}
void clear(){memset(a,0,sizeof(a));}
matrix(){clear();}
matrix operator * (matrix b){
static matrix ans;
ans.clear();
for(int i=0;i<3;i++)
for(int j=0;j<3;j++)
for(int k=0;k<3;k++)
ans.a[i][j]+=a[i][k]*b.a[k][j];
for(int i=0;i<3;i++)
for(int j=0;j<3;j++)
ans.a[i][j]%=m;
return ans;
}
matrix operator ^ (ll k){
static matrix ans,b;
ans.clear();
for(int i=0;i<3;i++)ans.a[i][i]=1;
memcpy(b.a,a,sizeof(a));
for(;k;k>>=1,b=b*b)if(k&1)ans=ans*b;
return ans;
}
}f,g;
int calc(ll n){
f.a[0][2]=1;
g.a[1][0]=g.a[1][1]=g.a[2][0]=g.a[2][1]=g.a[2][2]=1;
ll l=1,k=10%m;
while(true){
ll r=l*10-1;
if(r>n)r=n;
g.a[0][0]=k;
f=f*(g^(r-l+1));
if(r==n)break;
k=k*10%m;
l=r+1;
}
return f.a[0][0];
}
int main(){
//freopen("a.in","r",stdin);
ll n;scanf("%lld%d",&n,&m);
printf("%d\n",calc(n));
return 0;
}