标签：矩阵乘法
题目
分析
设f[n]表示Concatenate(1,n)
那么
\[f[i]=f[i-1]*10+(i-1)+1 \ \ (1\leq i\leq 9)\] \[f[i]=f[i-1]*100+(i-1)+1 \ \ (10\leq i\leq 99)\]可以用矩阵加速
code
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
#define lson (x<<1)
#define rson (x<<1|1)
#define mid ((l+r)>>1)
#define pi acos(-1)
#define mp make_pair
#define pb push_back
#define pa pair<int,int>
#define fi first
#define se second
#define lowbit(x) ((x)&(-x))
using namespace std;
inline ll read(){
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline ll getgcd(ll x,ll y){return !x?y:getgcd(y%x,x);}
inline ll getlcm(ll x,ll y){return x/getgcd(x,y)*y;}
inline ll mul(ll x,ll y,ll p){
ll re=0;
while(y){
if(y&1)re=(re+x)%p;
y>>=1,x=(x<<1)%p;
}
return re;
}
//**********head by yjjr**********
ll n,mod,a[4][4],b[4][4],t=10;
void Mul(ll a[4][4],ll b[4][4],ll s[4][4]){
ll tmp[4][4];
rep(i,1,3)
rep(j,1,3){
tmp[i][j]=0;
rep(k,1,3)tmp[i][j]=(tmp[i][j]+mul(a[i][k],b[k][j],mod))%mod;
}
rep(i,1,3)
rep(j,1,3)s[i][j]=tmp[i][j];
}
void cal(ll t,ll last){
mem(b,0);
b[1][1]=t;b[2][1]=b[2][2]=b[3][1]=b[3][2]=b[3][3]=1;
ll y=last-t/10+1;
while(y){
if(y&1)Mul(a,b,a);
Mul(b,b,b),y>>=1;
}
}
int main()
{
n=read(),mod=read();
rep(i,1,3)a[i][i]=1;
while(n>=t){
cal(t,t-1);
t*=10;
}
cal(t,n);
cout<<a[3][1]<<endl;
return 0;
}
