BZOJ2326 [HNOI2011]数学作业(分块矩阵快速幂)

题意：

定义函数Concatenate (1 ..N)是将所有正整数 1, 2, …, N 顺序连接起来得到的数，如concatenate(1..5)是12345，求concatenate(1...n)%m的值

思路：

矩阵快速幂，公式为

$$
\left[
\begin{matrix}
f(n)\\n\\1
\end{matrix}
\right]=
\left[
\begin{matrix}
10^k&1&1\\
0&1&1\\
0&0&1
\end{matrix}
\right]
\left[
\begin{matrix}
f(n-1)\\n-1\\1
\end{matrix}
\right]
$$

其中$10^k$可以分快处理，分n为0~9,10~99,100~999等

代码：

/**************************************************************
    Problem: 2326
    User: wrjlinkkkkkk
    Language: C++
    Result: Accepted
    Time:64 ms
    Memory:1292 kb
****************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
#include<list>

#define fst first
#define sc second
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))
#pragma Gcc optimize(2)

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const int maxn =  + ;
const int maxm = 5e3 + ;
const double eps = 1e-;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
int scan(){
    int res=,ch,flag=;
    if((ch=getchar())=='-')
        flag=;
    else if(ch>=''&&ch<='')
        res=ch-'';
    while((ch=getchar())>=''&&ch<='')
        res=res*+ch-'';
    return flag?-res:res;
}

ll n, m;

ll mul(ll x,ll y){
    ll s=;
    while(y)
    {
        if(y&)s=(s+x)%m;
        x=(x<<)%m;
        y>>=;
    }
    return s;
}
void mtpl(ll a[][], ll b[][],ll s[][]){
    ll tmp[][];
    for(int i = ; i <= ; i++){
        for(int j = ; j <= ; j++){
            tmp[i][j]=;
            for(int k = ; k <= ; k++){
                tmp[i][j] += mul(a[i][k] , b[k][j])%m;
            }

        }

    }
    for(int i = ; i <= ; i++){
        for(int j = ; j <= ; j++){
            s[i][j] = tmp[i][j];
        }
    }
    return;
}
ll b[][];
void Fast_power(ll t,ll last){
    ll a[][];
    mem(a, );
    a[][] = t;
    a[][] = a[][] = a[][] = a[][] = a[][] = ;
    ll y = last-t/+;
    while(y){
        if(y & ) mtpl(b, a, b);
        mtpl(a, a, a);
        y >>= ;
    }
    //prt(ans);
    return;
}
int main(){

    scanf("%lld %lld", &n, &m);

    mem(b, );
    for(int i = ; i <= ; i++)b[i][i] = ;
    ll t = ;
    while(t <= n){
        Fast_power(t, t-);
        t*=;

    }
    Fast_power(t,n);
    printf("%lld", b[][]%m);
    return ;
}
/*
8 3287492749272074
*/