CF575A Fibonotci

题意
\(s\{\}\) 是一个循环数列
循环节为 \(n\)，你可以改掉 \(m\) 项，这 \(m\) 项独立，且不影响循环节
考虑线段树维护矩阵，单点修改最多m次，每次矩阵快速幂就完事了

// powered by c++11
// by Isaunoya
#include
#define rep(i , x , y) for(register int i = (x) ; i <= (y) ; ++ i)
#define Rep(i , x , y) for(register int i = (x) ; i >= (y) ; -- i)
using namespace std ;
using db = double ;
using ll = long long ;
using uint = unsigned int ;
#define int long long
using pii = pair < int , int > ;
#define ve vector
#define Tp template
#define all(v) v.begin() , v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
#define fir first
#define sec second
// the cmin && cmax
Tp < class T > void cmax(T & x , const T & y) {
        if(x < y) x = y ;
}
Tp < class T > void cmin(T & x , const T & y) {
        if(x > y) x = y ;
}
// sort , unique , reverse
Tp < class T > void sort(ve < T > & v) {
        sort(all(v)) ;
}
Tp < class T > void unique(ve < T > & v) {
        sort(all(v)) ;
        v.erase(unique(all(v)) , v.end()) ;
}
Tp < class T > void reverse(ve < T > & v) {
        reverse(all(v)) ;
}
const int SZ = 0x191981 ;
struct FILEIN {
    ~ FILEIN () {} char qwq[SZ] , * S = qwq , * T = qwq , ch ;
    char GETC() {
        return (S == T) && (T = (S = qwq) + fread(qwq , 1 , SZ , stdin) , S == T) ? EOF : * S ++ ;
    }
    FILEIN & operator >> (char & c) {
        while(isspace(c = GETC())) ;
        return * this ;
    }
    FILEIN & operator >> (string & s) {
        while(isspace(ch = GETC())) ;
        s = "" + ch ;
        while(! isspace(ch = GETC())) s += ch ;
        return * this ;
    }
    Tp < class T > void read(T & x) {
            bool sign = 1 ;
            while((ch = GETC()) < 0x30) if(ch == 0x2d) sign = 0 ;
            x = (ch ^ 0x30) ;
            while((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30) ;
            x = sign ? x : -x ;
    }
    FILEIN & operator >> (int & x) {
        return read(x) , * this ;
    }
    FILEIN & operator >> (signed & x) {
        return read(x) , * this ;
    }
    FILEIN & operator >> (unsigned & x) {
        return read(x) , * this ;
    }
} in ;
struct FILEOUT {
    const static int LIMIT = 0x114514 ;
    char quq[SZ] , ST[0x114] ;
    signed sz , O ;
    ~ FILEOUT () {
        sz = O = 0 ;
    }
    void flush() {
        fwrite(quq , 1 , O , stdout) ;
        fflush(stdout) ;
        O = 0 ;
    }
    FILEOUT & operator << (char c) {
        return quq[O ++] = c , * this ;
    }
    FILEOUT & operator << (string str) {
        if(O > LIMIT) flush() ;
        for(char c : str) quq[O ++] = c ;
        return * this ;
    }
    Tp < class T > void write(T x) {
            if(O > LIMIT) flush() ;
            if(x < 0) {
                quq[O ++] = 0x2d ;
                x = -x ;
            }
            do {
                ST[++ sz] = x % 0xa ^ 0x30 ;
                x /= 0xa ;
            } while(x) ;
            while(sz) quq[O ++] = ST[sz --] ;
            return ;
    }
    FILEOUT & operator << (int x) {
        return write(x) , * this ;
    }
    FILEOUT & operator << (signed x) {
        return write(x) , * this ;
    }
    FILEOUT & operator << (unsigned x) {
        return write(x) , * this ;
    }
} out ;

int k , p , n , m ;

const int maxn = 5e4 + 45 ;
int s[maxn] ;
struct mat {
    int a[2][2] ;
    ~ mat () {
        clear() ;
    }
    void clear() {
        memset(a , 0 , sizeof(a)) ;
    }

    int * operator [] (int x) {
        return a[x] ;
    }
} ;

mat b[maxn] ;
mat operator * (mat a , mat b) {
    mat c ;
    for(int i = 0 ; i < 2 ; i ++)
        for(int j = 0 ; j < 2 ; j ++)
            c[i][j] = 0 ;
    for(int i = 0 ; i < 2 ; i ++)
        for(int j = 0 ; j < 2 ; j ++)
            for(int k = 0 ; k < 2 ; k ++)
                c[i][j] = (c[i][j] + a[i][k] * b[k][j] % p) % p ;
    return c ;
}

mat operator *= (mat & a , mat b) {
    return a = a * b ;
}

mat qpow(mat f , mat bs , int y) {
    for( ; y ; y >>= 1 , bs *= bs)
        if(y & 1) f *= bs ;
    return f ;
}

mat a[maxn] ;
mat sum[maxn << 2] ;
mat ans ;

void build(int l , int r , int rt) {
    if(l == r) {
        sum[rt][0][0] = 0 , sum[rt][0][1] = s[l - 1] ;
        sum[rt][1][0] = 1 , sum[rt][1][1] = s[l % n] ;
        b[l] = a[l] = sum[rt] ;
        return ;
    }
    int mid = l + r >> 1 ;
    build(l , mid , rt << 1) , build(mid + 1 , r , rt << 1 | 1) ;
    sum[rt] = sum[rt << 1] * sum[rt << 1 | 1] ;
}

void change(int l , int r , int rt , int x , mat v) {
    if(l == r) {
        sum[rt] = v ;
        return ;
    }
    int mid = l + r >> 1 ;
    (x <= mid) ? change(l , mid , rt << 1 , x , v)
    : change(mid + 1 , r , rt << 1 | 1 , x , v) ;
    sum[rt] = sum[rt << 1] * sum[rt << 1 | 1] ;
}

struct que {
    int x , y , k ;
} q[maxn << 1] ;

signed main() {
#ifdef _WIN64
    freopen("testdata.in" , "r" , stdin) ;
#else
    ios_base :: sync_with_stdio(false) ;
    cin.tie(nullptr) , cout.tie(nullptr) ;
#endif
// code begin.
    in >> k >> p >> n ;
    for(int i = 0 ; i < n ; i ++) in >> s[i] ;
    build(1 , n , 1) ;
    in >> m ;
    for(int i = 1 ; i <= m ; i ++)
        in >> q[i].x >> q[i].y ;
    for(int i = 1 ; i <= m ; i ++) {
        q[i].y %= p ;
        q[i].k = 1 ;
        q[i + m] = q[i] ;
        q[i + m].x ++ ;
        q[i + m].k -- ;
    }
    m <<= 1 ;
    sort(q + 1 , q + m + 1 ,
    [](const que & x , const que & y) {
        return x.x < y.x ;
    }) ;
    while(m > 0 && q[m].x > k) -- m ;
    ans.clear() ;
    ans[0][1] = 1 ;
    int lasbl = 0 ;
    for(int i = 1 , j = 1 ; i <= m ; i = ++ j) {
        int bl = (q[i].x - 1) / n ;
        while(j < m && bl == (q[j + 1].x - 1) / n) ++ j ;
        ans = qpow(ans , sum[1] , bl - lasbl) ;
        lasbl = bl ;
        for(int qwq = i ; qwq <= j ; qwq ++) {
            int d = (q[qwq].x - 1) % n + 1 ;
            b[d][q[qwq].k][1] = q[qwq].y ;
            change(1 , n , 1 , d , b[d]) ;
        }
        if(bl == k / n) break ;
        ans = ans * sum[1] ;
        lasbl = bl + 1 ;
        for(int qwq = i ; qwq <= j ; qwq ++) {
            int d = (q[qwq].x - 1) % n + 1 ;
            b[d] = a[d] ;
            change(1 , n , 1 , d , b[d]) ;
        }
    }
    int bl = k / n ;
    ans = qpow(ans , sum[1] , bl - lasbl) ;
    for(int i = 1 ; i <= k % n ; i ++)
        ans *= b[i] ;
    out << ans[0][0] << '\n' ;
    return out.flush() , 0 ;
// code end.
}