ASC7 Problem B. Cryptography

题目大意

给你$n$个矩阵，回答模$r$下$m$个区间内矩阵的乘积。

简要题解

分块傻题。

#include <bits/stdc++.h>
using namespace std;
namespace my_header {
#define pb push_back
#define mp make_pair
#define pir pair<int, int>
#define vec vector<int>
#define pc putchar
#define clr(t) memset(t, 0, sizeof t)
#define pse(t, v) memset(t, v, sizeof t)
#define bl puts("")
#define wn(x) wr(x), bl
#define ws(x) wr(x), pc(' ')
    const int INF = 0x3f3f3f3f;
    typedef long long LL;
    typedef double DB;
    inline char gchar() {
        char ret = getchar();
        for(; (ret == '
' || ret == '
' || ret == ' ') && ret != EOF; ret = getchar());
        return ret; }
    template<class T> inline void fr(T &ret, char c = ' ', int flg = 1) {
        for(c = getchar(); (c < '0' || '9' < c) && c != '-'; c = getchar());
        if (c == '-') { flg = -1; c = getchar(); }
        for(ret = 0; '0' <= c && c <= '9'; c = getchar())
            ret = ret * 10 + c - '0';
        ret = ret * flg; }
    inline int fr() { int t; fr(t); return t; }
    template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); }
    template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); }
    template<class T> inline char wr(T a, int b = 10, bool p = 1) {
        return a < 0 ? pc('-'), wr(-a, b, 0) : (a == 0 ? (p ? pc('0') : p) : 
            (wr(a/b, b, 0), pc('0' + a % b)));
    }
    template<class T> inline void wt(T a) { wn(a); }
    template<class T> inline void wt(T a, T b) { ws(a), wn(b); }
    template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); }
    template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); }
    template<class T> inline T gcd(T a, T b) {
        return b == 0 ? a : gcd(b, a % b); }
    template<class T> inline T fpw(T b, T i, T _m, T r = 1) {
        for(; i; i >>= 1, b = b * b % _m)
            if(i & 1) r = r * b % _m;
        return r; }
};
using namespace my_header;

const int MAXN = 30000 + 100;
const int NBLK = 200;
int r, n, m;

struct Matrix {
    int d[2][2];
    Matrix() {
        memset(d, 0, sizeof d);
    }
    int *operator [] (int a) {
        return d[a];
    }
    void read() {
        fr(d[0][0], d[0][1]);
        fr(d[1][0], d[1][1]);
    }
} mat[MAXN], blk[400];
Matrix operator * (Matrix a, Matrix b) {
    Matrix c;
    for (int i = 0; i < 2; ++i)
        for (int j = 0; j < 2; ++j)
            for (int k = 0; k < 2; ++k)
                (c[i][j] += a[i][k] * b[k][j]) %= r;
    return c;
}

int main() {
#ifdef lol
    freopen("B.in", "r", stdin);
    freopen("B.out", "w", stdout);
#else
    freopen("crypto.in", "r", stdin);
    freopen("crypto.out", "w", stdout);
#endif

    fr(r, n, m);
    int cb = 0;
    blk[cb][0][0] = blk[cb][1][1] = 1;
    for (int i = 1; i <= n; ++i) {
        if ((i - 1) % NBLK == 0) {
            ++cb;
            blk[cb][0][0] = blk[cb][1][1] = 1;
        }
        mat[i].read();
        blk[cb] = blk[cb] * mat[i];
    }
    while (m--) {
        int l, r;
        fr(l, r);
        Matrix ret;
        ret[0][0] = ret[1][1] = 1;
        for (int j = l; j <= r; ++j) {
            if (j % NBLK == 1 && j + 199 <= r) {
                ret = ret * blk[j / NBLK + 1];
                j += 199;
            } else {
                ret = ret * mat[j];
            }
        }
        wt(ret[0][0], ret[0][1]);
        wt(ret[1][0], ret[1][1]);
        bl;
    }

    return 0;
}