10pts求条

马蜂不是很好，记录

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll num[12][12];
ll ans[12][12];
ll tmp[12][12];
ll fin[12];
ll n, m, p, q, r, t, u, v, w, x, y, z;

ll qmul(ll a, ll b){
    ll res = 0;
    ll flag = 1;
    if(b < 0){
        flag = -1;
        b = -b;
    }
    while (b != 0){
        if (b & 1) res = (res + a) % m;
        a = (a << 1) % m;
        b >>= 1;
    }
    return res * flag;
}

void qpow(ll tme){
    for (int i = 1; i <= 11; i++){
        for (int j = 1; j <= 11; j++)
            ans[i][j] = 0;
        ans[i][i] = 1;
    }
    while (tme != 0){
        if (tme & 1){
            for (int i = 1; i <= 11; i++)
                for (int j = 1; j <= 11; j++){
                    tmp[i][j] = 0;
                    for (int k = 1; k <= 11; k++)
                        tmp[i][j] = (tmp[i][j] + qmul(ans[i][k], num[k][j]) % m) % m;
                }
            for (int i = 1; i <= 11; i++)
                for (int j = 1; j <= 11; j++)
                    ans[i][j] = tmp[i][j];
        }
        for (int i = 1; i <= 11; i++)
            for (int j = 1; j <= 11; j++){
                tmp[i][j] = 0;
                for (int k = 1; k <= 11; k++)
                    tmp[i][j] = (tmp[i][j] + qmul(num[i][k], num[k][j]) % m) % m;
            }
        for (int i = 1; i <= 11; i++)
            for (int j = 1; j <= 11; j++)
                num[i][j] = tmp[i][j];
        tme >>= 1;
    }
    return;
}

int main(){
    cin >> n >> m >> p >> q >> r >> t >> u >> v >> w >> x >> y >> z;
    if (n == 1){
        printf("nodgd 1\nCiocio 1\nNicole 1");
        return 0;
    }
    if (n == 2){
        printf("nodgd 3\nCiocio 3\nNicole 3");
        return 0;
    }
    num[1][1] = p; num[1][2] = q; num[1][3] = 1; num[1][5] = 1; num[1][7] = r; num[1][8] = t; num[1][11] = 1;
    num[2][1] = 1;
    num[3][1] = 1; num[3][3] = u; num[3][4] = v; num[3][5] = 1; num[3][9] = 1;
    num[4][3] = 1;
    num[5][1] = 1; num[5][3] = 1; num[5][5] = x; num[5][6] = y; num[5][8] = 1; num[5][10] = 1; num[5][11] = z;
    num[6][5] = 1;
    num[7][7] = 1; num[7][8] = 2; num[7][11] = 1;
    num[8][8] = 1; num[8][11] = 1;
    num[9][9] = w;
    num[10][10] = z;
    num[11][11] = 1;
    fin[1] = fin[3] = fin[5] = 3;
    fin[2] = fin[4] = fin[6] = fin[7] = fin[8] = fin[11] = 1;
    fin[9] = w;
    fin[10] = z;
    qpow(n - 3);
    ll ans1 = 0, ans2 = 0, ans3 = 0;
    for (int i = 1; i <= 11; i++){
        ans1 = (ans1 + fin[i] * num[1][i] % m) % m;
        ans2 = (ans2 + fin[i] * num[3][i] % m) % m;
        ans3 = (ans3 + fin[i] * num[5][i] % m) % m;
    }
    printf("nodgd %lld\nCiocio %lld\nNicole %lld", ans1, ans2, ans3);
    return 0;
}