HDU-4565

思路

这里

太难想了www

Code

#include <bits/stdc++.h>
using namespace std;
#define _u_u_ ios::sync_with_stdio(false), cin.tie(nullptr)
#define cf int _o_o_;cin>>_o_o_;for (int Case = 1; Case <= _o_o_;Case++)
#define SZ(x) (int)(x.size())
inline void _A_A_();
signed main() {_A_A_();return 0;}

using ll = long long;
#define int long long
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
const int N = 2, M = 5010;
const int inf = 0x3f3f3f3f;

int a, b, n, m;

struct mat
{
    int m[N][N];
};

mat operator * (const mat&a, const mat &b) {
    mat c;
    memset(c.m, 0, sizeof c.m);
    for (int i = 0;i < N;i++) {
        for (int j = 0;j < N;j++) {
            for(int k = 0;k < N;k++) {
                c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j] + m) % m;
            }
        }
    }
    return c;
}

mat qpow(mat aa, int bb) {
    mat c;
    memset(c.m, 0, sizeof c.m);
    for (int i = 0;i < N;i++) c.m[i][i] = 1;
    while (bb) {
        if (bb & 1) {
            c = c * aa;
        }
        aa = aa * aa;
        bb >>= 1;
    }
    return c;
}


inline void _A_A_() {
    #ifdef LOCAL
    freopen("in.in", "r", stdin);
    #endif
    _u_u_;
    mat s;
    while (cin >> a >> b >> n>> m) {
        memset(s.m, 0, sizeof s.m);
        s.m[0][0] = 2 * a % m;
        s.m[0][1] = 1 % m;
        s.m[1][0] = (b - a * a + m) % m;
        s = qpow(s, n);
        cout << (s.m[0][1] * 2 * a + 2 * s.m[1][1] + m ) % m << "\n";
    }
}