GukiZ and Binary Operations

显然我们要拆位， 因为每位都独立， 然后问题就变成能用dp求的东西，然后用矩阵快速幂优化一下。

注意mod为1的情况。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

ull n, k, l, m;
ull op[] = {1, 0, 1, 0};
struct Matrix {
    ull a[4][4];
    Matrix() {
        memset(a, 0, sizeof(a));
    }
    void init() {
        for(int i = 0; i < 4; i++)
            a[i][i] = 1;
    }
    Matrix operator * (const Matrix &B) const {
        Matrix C;
        for(int i = 0; i < 4; i++)
            for(int j = 0; j < 4; j++)
                for(int k = 0; k < 4; k++)
                    C.a[i][j] = (C.a[i][j] + a[i][k] * B.a[k][j]) % m;
        return C;
    }
    Matrix operator ^ (ull b) {
        Matrix C; C.init();
        Matrix A = (*this);
        while(b) {
            if(b & 1) C = C * A;
            A = A * A; b >>= 1;
        }
        return C;
    }
} M;

int main() {
    cin >> n >> k >> l >> m;
    ull ans = 1, ret0 = 0, ret1 = 0;

    M.a[0][0] = 1, M.a[0][1] = 0, M.a[0][2] = 1, M.a[0][3] = 0;
    M.a[1][0] = 0, M.a[1][1] = 1, M.a[1][2] = 0, M.a[1][3] = 1;
    M.a[2][0] = 1, M.a[2][1] = 0, M.a[2][2] = 0, M.a[2][3] = 0;
    M.a[3][0] = 0, M.a[3][1] = 1, M.a[3][2] = 1, M.a[3][3] = 1;
    Matrix tmp = M ^ (n - 1);
    for(int i = 0; i < 4; i += 2)
        for(int j = 0; j < 4; j++)
            ret0 = (ret0 + tmp.a[i][j] * op[j]) % m;
    for(int i = 1; i < 4; i += 2)
        for(int j = 0; j < 4; j++)
            ret1 = (ret1 + tmp.a[i][j] * op[j]) % m;
    for(int i = 0; i < l; i++) {
        if(k >> i & 1) ans = ans * ret1 % m, k ^= (1ll << i);
        else ans = ans * ret0 % m;
    }
    if(k) ans = 0;
    cout << ans % m << "\n";
    return 0;
}

/*
*/