地斗主

地斗主

思路

看到这非常大，感觉一定是个结论公式题，但是感觉又不像是排列组合，于是可以考虑矩阵快速幂了，所以关键就是得得到递推公式了。

我们将棋盘分成两部分我们假定显然对分别有种分法，对应到原来一整块的部分上也就是,并且后面的变化是由不断循环下去的，所以我们只要将即可得到递推式

接下来就是这么一个简单的矩阵乘法了

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

int mod, n;

struct matrix {
    ll a[4][4];

    matrix operator * (matrix & t) {
        matrix temp;
        for(int i = 0; i < 4; i++) {
            for(int j = 0; j < 4; j++) {
                temp.a[i][j] = 0;
                for(int k = 0; k < 4; k++) {
                    temp.a[i][j] = (temp.a[i][j] + a[i][k] * t.a[k][j]) % mod;
                }
            }
        }
        return temp;
    }
};

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    int T = read();
    while(T--) {
        n = read(), mod = read();
        if(n <= 4) {
            if(n == 1) printf("%d\n", 1);
            else if(n == 2) printf("%d\n", 5);
            else if(n == 3) printf("%d\n", 11);
            else printf("%d\n", 36);
            continue;
        }
        matrix ans = {36, 11, 5, 1};
        matrix a = {1, 1, 0, 0,
                    5, 0, 1, 0,
                    1, 0, 0, 1,
                    -1, 0, 0, 0};
        n -= 4;
        while(n) {
            if(n & 1) ans = ans * a;
            a = a * a;
            n >>= 1;
        }
        printf("%lld\n", (ans.a[0][0] % mod + mod) % mod);
    }
    return 0;
}