地斗主

思路

看到这$n$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$n-num,num$我们假定显然对$num=1,2,3,4,5$分别有$1,4,2,3,2,3$种分法，对应到原来一整块的部分上也就是$an{s}_n=an{s}_{n-1}+4an{s}_{n-2}+2an{s}_{n-3}+3an{s}_{n-4}\dots \dots$,并且后面的变化是由$2,3,2,3$不断循环下去的，所以我们只要将$an{s}_n-an{s}_{n-1}$即可得到递推式$an{s}_n=an{s}_{n-1}+5an{s}_{n-2}+an{s}_{n-3}-an{s}_{n-4}$

接下来就是这么一个简单的矩阵乘法了
$$\left[\begin{array}{cccc}{a}_4& a_3& a_2& a_1\end{array}\right]\left[\begin{array}{cccc}1& 1& 0& 0\\ 5& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right]$$

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 | 1using 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;
}