ICPC 2018 焦作网络预赛 L. Poor God Water

【题目链接】

题目意思

有肉，鱼，巧克力三种食物，对于连续的三个食物，
1.这三个食物不能都相同
2.若三种食物都有的情况，巧克力不能在中间
3.如果两边是巧克力，中间不能是肉或鱼，求方案数

样例输入
1
1 2
2 1
1
2
样例输出
0
1

解题思路

杜教BM线性递推！！！
杜教牛逼！！！
使用方法：暴力前几项，放入vector，直接交。

AC代码

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b)
{
	ll res = 1; a %= mod;
	assert(b >= 0); 
	for (; b; b >>= 1)
	{
		if (b & 1)
			res = res * a%mod;
		a = a * a%mod;
	}
	return res;
}
// head

int _, n;
namespace linear_seq 
{
	const int N = 10010;
	ll res[N], base[N], _c[N], _md[N];
	vector<int> Md;
	void mul(ll *a, ll *b, int k) 
	{
		rep(i, 0, k + k) _c[i] = 0;
		rep(i, 0, k) 
			if (a[i]) 
				rep(j, 0, k) 
					_c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
		for (int i = k + k - 1; i >= k; i--) 
			if (_c[i])
				rep(j, 0, SZ(Md)) 
					_c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
		rep(i, 0, k) a[i] = _c[i];
	}
	int solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d\n",SZ(b));
		ll ans = 0, pnt = 0;
		int k = SZ(a);
		assert(SZ(a) == SZ(b));
		rep(i, 0, k) 
			_md[k - 1 - i] = -a[i]; _md[k] = 1;
		Md.clear();
		rep(i, 0, k) 
			if (_md[i] != 0) Md.push_back(i);
		rep(i, 0, k) 
			res[i] = base[i] = 0;
		res[0] = 1;
		while ((1ll << pnt) <= n) pnt++;
		for (int p = pnt; p >= 0; p--) 
		{
			mul(res, res, k);
			if ((n >> p) & 1) 
			{
				for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; res[0] = 0;
				rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
			}
		}
		rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
		if (ans < 0) ans += mod;
		return ans;
	}
	VI BM(VI s) 
	{
		VI C(1, 1), B(1, 1);
		int L = 0, m = 1, b = 1;
		rep(n, 0, SZ(s)) 
		{
			ll d = 0;
			rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
			if (d == 0) ++m;
			else if (2 * L <= n) 
			{
				VI T = C;
				ll c = mod - d * powmod(b, mod - 2) % mod;
				while (SZ(C) < SZ(B) + m) C.pb(0);
				rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
				L = n + 1 - L; B = T; b = d; m = 1;
			}
			else 
			{
				ll c = mod - d * powmod(b, mod - 2) % mod;
				while (SZ(C) < SZ(B) + m) C.pb(0);
				rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
				++m;
			}
		}
		return C;
	}
	int gao(VI a, ll n) 
	{
		VI c = BM(a);
		c.erase(c.begin());
		rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
		return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
	}
};

int main() 
{
    int t;
    scanf("%d",&t);
	while (t--) 
	{
        long long n;
        scanf("%lld",&n);
		vector<int>v({3,9,20,46,106,244,560,1286,2956,6794});
		//VI{1,2,4,7,13,24}
		printf("%lld\n", linear_seq::gao(v, n - 1));
	}
}

//BM