[HDU6146]Pokémon GO

Problem

有一个2n的方格矩阵
在一个格子上可以往旁边8个方向走(如果有格子),求有多少方案把2
n走完

Solution

我们用Fi表示从一个角出发走遍所有格子回到这一列另外一点的方案数
显然,F1 = 1,Fn = 2 * Fn-1 = 2^(n-1)
[HDU6146]Pokémon GO_第1张图片
我们再用Gi表示从一个角出发,走遍所有格子的方案数
那么Gi = Fi + 2 * Gi-1 + 4 * Gi-2
[HDU6146]Pokémon GO_第2张图片
所以四个角出发的方案便为4Gn
我们再考虑从中间的列出发的方案:
每一列可以从上下两个格子出发,然后有2
(2 * Fi * Gn−i + 2 * Fn−i+1 * Gi−1)种方案

Notice

要考虑全所有情况

Code

#include
#include
#include
#include
#include
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 10000, mo = INF + 7;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
ll D[N + 5], E[N + 5];
int main()
{
    int T_T = read();
    D[1] = 1;
    rep(i, 2, N) D[i] = D[i - 1] * 2 % mo;
    E[0] = 0, E[1] = 1, E[2] = 6;
    rep(i, 3, N) E[i] = (D[i] + E[i - 1] * 2 + E[i - 2] * 4) % mo;
    rep(i, 1, T_T)
    {
        int n = read();
        if (n == 1)
        {
            puts("2");
            continue;
        }
        ll ans = E[n];
        rep(j, 2, n - 1) ans = (ans + D[j] * E[n - j] + D[n - j + 1] * E[j - 1]) % mo;
        write(ans * 4 % mo); puts("");
    }
}

转载于:https://www.cnblogs.com/WizardCowboy/p/7754952.html

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