"Hnoi Analog by YMD" move

Description

Set P (n) to the number of different paths from (0,0) to the point (n,0), moving in the following three ways: (x, Y), (x+1,y-1), (x, y), (X+1,y), (X+y), And the path cannot intersect with the fourth quadrant. P (n) is obtained and the 10^9+7 is modeled.

Input

The first line is an integer t that represents the number of data groups.

For each set of data, one line is an integer n.

Output

For each set of data, output the answer.

Data Range

20%:n≤10;
50%:n≤10000;
100%:n≤106,t≤10.

Solution

20%

Direct O (3^n*n) is a brute force enumeration.

50%

Consider the point of the first line y=0, assuming it is (i,0).

Then the path from (to i-1,1) is above the line Y=1, obviously there are P (i-2) species. Similarly, the path from (x,0) to (n,0) is above the line y=0, with the P (n-i) species.

So, all together you can get P (n) = Σni=1 P (i-2) p (n-i), wherein the specified p ( -1) =p (0) = 1.

In fact, you can make a table on the basis of 20%, the examination room has more than 5 people to write ...

100%

Assuming that there is a total of I (x, y) (x+1,y+1) in the path of the move, then there is bound to be an I (x, y) (x+1,y-1), n-2i (x, y) (x+1,y).

and if all (x, y)-x+1,y is not considered, then the number of paths is the number of Catalan, set to C_i. If you join (x, y) (X+1,y), which is to insert n-2i the same ball in 2i+1 gouges, the scheme number is C2iN.

So, P (n) = Σni=1 c_i C2in.

#include <algorithm>

#include <iostream>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <cmath>

using namespace Std;

typedef long Long LL;

#define N 1000010

#define MOD 1000000007

int t;

int n;

LL F[n],d[n];

inline ll Qpow (ll A,ll b)

{

LL Ans=1;

while (b)

{

if (b&1)

Ans= (1ll*ans*a)%mod;

b>>=1;

A= (1ll*a*a)%mod;

}

return ans;

}

inline ll C (ll X,ll y)

{

return f[x]*d[y]%mod*d[x-y]%mod;

}

inline ll Catalan (ll x)

{

Return (C (x<<1,x)-C (x<<1,x-1) +mod)%mod;

}

int main ()

{

Freopen ("Move.in", "R", stdin); Freopen ("Move.out", "w", stdout);

scanf ("%d", &t);

F[0]=1;

for (int i=1;i<=1000000;i++)

F[i]=f[i-1]*i%mod;

for (int i=0;i<=1000000;i++)

D[i]=qpow (f[i],mod-2)%mod;

while (t--)

{

scanf ("%d", &n);

LL ans=0;

for (int i=0;i<=n;i+=2)

Ans+=c (n,i) *catalan (i>>1), ans%=mod;

printf ("%lld\n", ans);

}

return 0;

}

2.1 20%
†O(3N∗N)Åqþ=œ "
2.2 50%
ä1˜g† 'y= 0: §b´(I,0)"
@ol(1,1)(I−1,1)ƒm´»þ3† 'y= 1Þ §w,kP(I−
2)«"Ón§l(x,0)(N,0)ƒm´»þ3† 'y= 0Þ §kP(N−
I)«"
¤±§üüå5œ±P(N) =Pn i=1P(I−2)P(N−I)§ù¥§5
½P(−1) =P(0) = 1"
2.3 100%
B£ä´»¥˜kI‡(x, y)−>(x+ 1, y+ 1)§@oò˜½¬
KI‡(x, y)−>(x+ 1, y−1)§N−2I‡(x, y)−>(x+ 1, y)"
Xjø ä¤k(x, y)−>(x+ 1, y)§@o´»«êò´
1I‡CatalanʧCI"Xj\\(x, y)−>(x+ 1, y)§ ò´32I+ 1‡
˜?¥\N−2IƑó¥§ Yê´C2I
N"
¤±§P(n) = pn i=1 Ci Cn 2i"

"Hnoi Analog by YMD" move