P2467 [SDOI2010]地精部落 DP

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;
typedef pair<ll,int>pli;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+9;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}
/*-----------------------showtime----------------------*/

            const int maxn = 15009;
            ll dp[3][maxn];
int main(){
          //  freopen("goblin.in", "r", stdin);
           // freopen("goblin.out", "w", stdout);
            int n,p;
            scanf("%d%d", &n, &p);
            int id = 0;
            if(n == 3) cout<<4 % p<<endl;
            else {
                    dp[id][1] = 0;  dp[id][2] = 1; dp[id][3] = 1;

                    for(int i=4; i<=n; i++){
                        for(int j=2; j<=i; j++){
                            dp[id^1][j] = (dp[id^1][j-1] + dp[id][i - j + 1])%p;
                        }
                        id = id ^ 1;
                    }
                    ll ans = 0;
                    for(int i=2; i<=n; i++) ans =(ans + dp[id][i])% p;
                    printf("%lld\n", ans * 2 % p);
            }
            return 0;
}