【HDOJ6630】permutation 2（递推）

题意：给定x，y，n，有标号从1到n的n个数组，求合法的排列个数模998244353使得

1：p[1]=x

2：p[n]=y

3：相邻两项的差的绝对值<=2

n<=1e5

思路：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
#define N  1100000
#define M  4100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=998244353,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;

int f[110000];

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}



int main()
{
    int cas=read();
    f[1]=f[2]=f[3]=1;
    rep(i,4,100000) f[i]=(f[i-1]+f[i-3])%MOD;
    while(cas--)
    {
        int n=read(),x=read(),y=read();
        if(x>1) x++;
        if(y<n) y--;
        printf("%d
",f[y-x+1]);
    }

    return 0;
}