2019HDU多校训练第五场1007-permutation 2

permutation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

 

Problem Description

You are given three positive integers N,x,y.
Please calculate how many permutations of 1∼N satisfies the following conditions (We denote the i-th number of a permutation by pi):

1. p1=x

2. pN=y

3. for all 1≤i<N, |pi−pi+1|≤2

 

Input

The first line contains one integer T denoting the number of tests.

For each test, there is one line containing three integers N,x,y.

* 1≤T≤5000

* 2≤N≤105

* 1≤x<y≤N

 

 

Output

For each test, output one integer in a single line indicating the answer modulo 998244353.

 

Sample Input

3

4 1 4

4 2 4

100000 514 51144

 

Sample Output

2

1

253604680

 

 emmm代码很短，应该很容易看懂。。。。

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 #define maxn 100005
 5 #define mod 998244353
 6 ll fac[maxn];
 7 int main()
 8 {
 9     fac[0]=0;
10     fac[1]=1;
11     fac[2]=1;
12     for(int i=3; i<=maxn-2; i++)
13     {
14         fac[i]=(fac[i-1]+fac[i-3])%mod;
15     }
16     int t;
17     scanf("%d",&t);
18     while(t--)
19     {
20         int n,a,b;
21         scanf("%d%d%d",&n,&a,&b);
22         int ans=b-a;
23         if(a==1&&b==n)ans++;
24         else if(a!=1&&b!=n)ans--;
25         printf("%lld\n",fac[ans]);
26     }
27     return 0;
28 }
29 /*
30 3
31 4 1 4
32 4 2 4
33 100000 514 51144
34 */

 

转载于:https://www.cnblogs.com/CharlieWade/p/11305247.html