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2019HDU多校训练第五场1007-permutation 2

时间:2019-08-05 20:33:43      阅读:344      评论:0      收藏:0      [点我收藏+]

标签:review   each   names   out   ali   scanf   i++   ber   contain   

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 1N satisfies the following conditions (We denote the i-th number of a permutation by pi):

1. p1=x

2. pN=y

3. for all 1i<N|pipi+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.

1T5000

2N105

1x<yN
 

 

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
 
 
 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 */

 

2019HDU多校训练第五场1007-permutation 2

标签:review   each   names   out   ali   scanf   i++   ber   contain   

原文地址:https://www.cnblogs.com/CharlieWade/p/11305247.html

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