Codeforces 1312 D. Count the Arrays (combinatorics)

E. Array Shrinking
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array a1,a2,…,an. You can perform the following operation any number of times:

Choose a pair of two neighboring equal elements ai=ai+1 (if there is at least one such pair).
Replace them by one element with value ai+1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?

Input
The first line contains the single integer n (1≤n≤500) — the initial length of the array a.

The second line contains n integers a1,a2,…,an (1≤ai≤1000) — the initial array a.

Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.

Examples
inputCopy
5
4 3 2 2 3
outputCopy
2
inputCopy
7
3 3 4 4 4 3 3
outputCopy
2
inputCopy
3
1 3 5
outputCopy
3
inputCopy
1
1000
outputCopy
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.

In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.

In the third and fourth tests, you can’t perform the operation at all.

The meaning of problems:
in which m number of pick number n, the number n of first decremented is incremented, and must have the same number of two, the remaining number varies.

Ideas:

1. It is clear that a combination of mathematics.
2. First, pick which m is the number of n-1 taken Cm n-1.
3. Then is to ensure that the largest element is not on both sides, the remaining n-2 n-3 elements have different elements can be placed on the left or right of the largest element, so is the (n-2) * qpow (2, n- 3).
4. Because the range where n is greater than or equal to 2, the power can not be processed quickly negative power, so there may be employed Laid judged treated with n = 2 or the inverse, see the codes.
5. There is a fast power can not bring fast ride, I do not know the specific reasons, when the game was quickly multiply out of overtime, rewrite fast power just over.
6. If there is any talk of unclear welcome to welcome the exchange ~

Code:

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
#define fi first
#define se second
#define lowbit(x) (x&(-x))
#define mp make_pair
#define pb push_back
#define ins insert
#define si size()
#define E exp(1.0)
#define fixed cout.setf(ios::fixed)
#define fixeds(x) setprecision(x)
#pragma GCC optimize(2)
using namespace std;
inline ll read(){ll s=0,w=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;}void put1(){ puts("YES") ;}void put2(){ puts("NO") ;}
void put3(){ puts("-1"); }//��=acos(L/2R);
//void debug1(){ cout<<"I CAN AC"<<endl;}//void debug2(){printf("T A T\n");}
ll qp(ll a,ll b, ll p){ll ans = 1;while(b){if(b&1){
ans = (ans*a)%p;--b;}a = (a*a)%p;b >>= 1;}return ans%p;}
const int manx=1e3+5;

ll Inv(ll x){
	return qp(x,mod-2,mod);
}
ll Cal(ll n,ll m){
	if (m>n) return 0;
	ll ans = 1;
	for (int i = 1; i <= m; ++i) ans=ans*Inv(i)%mod*(n-i+1)%mod;
	return ans%mod;
}
int main()
{
    ll n=read(),m=read();
    ll ans=Cal(m,n-1);
    ans%=mod;
    /*特判
    if(n>2) ans=ans*qp(2,n-3,mod)%mod*(n-2)%mod;
    else ans=0;
    */
    //逆元
    ans=ans*qp(2,n-2,mod)%mod*(n-2)%mod*Inv(2);
    ans=(ans+mod)%mod;
    printf("%lld\n",ans);
    return 0;
}