Array Without Local Maximums

题目描述

Ivan unexpectedly saw a present from one of his previous birthdays. It is array of n numbers from 1 to 200. Array is old and some numbers are hard to read. Ivan remembers that for all elements at least one of its neighbours ls not less than it, more formally:
a1≤a2,
an≤an−1 and
ai≤max(ai−1,ai+1) for all i from 2 to n−1.
Ivan does not remember the array and asks to find the number of ways to restore it. Restored elements also should be integers from 1 to 200. Since the number of ways can be big, print it modulo 998244353.

 

输入

First line of input contains one integer n (2≤n≤105) — size of the array.

Second line of input contains n integers ai — elements of array. Either ai=−1 or 1≤ai≤200. ai=−1 means that i-th element can't be read.

 

输出

Print number of ways to restore the array modulo 998244353.

 

样例输入

复制样例数据

3
1 -1 2

样例输出

参考：https://www.luogu.org/problemnew/solution/CF1067A

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll mod=1e9+7;
const int modp=998244353;
const int maxn=1e6+50;
const double eps=1e-6;
#define lowbit(x)  x&(-x)
#define INF 0x3f3f3f3f
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
int dcmp(double x)
{
    if(fabs(x)<eps)return 0;
    return (x>0)?1:-1;
}
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
ll qmod(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=(ans*a)%mod;
        }
        b>>=1;
        a=(a*a)%mod;
    }
    return ans;
}
ll a[maxn],dp[2][205][3];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        a[i]=read();
    }

    int k=0;
    if(a[1]==-1){
        for(int i=1;i<=200;i++){
            dp[k][i][0]=1;
        }
    }
    else{
        dp[k][a[1]][0]=1;
    }
    ll sum=0;
    for(int i=2;i<=n;i++){
        //1
        sum=0;
        for(int j=1;j<=200;j++){
            if(a[i]==-1||a[i]==j){
                dp[k^1][j][0]=sum;
            }
            else{
                dp[k^1][j][0]=0;
            }
            sum+=(dp[k][j][0]+dp[k][j][1]+dp[k][j][2])%modp;
            sum%=modp;
        }
        //2
        for(int j=1;j<=200;j++){
            if(a[i]==-1||a[i]==j){
                dp[k^1][j][1]=(dp[k][j][0]+dp[k][j][1]+dp[k][j][2])%modp;
            }
            else{
                dp[k^1][j][1]=0;
            }
        }
        //3
        sum=0;
        for(int j=200;j>=1;j--){
            if(a[i]==-1||a[i]==j){
                dp[k^1][j][2]=sum;
            }
            else{
                dp[k^1][j][2]=0;
            }
            sum+=(dp[k][j][1]+dp[k][j][2])%modp;
            sum%=modp;
        }

        k^=1;
    }
    ll ans=0;
    for(int i=1;i<=200;i++){
        ans+=(dp[k][i][1]+dp[k][i][2])%modp;
        ans%=modp;
    }
    printf("%lld\n",ans);
    return 0;
}