Codeforces Round #518 Div. 1没翻车记

　　A：设f[i][j][0/1]为前i个数第i位为j且第i位未满足/已满足限制的方案数。大力dp前缀和优化即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 100010
#define P 998244353
#define K 200
int n,a[N],f[N][K+1][2],g[N][K+1][2];//0 there is limit 
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    f[0][0][0]=1;
    for (int i=0;i<=K;i++) g[0][i][0]=1;
    for (int i=1;i<=n;i++)
    {
    if (a[i]!=-1)
    {
        inc(f[i][a[i]][1],(g[i-1][K][1]-g[i-1][a[i]-1][1]+P)%P);
        inc(f[i][a[i]][1],f[i-1][a[i]][0]);
        inc(f[i][a[i]][0],g[i-1][a[i]-1][0]),inc(f[i][a[i]][0],g[i-1][a[i]-1][1]);
    }
    else
    {
        for (int j=1;j<=K;j++)
        {
            inc(f[i][j][1],(g[i-1][K][1]-g[i-1][j-1][1]+P)%P);
            inc(f[i][j][1],f[i-1][j][0]);
            inc(f[i][j][0],g[i-1][j-1][0]),inc(f[i][j][0],g[i-1][j-1][1]);
        }
    }
    for (int j=1;j<=K;j++)
    g[i][j][0]=(f[i][j][0]+g[i][j-1][0])%P,g[i][j][1]=(f[i][j][1]+g[i][j-1][1])%P;
    }
    if (a[n]!=-1) cout<<f[n][a[n]][1];
    else
    {
        int ans=0;
        for (int i=1;i<=200;i++) inc(ans,f[n][i][1]);
        cout<<ans;
    }
    return 0;
}

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