ICPC2008哈尔滨-A-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

3
1 -1 2

样例输出

1

题意
构造一个长度为n的序列，有些位置是-，可以填1-200的数字，要使得每个位置都比它左右两侧的最大值小，求方案数

思路
dp
f[i][j][//]表示到第i位，当前数为j，从i-1到i是上升/相等/下降的方案数
显然
f[i][j][]=f[i-][k][]+f[i-][k][]+f[i-][k][]; k<j;
f[i][j][]=f[i-][k][]+f[i-][k][]+f[i-][k][]; k=j
f[i][j][]=f[i-][k][]+f[i-][k][];

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int P=;
const int N=1e5+;
ll f[N][][];
ll sum[][][];
int a[N];
int n;
int main(){
    scanf("%d",&n);
    for (int i=;i<=n;i++) scanf("%d",&a[i]);

    if (a[]==-)
    {
        for (int i=;i<=;i++) f[][i][]=;
    } else f[][a[]][]=;

    for(int i = ; i <= ; i++) sum[][i][] = (sum[][i-][] + f[][i][])%P;

    for (int i=;i<=n;i++) {
        //sum[!(i&1)][0][0] = sum[!(i&1)][0][1] = sum[!(i&1)][0][2] = 0;
        for (int j=;j<=;j++) {
        if (a[i]==- ||  a[i]==j)
        {
            //f[i][j][0]=f[i-1][k][0]+f[i-1][k][1]+f[i-1][k][2]; k<j;
            f[i][j][]=((sum[i&][j-][]+sum[i&][j-][])%P+sum[i&][j-][])%P;
            //f[i][j][1]=f[i-1][k][0]+f[i-1][k][1]+f[i-1][k][2]; k=j
            f[i][j][]=(f[i-][j][]+f[i-][j][]+f[i-][j][])%P;
            //f[i][j][2]=(f[i][j][2]+f[i-1][k][1]+f[i-1][k][2])%p;
            f[i][j][]=((sum[i&][][] - sum[i&][j][] +P)%P + (sum[i&][][] - sum[i&][j][]+P)%P)%P;

        }
        sum[!(i&)][j][] = (sum[!(i&)][j-][] + f[i][j][])%P;
        sum[!(i&)][j][] = (sum[!(i&)][j-][] + f[i][j][])%P;
        sum[!(i&)][j][] = (sum[!(i&)][j-][] + f[i][j][])%P;
        }
    }
   // cout<<f[1][a[1]][0]<<' '<<f[1][a[1]][1]<<' '<<f[1][a[1]][2]<<endl;
    ll ans=;
    if (a[n]==-)
    {
        for (int i=;i<=;i++)
        {
           // printf("f[3][%d][0]=%lld,f[3][%d][1]=%lld,f[3][%d][2]=%lld\n",i,f[3][i][0],i,f[3][i][1],i,f[3][i][2]);
            ans=(ans+f[n][i][]+f[n][i][])%P;
        }
    } else ans=(f[n][a[n]][]+f[n][a[n]][])%P;
    printf("%lld\n",ans);
    return ;
}

k>j
枚举k的话是200**n，所以要前缀和优化……但可能写的过于诡异