题目链接：https://www.nowcoder.com/questionTerminal/49c5284278974cbda474ec13d8bd86a9
题目大意
略
分析1
为了兼容题目要求，我在 0 位置和 n + 1 位置设置了值为 1 的哨兵，如此一来，前两个条件都可以无视，只需要关注第 3 个条件即可。
我首先想到的第一个DP是从后往前递推（详细见代码注释），不过只能过60%的案例，贴在这里给自己看看，正解在分析2。
代码如下（失败的DP，TLE）
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (n); ++i) 6 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 21 #define ms0(a) memset(a,0,sizeof(a)) 22 #define msI(a) memset(a,inf,sizeof(a)) 23 #define msM(a) memset(a,-1,sizeof(a)) 24 25 #define MP make_pair 26 #define PB push_back 27 #define ft first 28 #define sd second 29 30 template<typename T1, typename T2> 31 istream &operator>>(istream &in, pair<T1, T2> &p) { 32 in >> p.first >> p.second; 33 return in; 34 } 35 36 template<typename T> 37 istream &operator>>(istream &in, vector<T> &v) { 38 for (auto &x: v) 39 in >> x; 40 return in; 41 } 42 43 template<typename T1, typename T2> 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 45 out << "[" << p.first << ", " << p.second << "]" << "\n"; 46 return out; 47 } 48 49 inline int gc(){ 50 static const int BUF = 1e7; 51 static char buf[BUF], *bg = buf + BUF, *ed = bg; 52 53 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 54 return *bg++; 55 } 56 57 inline int ri(){ 58 int x = 0, f = 1, c = gc(); 59 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 60 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 61 return x*f; 62 } 63 64 typedef long long LL; 65 typedef unsigned long long uLL; 66 typedef pair< double, double > PDD; 67 typedef pair< int, int > PII; 68 typedef pair< string, int > PSI; 69 typedef set< int > SI; 70 typedef vector< int > VI; 71 typedef vector< PII > VPII; 72 typedef map< int, int > MII; 73 typedef pair< LL, LL > PLL; 74 typedef vector< LL > VL; 75 typedef vector< VL > VVL; 76 const double EPS = 1e-10; 77 const LL inf = 0x7fffffff; 78 const LL infLL = 0x7fffffffffffffffLL; 79 const LL mod = 998244353; 80 const int maxN = 1e4 + 7; 81 const LL ONE = 1; 82 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 83 const LL oddBits = 0x5555555555555555; 84 85 int n, a[maxN]; 86 // dp[x][y][z]表示x位置的后一个元素为y，再后一个元素为z，这种情况下前x+2长度一共有多少种 87 LL dp[2][207][207]; 88 LL ans = 0; 89 90 void add_mod(LL &a, LL b) { 91 a = (a + b) % mod; 92 } 93 94 /* 95 // x：当前处理到的位置 96 // y：第 x + 1 位置元素的值 97 // z：第 x + 2 位置元素的值 98 unordered_map< LL, int > dp; 99 inline LL dfs(LL x, int y, int z) { 100 if(x == 0) return y <= max(z, a[x]); 101 LL tmp = (x << 16) + (y << 8) + z; 102 if(dp.find(tmp) != dp.end()) return dp[tmp]; 103 LL ret = 0; 104 int s = 1, t = 200 105 if(a[x]) s = t = a[x]; 106 107 For(i, s, t) if(y <= max(z, i)) add_mod(ret, dfs(x - 1, i, y)); 108 109 return dp[tmp] = ret % mod; 110 } 111 */ 112 113 114 115 int main(){ 116 INIT(); 117 cin >> n; 118 For(i, 1, n) cin >> a[i]; 119 a[0] = 1; 120 For(y, 1, 200) For(z, 1, 200) dp[0][y][z] = y <= max(z, a[0]); 121 122 int now = 0; 123 For(x, 1, n) { 124 now = !now; 125 ms0(dp[now]); 126 For(y, 1, 200) { 127 For(z, 1, 200) { 128 int s = 1, t = 200; 129 if(a[x]) s = t = a[x]; 130 For(i, s, t) if(y <= max(z, i)) add_mod(dp[now][y][z], dp[!now][i][y]); 131 } 132 } 133 } 134 135 cout << dp[now][1][1] << endl; 136 return 0; 137 }