考虑从最后的队形开始依次还原最初的队形。
对于当前的队形，要么选最左边的，要么选最右边的。 如果选了左边的，那么下次选择的一定是大于它的。右边的同理。
所以定义dp[mark][l][r]为区间[l,r]的选择状态为mark的方法数。
然后记忆化搜索一下就可以了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-9 # define MOD 19650827 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=1005; //Code begin... int dp[2][N][N], a[N]; int dfs(int mark, int l, int r){ if (~dp[mark][l][r]) return dp[mark][l][r]; if (l==r) return dp[mark][l][r]=1; int res=0; if (mark) { if (a[l]<a[r]) res+=dfs(0,l,r-1); if (r-1!=l&&a[r-1]<a[r]) res+=dfs(1,l,r-1); } else { if (a[r]>a[l]) res+=dfs(1,l+1,r); if (l+1!=r&&a[l+1]>a[l]) res+=dfs(0,l+1,r); } return dp[mark][l][r]=res%MOD; } int main () { mem(dp,-1); int n; scanf("%d",&n); FOR(i,1,n) scanf("%d",a+i); printf("%d ",(dfs(0,1,n)+dfs(1,1,n))%MOD); return 0; }