題意：

給出一個長度為$n - 1的字符串，s_i為‘I‘, ‘D‘, ‘?‘$

$‘I‘表示a_i < a_{i + 1}$

$‘D‘表示 a_i > a_{i + 1}$

$‘?‘ 表示 a_i 和 a_{i + 1} 的大小關系任意$

思路：

$dp[i][j] 表示考慮到第i位，當前位的數字位j的方案數$

$s[i - 1] == ‘?‘ 那麽 dp[i][j] = \sum_{k = 1}^{i - 1} dp[i - 1][k]$

$s[i - 1] == ‘I‘ 那麽 dp[i][j] = \sum_{k = 1}^{j - 1} dp[i - 1][k]$

$s[i - 1] == ‘D‘ 那麽 dp[i][j] = \sum_{k = j}^{k = i - 1} dp[i - 1][k]$

$第三個轉移可以這樣考慮，比如說之前有一個排列{1, 3, 2} 那麽我當前後面插入一位2$

$我們把{1, 3, 2} 中 >= 2的數字都加1 再加入2 就變成 {1, 4, 3, 2}$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 #define N 1010
 6 const ll MOD = (ll)1e9 + 7;
 7 char s[N];
 8 ll f[N][N];
 9 
10 int main()
11 {
12     while
 
 (scanf("%s", s + 2) != EOF)
13     {
14         f[1][1] = 1;
15         int n = strlen(s + 2) + 1;
16         for (int i = 2; i <= n; ++i)
17         {
18             if (s[i] == ‘I‘)
19             {
20                 for (int j = 1; j <= i; ++j) 
21                     f[i][j] = (f[i][j - 1
 
] + f[i - 1][j - 1]) % MOD;
22             }
23             else if (s[i] == ‘D‘)
24             {
25                 for (int j = i; j >= 1; --j)
26                     f[i][j] = (f[i][j + 1] + f[i - 1][j]) % MOD;   
27             }
28             else
29             {
30                 ll sum = 0;
31                 for (int j = 1; j < i; ++j)
32                     sum = (sum + f[i - 1][j]) % MOD;
33                 for (int j = 1; j <= i; ++j) 
34                     f[i][j] = sum;  
35             }
36         }
37         ll res = 0;
38         for (int i = 1; i <= n; ++i) res = (res + f[n][i]) % MOD;
39         printf("%lld\n", res);
40     }
41     return 0;
42 }

HDU 4055 Number String