51nod1009 数字1的数量

思路：
数位dp。
实现：
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 class Solution
 4 {
 5 public:
 6     vector<int> preprocess(int x)
 7     {
 8         vector<int> res;
 9         while (x) { res.push_back(x % 10); x /= 10; }
10         return res;
11     }
12     int dfs(vector<int>& a, int p, int cnt, bool u, vector<vector<int>>& dp)
13     {
14         if (p == -1) return cnt;
15         if (!u && dp[p][cnt] != -1) return dp[p][cnt];
16         int res = 0;
17         int h = u ? a[p] : 9;
18         for (int i = 0; i <= h; i++)
19         {
20             res += dfs(a, p - 1, cnt + (i == 1), u && (i == h), dp);
21         }
22         return u ? res : dp[p][cnt] = res;
23     }
24     int NumberOf1Between1AndN_Solution(int n)
25     {
26         vector<int> a = preprocess(n);
27         int l = a.size();
28         vector<vector<int>> dp(l, vector<int>(l + 1, -1));
29         return dfs(a, l - 1, 0, true, dp);
30     }
31 };
32 
33 int main()
34 {
35     int n;
36     while (cin >> n)
37     {
38         cout << Solution().NumberOf1Between1AndN_Solution(n) << endl;
39     }
40     return 0;
41 }