51nod1009 数字1的数量

思路：

数位dp。

实现：

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    vector<int> preprocess(int x)
    {
        vector<int> res;
        while (x) { res.push_back(x % 10); x /= 10; }
        return res;
    }
    int dfs(vector<int>& a, int p, int cnt, bool u, vector<vector<int>>& dp)
    {
        if (p == -1) return cnt;
        if (!u && dp[p][cnt] != -1) return dp[p][cnt];
        int res = 0;
        int h = u ? a[p] : 9;
        for (int i = 0; i <= h; i++)
        {
            res += dfs(a, p - 1, cnt + (i == 1), u && (i == h), dp);
        }
        return u ? res : dp[p][cnt] = res;
    }
    int NumberOf1Between1AndN_Solution(int n)
    {
        vector<int> a = preprocess(n);
        int l = a.size();
        vector<vector<int>> dp(l, vector<int>(l + 1, -1));
        return dfs(a, l - 1, 0, true, dp);
    }
};

int main()
{
    int n;
    while (cin >> n)
    {
        cout << Solution().NumberOf1Between1AndN_Solution(n) << endl;
    }
    return 0;
}