題目連結

很明顯的，有“大於”和“小於等於”兩種情況，且一旦“大於”了，只能繼續“大於”下去，如果“小於等於”，還可以選擇任意一種情況。於是乎，直接數位dp，並且記錄這一位是“大於狀態”還是“小於等於狀態”，就可以進行記憶化了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <string>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <limits>
 8
 
 #include <vector>
 9 #include <stack>
10 #include <queue>
11 #include <set>
12 #include <map>
13 #include <bitset>
14 #include <unordered_map>
15 #include <unordered_set>
16 #define lowbit(x) ( x&(-x) )
17 #define pi 3.141592653589793
18 #define e 2.718281828459045
19
 
 #define INF 0x3f3f3f3f
20 #define HalF (l + r)>>1
21 #define lsn rt<<1
22 #define rsn rt<<1|1
23 #define Lson lsn, l, mid
24 #define Rson rsn, mid+1, r
25 #define QL Lson, ql, qr
26 #define QR Rson, ql, qr
27 #define myself rt, l, r
28 #define pii pair<int, int>
29 #define MP(a, b) make_pair(a, b)
30
 
 using namespace std;
31 typedef unsigned long long ull;
32 typedef unsigned int uit;
33 typedef long long ll;
34 const int maxN = 1e2 + 7;
35 const ll mod = 1e9 + 7;
36 void MOD(ll & x) { x >= mod ? x %= mod : x; }
37 char s[maxN];
38 int N, dig[maxN];
39 ll dp[maxN][10][2];
40 ll dfs(int pos, int x, bool op, bool top, bool zero)
41 {
42     if(pos == 1) return !zero;
43     if(!top && (~dp[pos][x][op])) return dp[pos][x][op];
44     ll sum = 0;
45     int u = top ? dig[pos - 1] : 9;
46     for(int i = 0; i <= u; i ++)
47     {
48         if(op)
49         {
50             if(i < x) continue;
51             sum += dfs(pos - 1, i, true, top && i == u, zero && (!i));
52             MOD(sum);
53         }
54         else
55         {
56             sum += dfs(pos - 1, i, !zero && i > x, top && i == u, zero && (!i));
57             MOD(sum);
58         }
59     }
60     if(!top && !zero) dp[pos][x][op] = sum;
61     return sum;
62 }
63 int main()
64 {
65     int T; scanf("%d", &T);
66     while(T --)
67     {
68         scanf("%s", s);
69         N = (int)strlen(s);
70         for(int i = 0; i < N; i ++) dig[N - i] = s[i] - '0';
71         memset(dp, -1, sizeof(dp));
72         printf("%lld\n", dfs(N + 1, 0, false, true, true));
73     }
74     return 0;
75 }