数位DP。
需要注意的是需要特殊处理前导0，另外连续的==匹配，不要计重了，尽量贪心的匹配掉。
1 /* 3886 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 const int mod = 100000000;
45 const int maxl = 105;
46 const int maxn = 105;
47 char ps[maxl], ps_[maxl];
48 char sa[maxl], sb[maxl];
49 int a[maxl], plen;
50 int dp[maxl][maxn][2][10];
51 bool dp_[maxl][maxn];
52
53 void f(char *s, int& l) {
54 int len = strlen(s);
55 int i = 0;
56
57 l = 0;
58 while (i<len-1 && s[i]==‘0‘)
59 ++i;
60 while (i < len) {
61 s[l++] = s[i++]-‘0‘;
62 }
63 s[l] = ‘\0‘;
64 }
65
66 bool check(char ch, int d, int dd) {
67 if (ch == ‘/‘)
68 return d < dd;
69 if (ch == ‘-‘)
70 return d == dd;
71 if (ch == ‘\\‘)
72 return d > dd;
73 return false;
74 }
75
76 bool judge(char *s, int len) {
77 if (len==1 || check(ps[1], s[0], s[1])==false)
78 return false;
79
80 memset(dp_, false, sizeof(dp_));
81 dp_[1][1] = true;
82
83 rep(j, 1, plen) {
84 rep(i, 1, len-1) {
85 if (!dp_[j][i])
86 continue;
87
88 if (check(ps[j], s[i], s[i+1]))
89 dp_[j][i+1] = true;
90 if (check(ps[j+1], s[i], s[i+1]))
91 dp_[j+1][i+1] = true;
92 }
93 }
94
95 return dp_[plen][len-1];
96 }
97
98 int cal(char *s, int len) {
99 if (len <= 1)
100 return 0;
101
102 memset(dp, -1, sizeof(dp));
103 rep(k, 0, s[0]) {
104 rep(kk, 0, 10) {
105 if (k == 0) {
106 if (dp[0][1][0][kk] == -1)
107 dp[0][1][0][kk] = 1;
108 else
109 ++dp[0][1][0][kk];
110 continue;
111 }
112
113 if (check(ps[1], k, kk)) {
114 if (dp[1][1][0][kk] == -1)
115 dp[1][1][0][kk] = 1;
116 else
117 ++dp[1][1][0][kk];
118 }
119 }
120 }
121 rep(kk, 0, s[1]+1) {
122 int at = kk==s[1];
123 if (check(ps[1], s[0], kk)) {
124 if (dp[1][1][at][kk] == -1)
125 dp[1][1][at][kk] = 1;
126 else
127 ++dp[1][1][at][kk];
128 }
129 }
130
131 rep(i, 0, plen+1) {
132 int ii = i + 1;
133 rep(j, 1, len-1) {
134 int jj = j + 1;
135
136 // consider boundary
137 if (dp[i][j][1][s[j]] >= 0) {
138 rep(k, 0, s[j+1]+1) {
139 int at = k==s[j+1];
140
141 if (check(ps[ii], s[j], k)) {
142 if (dp[ii][jj][at][k] >= 0) {
143 dp[ii][jj][at][k] = (dp[ii][jj][at][k] + dp[i][j][1][s[j]]) % mod;
144 } else {
145 dp[ii][jj][at][k] = dp[i][j][1][s[j]];
146 }
147 } else if (check(ps[i], s[j], k)) {
148 if (dp[i][jj][at][k] >= 0) {
149 dp[i][jj][at][k] = (dp[i][jj][at][k] + dp[i][j][1][s[j]]) % mod;
150 } else {
151 dp[i][jj][at][k] = dp[i][j][1][s[j]];
152 }
153 }
154 }
155 }
156
157 // consider < boundary
158 rep(k, 0, 10) {
159 if (dp[i][j][0][k] < 0)
160 continue;
161 rep(kk, 0, 10) {
162 if (i == 0) {
163 if (k == 0) {
164 if (dp[i][jj][0][kk] >= 0) {
165 dp[i][jj][0][kk] = (dp[i][jj][0][kk] + dp[i][j][0][k]) % mod;
166 } else {
167 dp[i][jj][0][kk] = dp[i][j][0][k];
168 }
169 } else {
170 if (check(ps[ii], k, kk)) {
171 if (dp[ii][jj][0][kk] >= 0) {
172 dp[ii][jj][0][kk] = (dp[ii][jj][0][kk] + dp[i][j][0][k]) % mod;
173 } else {
174 dp[ii][jj][0][kk] = dp[i][j][0][k];
175 }
176 }
177 }
178 continue;
179 }
180
181 if (check(ps[ii], k, kk)) {
182 if (dp[ii][jj][0][kk] >= 0) {
183 dp[ii][jj][0][kk] = (dp[ii][jj][0][kk] + dp[i][j][0][k]) % mod;
184 } else {
185 dp[ii][jj][0][kk] = dp[i][j][0][k];
186 }
187 } else if (check(ps[i], k, kk)) {
188 if (dp[i][jj][0][kk] >= 0) {
189 dp[i][jj][0][kk] = (dp[i][jj][0][kk] + dp[i][j][0][k]) % mod;
190 } else {
191 dp[i][jj][0][kk] = dp[i][j][0][k];
192 }
193 }
194 }
195 }
196 }
197 }
198
199 int ret = 0;
200
201 rep(k, 0, 10) {
202 if (dp[plen][len-1][1][k] >= 0)
203 ret = (ret + dp[plen][len-1][1][k]) % mod;
204 if (dp[plen][len-1][0][k] >= 0)
205 ret = (ret + dp[plen][len-1][0][k]) % mod;
206 }
207
208 return ret;
209 }
210
211 void solve() {
212 int alen, blen;
213 int ans = 0, tmp;
214
215 plen = strlen(ps+1);
216 f(sa, alen);
217 f(sb, blen);
218
219 tmp = cal(sb, blen);
220 ans += tmp;
221 tmp = cal(sa, alen);
222 ans -= tmp;
223 if (judge(sa, alen))
224 ++ans;
225
226 ans = (ans + mod) % mod;
227 printf("%08d\n", ans);
228 }
229
230 int main() {
231 ios::sync_with_stdio(false);
232 #ifndef ONLINE_JUDGE
233 freopen("data.in", "r", stdin);
234 freopen("data.out", "w", stdout);
235 #endif
236
237 while (scanf("%s", ps+1)!=EOF) {
238 scanf("%s %s", sa, sb);
239 solve();
240 }
241
242 #ifndef ONLINE_JUDGE
243 printf("time = %d.\n", (int)clock());
244 #endif
245
246 return 0;
247 }
时间： 2024-10-19 11:00:12