BCD Code zoj3494
都是泪，看来以后要多做做数位DP题了，看了《浅谈数位类统计问题》感觉数位DP转化成树的思想真心不错，能简化思维的难度，可以很方便的处理一些细节问题（主要就是对于前导0的处理），用记忆化搜索实现起来比较方便，而且效率也比较高。
记忆化搜索:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>
#include <map>
#include <string>
#include <climits>
#include <set>
#include <string>
#include <sstream>
#include <utility>
#include <ctime>
using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::istringstream;
using std::make_pair;
using std::getline;
using std::greater;
using std::endl;
using std::multimap;
using std::deque;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PAIR;
typedef multimap<int, int> MMAP;
const int MAXN(2010);
const int SIGMA_SIZE(2);
const int MAXM(110);
const int MAXE(300010);
const int MAXH(18);
const int INFI((INT_MAX-1) >> 1);
const int MOD(1000000009);
const ULL LIM(1000000000000000ull);
struct AC
{
int ch[MAXN][SIGMA_SIZE];
int f[MAXN];
bool val[MAXN];
int size;
void init()
{
memset(ch[0], 0, sizeof(ch[0]));
f[0] = 0;
val[0] = false;
size = 1;
}
inline int idx(char temp)
{
return temp-'0';
}
void insert(char *S)
{
int u = 0, id;
for(; *S; ++S)
{
id = idx(*S);
if(!ch[u][id])
{
memset(ch[size], 0, sizeof(ch[size]));
val[size] = 0;
ch[u][id] = size++;
}
u = ch[u][id];
}
val[u] = true;
}
int que[MAXN];
int front, back;
void construct()
{
front = back = 0;
int cur, u;
for(int i = 0; i < SIGMA_SIZE; ++i)
{
u = ch[0][i];
if(u)
{
que[back++] = u;
f[u] = 0;
}
}
while(front < back)
{
cur = que[front++];
for(int i = 0; i < SIGMA_SIZE; ++i)
{
u = ch[cur][i];
if(u)
{
que[back++] = u;
f[u] = ch[f[cur]][i];
val[u] |= val[f[u]];
}
else
ch[cur][i] = ch[f[cur]][i];
}
}
}
};
AC ac;
int hash[10][4] = {
{0, 0, 0, 0},
{0, 0, 0, 1},
{0, 0, 1, 0},
{0, 0, 1, 1},
{0, 1, 0, 0},
{0, 1, 0, 1},
{0, 1, 1, 0},
{0, 1, 1, 1},
{1, 0, 0, 0},
{1, 0, 0, 1}
};
bool move(int &u, int id)
{
for(int i = 0; i < 4; ++i)
{
u = ac.ch[u][hash[id][i]];
if(ac.val[u])
return false;
}
return true;
}
int table[MAXN][210];
int digit[210];
int dfs(int u, int len, bool bound, bool zero)
{
if(len == 0)
return 1;
if(table[u][len] != -1 && !bound && !zero)
return table[u][len];
int ans = 0, tu;
int up = bound? digit[len]: 9;
if(zero && len > 1)
ans = (ans+dfs(u, len-1, bound && up == 0, true))%MOD;
else
{
tu = u;
if(move(tu, 0))
ans = (ans+dfs(tu, len-1, bound && up == 0, false))%MOD;
}
for(int i = 1; i <= up; ++i)
{
tu = u;
if(move(tu, i))
ans = (ans+dfs(tu, len-1, bound && i == up, false))%MOD;
}
if(!bound && !zero)
table[u][len] = ans;
return ans;
}
char num1[210], num2[210];
void solve()
{
memset(table, -1, sizeof(table));
int len = strlen(num2);
for(int i = len; i >= 1; --i)
digit[i] = num2[len-i]-'0';
int tans = dfs(0, len, true, true);
len = strlen(num1);
for(int i = len; i >= 1; --i)
digit[i] = num1[len-i]-'0';
tans = (tans-dfs(0, len, true, true)+MOD)%MOD;
printf("%d\n", tans);
}
int main()
{
int TC;
scanf("%d", &TC);
while(TC--)
{
int n;
ac.init();
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
scanf("%s", num1);
ac.insert(num1);
}
ac.construct();
scanf("%s%s", num1, num2);
for(int i = strlen(num1)-1; i >= 0; --i) //左界-1
if(num1[i] == '0')
num1[i] = '9';
else
{
--num1[i];
break;
}
solve();
}
return 0;
}
//预处理，按位统计
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>
#include <map>
#include <string>
#include <climits>
#include <set>
#include <string>
#include <sstream>
#include <utility>
#include <ctime>
using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::istringstream;
using std::make_pair;
using std::getline;
using std::greater;
using std::endl;
using std::multimap;
using std::deque;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PAIR;
typedef multimap<int, int> MMAP;
const int MAXN(2010);
const int SIGMA_SIZE(2);
const int MAXM(110);
const int MAXE(300010);
const int MAXH(18);
const int INFI((INT_MAX-1) >> 1);
const int MOD(1000000009);
const ULL LIM(1000000000000000ull);
struct AC
{
int ch[MAXN][SIGMA_SIZE];
int f[MAXN];
bool val[MAXN];
int size;
void init()
{
memset(ch[0], 0, sizeof(ch[0]));
f[0] = 0;
val[0] = false;
size = 1;
}
inline int idx(char temp)
{
return temp-'0';
}
void insert(char *S)
{
int u = 0, id;
for(; *S; ++S)
{
id = idx(*S);
if(!ch[u][id])
{
memset(ch[size], 0, sizeof(ch[size]));
val[size] = 0;
ch[u][id] = size++;
}
u = ch[u][id];
}
val[u] = true;
}
int que[MAXN];
int front, back;
void construct()
{
front = back = 0;
int cur, u;
for(int i = 0; i < SIGMA_SIZE; ++i)
{
u = ch[0][i];
if(u)
{
que[back++] = u;
f[u] = 0;
}
}
while(front < back)
{
cur = que[front++];
for(int i = 0; i < SIGMA_SIZE; ++i)
{
u = ch[cur][i];
if(u)
{
que[back++] = u;
f[u] = ch[f[cur]][i];
val[u] |= val[f[u]];
}
else
ch[cur][i] = ch[f[cur]][i];
}
}
}
};
AC ac;
int table[MAXN][210]; //table[i][j]表示在trie图中点i剩下的数字个数为j能得到的合法数字个数
int table2[210]; //对于前导为0的情况需要特殊处理
int hash[10][4] = {
{0, 0, 0, 0},
{0, 0, 0, 1},
{0, 0, 1, 0},
{0, 0, 1, 1},
{0, 1, 0, 0},
{0, 1, 0, 1},
{0, 1, 1, 0},
{0, 1, 1, 1},
{1, 0, 0, 0},
{1, 0, 0, 1}
};
int mlen;
bool move(int &u, int id)
{
for(int i = 0; i < 4; ++i)
{
u = ac.ch[u][hash[id][i]];
if(ac.val[u])
return false;
}
return true;
}
int dfs(int u, int left)
{
if(table[u][left] != -1)
return table[u][left];
int &cur = table[u][left];
int tu;
cur = 0;
for(int i = 0; i <= 9; ++i)
{
tu = u;
if(move(tu, i))
cur = (cur+dfs(tu, left-1))%MOD;
}
return cur;
}
void pre()
{
memset(table, -1, sizeof(table[0])*ac.size);
for(int i = 0; i < ac.size; ++i)
table[i][0] = 1;
for(int i = 1; i <= mlen; ++i)
for(int j = 0; j < ac.size; ++j)
dfs(j, i);
int u = 0;
table2[1] = move(u, 0)? 1: 0;
for(int i = 2; i <= mlen; ++i)
{
table2[i] = table2[i-1];
for(int j = 1; j <= 9; ++j)
{
u = 0;
if(move(u, j))
table2[i] = (table2[i]+table[u][i-2])%MOD;
}
}
}
int fun(char *S)
{
int ret = 0, u, tu, id;
int i = 0;
while(i < mlen && S[i] == '0')
++i;
if(i == mlen)
return table2[1];
id = S[i]-'0';
bool flag(true);
ret = (ret+table2[mlen-i])%MOD;
for(int j = 1; j < id; ++j)
{
u = 0;
if(move(u, j))
ret = (ret+table[u][mlen-i-1])%MOD;
}
u = 0;
while(i < mlen-1)
{
flag = move(u, id);
if(!flag)
break;
id = S[i+1]-'0';
for(int j = 0; j < id; ++j)
{
tu = u;
if(move(tu, j))
ret = (ret+table[tu][mlen-i-2])%MOD;
}
++i;
}
if(flag)
{
flag = move(u, S[mlen-1]-'0');
if(flag)
ret = (ret+1)%MOD;
}
return ret;
}
char num1[210], num2[210];
void solve()
{
printf("%d\n", (fun(num2)-fun(num1)+MOD)%MOD);
}
int main()
{
int TC;
scanf("%d", &TC);
while(TC--)
{
int n;
ac.init();
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
scanf("%s", num1);
ac.insert(num1);
}
ac.construct();
scanf("%s%s", num1, num2);
mlen = strlen(num2);
int len = strlen(num1);
for(int i = len-1; i >= 0; --i) //左界-1
if(num1[i] == '0')
num1[i] = '9';
else
{
--num1[i];
break;
}
int temp = mlen-len;
for(int i = len; i >= 0; --i) //为了统一处理，添加前导0
num1[i+temp] = num1[i];
for(int i = 0; i < temp; ++i)
num1[i] = '0';
pre();
solve();
}
return 0;
}