# Codeforces Round #157 (Div. 1) B digital dp

Source: Internet
Author: User

Codeforces Round #157 (Div. 1) B digital dp
// There are several enumerated values (7 or 4). The number of I (7 or 4) is searched with the memory of the digital dp.
// Select the number of brute-force searches in the I field.
# Include
# Include
# Include
Using namespace std;
Const int mod = 1e9 + 7;
Int dp [20] [20]; // Number of j in the I-th digit (7 or 4)
Int bit [20];
Int temp [20];
Int luck [20];
Int dfs (int pos, int flag, int max_flag, int lim)
{
If (pos = 0)
Return flag = max_flag;
If (! Lim & dp [pos] [flag]! =-1)
Return dp [pos] [flag];
Int num = lim? Bit [pos]: 9;
Int ans = 0;
For (int I = 0; I <= num; I ++)
{
Int flag_x = flag;
If (I = 7 | I = 4)
Flag_x ++;
Ans + = dfs (pos-1, flag_x, max_flag, lim & (I = num ));
}
If (! Lim) dp [pos] [flag] = ans;
Return ans;
}
// _ Int64 ans_ans = 0;
_ Int64 dfs_s (int res, _ int64 ans, int pos, int num, int max_flag)
{
If (num> = max_flag)
Return 0;
If (res = 0)
Return (ans * (_ int64) temp [num + 1]) % mod;
_ Int64 sum = 0;
For (int I = 0; I <= pos; I ++)
{
If (luck [I]> 0)
{
_ Int64 ans_x = (ans * (_ int64) luck [I]) % mod );
Luck [I]-= 1;
Sum + = dfs_s (res-1, ans_x, pos, num + I, max_flag );
Luck [I] + = 1;
}
}
Return sum % mod;
}
_ Int64 solve (int n)
{
If (n = 7) return 0;
Int t = n;
Int pos = 0;
While (t)
{
Bit [++ pos] = t % 10;
T/= 10;
}
Memset (luck, 0, sizeof (luck ));
Memset (temp, 0, sizeof (temp ));
For (int I = pos; I> = 0; I --)
{
Memset (dp,-1, sizeof (dp ));
Luck [I] = dfs (pos, 0, I, 1 );
Temp [I] = temp [I + 1] + luck [I];
}
Luck [0]-= 1;
Return dfs_s (6, 1, pos, 0, luck [pos]? Pos: pos-1 );
}
Int main ()
{
// Freopen ("input.txt", "r", stdin );
Int n;
While (~ Scanf ("% d", & n ))
{
Printf ("% I64d \ n", solve (n) % mod );
}
Return 0;
}

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