D. Magic Numbers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Consider the decimal presentation of an integer. Let's call a number d-magic if digit d appears in decimal presentation of the number on even positions and nowhere else.

For example, the numbers 1727374, 17, 1 are 7-magic but 77, 7, 123, 34, 71 are not 7-magic. On the other hand the number 7 is 0-magic, 123 is 2-magic, 34 is 4-magic and 71 is 1-magic.

Find the number of d-magic numbers in the segment [a, b] that are multiple of m. Because the answer can be very huge you should only find its value modulo 109 + 7 (so you should find the remainder after dividing by 109 + 7).

Input

The first line contains two integers m, d (1 ≤ m ≤ 2000, 0 ≤ d ≤ 9) — the parameters from the problem statement.

The second line contains positive integer a in decimal presentation (without leading zeroes).

The third line contains positive integer b in decimal presentation (without leading zeroes).

It is guaranteed that a ≤ b, the number of digits in a and b are the same and don't exceed 2000.

Output

Print the only integer a — the remainder after dividing by 109 + 7 of the number of d-magic numbers in segment [a, b] that are multiple of m.

Examples
input 
2 6
10
99
output 
8
input 
2 0
1
9
output 
4
input 
19 7
1000
9999
output 
6
Note

The numbers from the answer of the first example are 16, 26, 36, 46, 56, 76, 86 and 96.

The numbers from the answer of the second example are 2, 4, 6 and 8.

The numbers from the answer of the third example are 1767, 2717, 5757, 6707, 8797 and 9747.
思路:偶数位置只能有d,奇数位置 不能有d

数位DP....  肝


#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define bug1 cout <<"bug1"<<endl
#define bug2 cout <<"bug2"<<endl
#define bug3 cout <<"bug3"<<endl
using namespace std;
typedef long long ll;

const int MAX_N=2e5+5;
int m,d;
char a[2300],b[2300];
int bit[2300];
int dp[2300][4000];
int len;

ll dfs(int pos,int mod,bool lead,bool limit){
    if(pos==len+1) return (mod==0);
    if(!limit && dp[pos][mod]!=-1) return dp[pos][mod];
    int up=limit ? bit[pos] : 9 ;
    ll ans=0;
    for(int i=0;i<=up;i++){
        int tempmod=(mod*10+i)%m; //模拟除法
        if(pos%2==1 && i==d)    continue;
        if(pos%2==0 && i!=d)    continue;
        ans+=dfs(pos+1,tempmod,lead&&i==0,limit && bit[pos]==i);
    }
    ans=ans%MOD;
    if(!limit)  dp[pos][mod]=ans;
    return ans;
}

ll solve(const char *s){
    memset(dp,-1,sizeof(dp));
    int pos=1;
    len=strlen(s+1);
    while(pos<=len){
        bit[pos]=s[pos]-'0'; ///bit存的是数字而不是字符
        pos++;

    }
//    cout <<"***********"<<endl;
//    printf("%s\n",bit+1);
    return dfs(1,0,true,true);
}

bool judge(const char*s){
    int mod=0;
    len=strlen(s+1);
    for(int i=1;i<=len;i++){
        mod=(mod*10+s[i]-'0')%m;
        if(i%2==1&&s[i]-'0'==d)    return false;///是s[i]和d的关系
        if(i%2==0&&s[i]-'0'!=d)   return false;
    }
    return mod==0;
}

int main(void){
    cin >> m>>d>>a+1>>b+1;
    printf("%lld",((solve(b)-solve(a)+judge(a))%MOD+MOD)%MOD);
}