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).

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.

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.

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.

题意：给你一个区间[a,b]，让你找到这个区间内满足没有前导零且偶数位都是d，奇数位不出现d，并且这个数能被m整除的数的个数。

思路：用dp[pos][yushu][oushu]表示pos位前面的位形成的数modm后余数为yushu，且当前位是否是偶数的方案数，要注意前导零。

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<math.h>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#include<bitset>

#include<algorithm>

using namespace std;

typedef long long ll;

#define inf 99999999

#define MOD 1000000007

char s1[2005],s2[2005];

int wei[2005];

ll dp[2005][2005][2];

int m,d;

void add(ll& x,ll y) {

    x += y;

    if(x>=MOD) x-=MOD;

}

ll dfs(int pos,int yushu,int oushu,int flag,int zero)

{

    int i,j;

    if(pos==-1){

        if(zero==1)return 0;

        if(yushu==0)return 1;

        else return 0;

    }

    if(!flag && !zero && dp[pos][yushu][oushu]!=-1){

        return dp[pos][yushu][oushu];

    }

    int ed=flag?wei[pos]:9;

    ll ans=0;

    if(zero==1){

        add(ans,dfs(pos-1,yushu,oushu,0,1));

        for(i=1;i<=ed;i++){

            if(i!=d)add(ans,dfs(pos-1,(yushu*10+i)%m,1^oushu,flag&&wei[pos]==i,0) );

        }

    }

    else{

        if(oushu){

            if(d<=ed)add(ans,dfs(pos-1,(yushu*10+d)%m,1^oushu,flag&&wei[pos]==d,0) );

        }

        else{

            for(i=0;i<=ed;i++){

                if(i!=d)add(ans,dfs(pos-1,(yushu*10+i)%m,1^oushu,flag&&wei[pos]==i,0) );

            }

        }

    }

    if(!flag && !zero){

        dp[pos][yushu][oushu]=ans;

    }

    return ans;

}

ll solve(char s[])

{

    int len,i,j;

    len=strlen(s);

    for(i=len-1;i>=0;i--){

        wei[i]=s[i]-'0';

    }

    return dfs(len-1,0,0,1,1);

}

int main()

{

    int n,i,j,len1,len2;

    while(scanf("%d%d",&m,&d)!=EOF)

    {

        scanf("%s%s",s1,s2);

        len1=strlen(s1);

        reverse(s1,s1+len1);

        for(i=0;i<len1;i++){

            if(s1[i]=='0'){

                s1[i]='9';

            }

            else{

                s1[i]--;break;

            }

        }

        if(s1[len1-1]=='0'){

            s1[len1-1]='\0';

            len1--;

        }

        len2=strlen(s2);

        reverse(s2,s2+len2);

        memset(dp,-1,sizeof(dp));

        ll num1=solve(s1);

        ll num2=solve(s2);

        printf("%I64d\n",((num2-num1)%MOD+MOD)%MOD );

    }

    return 0;

}