628D - Magic Numbers(数位dp)

男娘i 2022-05-28 08:35 182阅读 0赞

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

Copy

2 6
    10
    99

output

input

Copy

2 0
    1
    9

output

input

Copy

19 7
    1000
    9999

output

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：自我感觉不错的一个数位dp，筛选条件是偶数位必须是d 奇数位不能是d，有个坑点就是如果b太大导致取摸之后数比a小， 这时候需要先加上mod再取模。

#include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    typedef long long ll;
    const int maxn=2010;
    const ll mod=1e9+7;
    ll dp[maxn][maxn];
    char a[maxn], b[maxn];
    int wei[maxn], m, d;
    
    ll dfs(int pos, ll sum, bool limit, int len){
    	if(pos==-1) return sum==0;
    	if(!limit && dp[pos][sum]!=-1) return dp[pos][sum];
    	
    	ll ans=0; 
    	int up= limit ? wei[pos] : 9;
    	for(int i=0; i<=up; i++)
    	{
    		if((len-pos)%2==0 && i!=d) continue;
    		if((len-pos)%2==1 && i==d) continue;
    		ans=(ans%mod+dfs(pos-1, (sum*10+i)%m, limit&&(i==wei[pos]), len)%mod)%mod;
    	}
    	
    	return limit ? ans : dp[pos][sum]=ans;
    }
    
    ll solve(char *s){
    	int len=strlen(s);
    	
    	for(int i=0; i<len; i++)
    	{
    		wei[i]=s[len-i-1]-'0';
    	}
    	
    	return dfs(len-1, 0, true, len);
    }
    
    bool check(){
    	int len=strlen(a);
    	ll s=0;
    	for(int i=0; i<len; i++)
    	{
    		s=(s*10+a[i]-'0')%m;
    		if((i+1)%2==0 && a[i]-'0'!=d || (i+1)%2==1 && a[i]-'0'==d)
    			return false;
    	}
    	if(!s) 
    		return true;
    	else
    		return false;
    }
    
    int  main(){
    	memset(dp, -1, sizeof(dp));
    	
    	scanf("%d%d", &m, &d);
    	scanf("%s%s", a, b);
    	
    	ll ans=solve(b)-solve(a);
    
    	if(check()){//判断a是否合格
    		ans++;
    	}
    	ans=(ans+mod)%mod;
    	printf("%lld\n", ans);
    	return 0;
    }