Problem:
Score : 100 points

Problem Statement

Find the number of integers between 1 and K (inclusive) satisfying the following condition, modulo 10^9 + 7:

The sum of the digits in base ten is a multiple of D.

Constraints

All values in input are integers.

1 \leq K < 10^{10000}

1 \leq D \leq 100

Input

Input is given from Standard Input in the following format:

K
D

Output

Print the number of integers satisfying the condition, modulo 10^9 + 7.

Sample Input 1

30
4

Sample Output 1

6

Those six integers are: 4, 8, 13, 17, 22 and 26.

Sample Input 2

1000000009
1

Sample Output 2

2

Be sure to print the number modulo 10^9 + 7.

Sample Input 3

98765432109876543210
58

Sample Output 3

635270834

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
typedef long long ll;
char a[10005];
int d;
ll dp[10005][105][3];

ll solve(int idx,int sum,int pre){

    if(idx == strlen(a)){
        if(sum == 0) return dp[idx][sum][pre] = 1;
        return dp[idx][sum][pre] = 0;
    }
    if(dp[idx][sum][pre] != -1) return dp[idx][sum][pre];
    
    ll ret = 0;
    if(idx == 0 || pre == 1){
    for(int i = 0;i <= (a[idx]-'0'); i++){
        if(i == a[idx]-'0')
        ret = (ret % mod + solve(idx+1,(sum + i)%d, 1) % mod) % mod;
        else
        ret = (ret % mod + solve(idx+1,(sum+i)%d,0) % mod) % mod;
        }
    }
    else{
        for(int i=0;i < 10; i++){
        ret = (ret % mod + solve(idx+1,(sum + i)%d, pre) % mod) % mod;
        }
    }
    return dp[idx][sum][pre] = ret;
}

int main()
{
    cin>>a>>d;
    memset(dp,-1,sizeof(dp));
    cout<<(((solve(0,0,0)-1) + mod)%mod)%mod;

    return 0;
}