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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
char s[100005];
int n,d;
int a[100005],f[10005][105][2];
void add(int &x,int v)
{
	x+=v;
	if(x>=mod) x-=mod;
}
int main()
{
	scanf("%s%d",s+1,&d);
	int n=strlen(s+1);
	for(int i=1;i<=n;i++) a[i]=s[i]-'0';
	f[0][0][1]=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<d;j++)
		{
			for(int k=0;k<=9;k++)
			{
				int p=j-k;
				while(p<0) p+=d;
				//printf("j=%d k=%d p=%d\n",j,k,p);
				add(f[i][j][0],f[i-1][p][0]);
				if(k==a[i]) add(f[i][j][1],f[i-1][p][1]);
				else if(k<a[i]) add(f[i][j][0],f[i-1][p][1]);
			}
			//printf("f(%d,%d)=(%d,%d)\n",i,j,f[i][j][0],f[i][j][1]);
		}
	}
	int ans=(f[n][0][0]+f[n][0][1]-1)%mod;
	printf("%d\n",ans);
	return 0;
}