【Atcoder S Digit Sum】

Atc S
题意 给你 一个 n 问小于等于 n 的数的 各数字之和 能整除 D 的个数
数位dp 统计一下

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<#define dbg3(x1,x2,x3) cout<<#x1<<" = "<#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)
#define MOD 1000000007

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 10025;
ll dp[MAX_N][105];
int arr[MAX_N],len,D;
char str[MAX_N];

ll dfs(int pos,int now,bool limit)
{
    if(pos==len+1)
        return dp[pos][now] = (now==0);
    if(!limit && dp[pos][now] != -1)
        return dp[pos][now];
    int up = limit?arr[pos]:9;
    ll sum = 0;
    for(int i = 0;i<=up;++i)
    {
        sum+=dfs(pos+1,(now+i)%D,limit&&i==up);
        if(sum>=MOD) sum-=MOD;
    }
    if(!limit)
        dp[pos][now] = sum;
    return sum;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    scanf("%s",str+1);
    scanf("%d",&D);
    len = strlen(str+1);
    for(int i = 1;i<=len;++i) arr[i] = str[i] - '0';
    memset(dp,-1,sizeof(dp));
    printf("%lld\n",(dfs(1,0,true)-1+MOD)%MOD);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}