light oj 1068 数位dp

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <climits>
#include <queue>
#define ll long long

using namespace std;
const int N = 1e4+1000;

int digit[25],l,k;
ll dp[25][100][100];

void init()
{
    memset(dp,-1,sizeof(dp));
}

ll dfs( int cur, int n, int sum, int f)
{
    if(cur == -1)
    {
        return (n == 0 && sum == 0) ?1:0;
    }

    if(!f && dp[cur][n][sum] != -1)
        return dp[cur][n][sum];
        
    ll ans = 0;
    int len = f?digit[cur]:9;
    for(int i = 0; i <= len; i++)
    {
        ans += dfs(cur-1, (n*10 + i)%k, (sum+i)%k, f&(i == len));
    }
    if(!f)
        return dp[cur][n][sum] = ans;
    else
        return ans;
}

ll work(int num,int mod)
{
    l = 0;
    while(num > 0)
    {
        digit[l++] = num%10;
        num /= 10;
    }
    ll ans = dfs(l-1, 0, 0, 1);
    return ans;
}

void solve()
{
    int a,b;

    init();
    scanf("%d %d %d",&a,&b,&k);
    if(k > 90)
    {
        printf("0
");
        return ;
    }
    printf("%lld
", work(b,k) - work(a-1,k));

}

int main(void)
{

    int t,cnt = 0;
    scanf("%d",&t);
    while(t--)
    {
        printf("Case %d: ",++cnt);
        solve();
    }
    return 0;
}