2018上海大都会 J-Beautiful Numbers（数位dp-模未知）

J-Beautiful Numbers

链接：https://www.nowcoder.com/acm/contest/163/J

来源：牛客网

时间限制：C/C++ 8秒，其他语言16秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

NIBGNAUK is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by the sum of its digits.

We will not argue with this and just count the quantity of beautiful numbers from 1 to N.

输入描述:

The first line of the input is T(1≤ T ≤ 100), which stands for the number of test cases you need to solve.
Each test case contains a line with a positive integer N (1 ≤ N ≤ 10

¹²

).

输出描述:

For each test case, print the case number and the quantity of beautiful numbers in [1, N].

输入例子:

2
10
18

输出例子:

Case 1: 10
Case 2: 12

-->

示例1

输入

复制

2
10
18

输出

复制

Case 1: 10
Case 2: 12

对于模未知的数位dp，暴力枚举模即可。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#define MAX 13
#define INF 0x3f3f3f3f
using namespace std;

typedef long long ll;

int a[MAX];
int poss,MOD;
ll dp[MAX][][][];
ll dfs(int pos,ll sta,int sum,bool limit){
    int i;
    if(pos==-) return sum==MOD&&sta==;
    if(!limit&&dp[pos][sta][sum][MOD]!=-) return dp[pos][sta][sum][MOD];
    int up=limit?a[pos]:;
    ll cnt=;
    for(i=;i<=up;i++){
        cnt+=dfs(pos-,(sta*+i)%MOD,sum+i,limit&&i==a[pos]);
    }
    if(!limit) dp[pos][sta][sum][MOD]=cnt;
    return cnt;
}
ll solve(ll x){
    int pos=;
    while(x){
        a[pos++]=x%;
        x/=;
    }
    poss=pos-;
    ll ans=;
    for(MOD=;MOD<=;MOD++){
        ans+=dfs(pos-,,,true);
    }
    return ans;
}
int main()
{
    int t,tt=;
    ll r;
    scanf("%d",&t);
    memset(dp,-,sizeof(dp));
    while(t--){
        scanf("%lld",&r);
        printf("Case %d: %lld\n",++tt,solve(r));
    }
    return ;
}