J- Beautiful Numbers

最新推荐文章于 2020-10-26 07:42:17 发布

原创最新推荐文章于 2020-10-26 07:42:17 发布 · 232 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#DP #数位DP

本文介绍了一种使用数位DP算法解决特殊美丽数计数问题的方法。美丽数是指那些能够被自身各位数字之和整除的正整数。通过递归状态压缩DP的方式，文章详细解释了如何高效地计算出在1到N范围内满足条件的美丽数的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 ≤ 1012).

输出描述:

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

示例1

输入

2
10
18

输出

Case 1: 10
Case 2: 12

思路：数位DP，做的太少，码上，还需理解。

#include<bits/stdc++.h>
using namespace std;
long long n;
int a[20],mod;
long long dp[20][200][200];

long long dfs(int pos,int state,int tmod,bool limit){
    if(!pos) return (state==mod)&&(!tmod);
    if(state>mod) return 0;
    if(!limit&&dp[pos][state][tmod]!=-1) return dp[pos][state][tmod];
    int up=limit?a[pos]:9;
    long long cnt=0;
    for(int i=0;i<=up;i++) cnt+=dfs(pos-1,state+i,(tmod*10+i)%mod,limit&&i==a[pos]);
    return limit?cnt:dp[pos][state][tmod]=cnt;
}



long long cal(long long n){
    int len=0;
    long long nn=n;
    while(nn){
        a[++len]=nn%10;
        nn/=10;

    }
    long long ans=0;
    for(int i=1;i<=9*len;i++){
        mod=i;
        memset(dp,-1,sizeof(dp));
        ans+=dfs(len,0,0,true);
    }
    return ans;
}

int main(){
    int t,cas=0;
    scanf("%d",&t);
    while(t--){
        scanf("%lld",&n);
        printf("Case %d: %lld\n",++cas,cal(n));
    }
}