【hdu-2089】不要62(数位DP入门)

本文通过一道具体的题目,介绍了数位动态规划(数位DP)的实现方法,并使用记忆化搜索来避免重复计算,有效解决了在给定区间内寻找不包含特定数字的问题。

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题目链接:点击打开链接


【题意】

给定一段区间(0~1000000),求这段区间内不含62和4的个数


【个人感悟】
第一次做数位DP,查阅了好多资料,有用递推做的,也有用记忆化搜索做的,我主要针对记忆化搜索做谈谈我对数位dp的拙见。当然这道题是0~1000000,打表也是可以过的。

数位dp之所以不会超时,就在于他在数位搜索中添加了记忆化功能,即为mem数组,mem[i][j],i代表i位数,j代表上一位是否为6两种状态,当搜索过相同的mem[i][j]时,return 即可,但是有一个前提条件limit != 0,意思就是假如要搜索256以内的数,对于相同的56,百位为1和为2有不同的结果,因为百位为2有limit限制。这篇题解讲得很好


#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int bit[8], mem[8][2];
//dfs返回结果,即一个len位的符合要求的数的个数
int dfs(int len, bool lim, bool preis_6)
{
    if(len == 0) return 1;//递归一定先考虑退出条件
    if(!lim && mem[len][preis_6] != -1) return mem[len][preis_6];
    int up = lim? bit[len] : 9;
    int ans = 0;
    for(int i = 0; i <= up; i++)
    {
        if(i == 4 || preis_6 && i == 2) continue;
        ans += dfs(len - 1, lim && i == up, i == 6);//直接把dfs看成返回的结果数
    }
    if(!lim) mem[len][preis_6] = ans;
    return ans;
}
int sol(int n)
{
    int len = 0;
    while(n)
    {
        bit[++len] = n % 10;
        n /= 10;
    }
    return dfs(len, 1, 0); 
}
int main()
{
    int n, m;
    while(scanf("%d%d", &m, &n),m+n)
    {
        memset(mem, -1, sizeof mem);
        printf("%d\n", sol(n) - sol(m-1));//因为搜索是包括的数本身的,所以要m-1
    }
    return 0;
}


### HDU OJ 2089 Problem Solution and Description The problem titled "不高兴的津津" (Unhappy Jinjin) involves simulating a scenario where one needs to calculate the number of days an individual named Jinjin feels unhappy based on certain conditions related to her daily activities. #### Problem Statement Given a series of integers representing different aspects of Jinjin's day, such as homework completion status, weather condition, etc., determine how many days she was not happy during a given period. Each integer corresponds to whether specific events occurred which could affect her mood positively or negatively[^1]. #### Input Format Input consists of multiple sets; each set starts with two positive integers n and m separated by spaces, indicating the total number of days considered and types of influencing factors respectively. Following lines contain details about these influences over those days until all cases are processed when both numbers become zero simultaneously. #### Output Requirement For every dataset provided, output should be formatted according to sample outputs shown below: ```plaintext Case k: The maximum times of appearance is x, the color is c. ``` Where `k` represents case index starting from 1, while `x` stands for frequency count and `c` denotes associated attribute like colors mentioned earlier but adapted accordingly here depending upon context i.e., reasons causing unhappiness instead[^2]. #### Sample Code Implementation Below demonstrates a simple approach using Python language to solve this particular challenge efficiently without unnecessary complexity: ```python def main(): import sys input = sys.stdin.read().strip() datasets = input.split('\n\n') results = [] for idx, ds in enumerate(datasets[:-1], start=1): data = list(map(int, ds.strip().split())) n, m = data[:2] if n == 0 and m == 0: break counts = {} for _ in range(m): factor_counts = dict(zip(data[2::2], data[3::2])) for key, value in factor_counts.items(): try: counts[key] += value except KeyError: counts[key] = value max_key = max(counts, key=lambda k:counts[k]) result_line = f'Case {idx}: The maximum times of appearance is {counts[max_key]}, the reason is {max_key}.' results.append(result_line) print("\n".join(results)) if __name__ == '__main__': main() ```
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