HDU - 2089 不要62 ——数位DP入门

本文详细介绍了使用数位DP解决特定计数问题的两种方法:递推写法与递归写法,并提供了完整的C++代码实现。通过对不同状态的转移来避免重复计算,实现了高效求解。

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vj题目链接:https://cn.vjudge.net/problem/HDU-2089

HDU题目链接:acm.hdu.edu.cn/showproblem.php?pid=2089

借鉴博客:https://www.cnblogs.com/wenruo/p/4725005.html

递推写法:

#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
int x,y,dp[15][15],d[15];
void init()
{
    dp[0][0]=1;
    for(int i=1; i<10; i++)
    {
        for(int j=0; j<10; j++)
        {
            for(int k=0; k<10; k++)
            {
                if(j!=4&&!(j==6&&k==2))
                    dp[i][j]+=dp[i-1][k];
            }
        }
    }
}
int f(int x)
{
    int len=0,ans=0;
    while(x)
    {
        d[++len]=x%10;
        x/=10;
    }
    d[len+1]=0;
    for(int i=len; i>0; i--)
    {
        for(int j=0; j<d[i]; j++)
        {
            if(!(d[i+1]==6&&j==2))
            {
                ans+=dp[i][j];
            }
        }
        if(d[i]==4||(d[i+1]==6&&d[i]==2))
            break;
    }
    return ans;
}
int main()
{
    init();
    while(~scanf("%d%d",&x,&y),x||y)
    {
        printf("%d\n",f(y+1)-f(x));
    }
}

 

递归写法:

#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;
int n,m;
int w[15],dp[20][15];
int dfs(int pos,int pre,int lim)
{
    if(pos==0) return 1;
    if(!lim&&dp[pos][pre]!=-1) return dp[pos][pre];
    int up=lim?w[pos]:9;
    int ans=0;
    for(int i=0;i<=up;i++)
    {
        if((pre==6&&i==2)||i==4)continue;
        ans+=dfs(pos-1,i,lim&&i==w[pos]);
    }
    if(!lim) dp[pos][pre]=ans;
    return ans;
}
int get(int x)
{
    int pos=0;
    while(x)
        w[++pos]=x%10,x/=10;
    return dfs(pos,0,1);
}
int main()
{
    memset(dp,-1,sizeof(dp));
    while(~scanf("%d%d",&n,&m),n||m)
    {
        printf("%d\n",get(m)-get(n-1));
    }
}

 

### 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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