hdu 2089 不要62

不要62

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29113    Accepted Submission(s): 10215


Problem Description
杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer)。
杭州交通管理局经常会扩充一些的士车牌照,新近出来一个好消息,以后上牌照,不再含有不吉利的数字了,这样一来,就可以消除个别的士司机和乘客的心理障碍,更安全地服务大众。
不吉利的数字为所有含有4或62的号码。例如:
62315 73418 88914
都属于不吉利号码。但是,61152虽然含有6和2,但不是62连号,所以不属于不吉利数字之列。
你的任务是,对于每次给出的一个牌照区间号,推断出交管局今次又要实际上给多少辆新的士车上牌照了。
 

Input
输入的都是整数对n、m(0<n≤m<1000000),如果遇到都是0的整数对,则输入结束。
 

Output
对于每个整数对,输出一个不含有不吉利数字的统计个数,该数值占一行位置。
 

Sample Input
1 100 0 0
 

Sample Output
80
本题有两种解法,一种是暴力破解,第二种是数位dp。
详细解释请看如下:
一 暴力破解。
#include <iostream> #include <cstdio> //暴力破解; using namespace std; int s[1000003]={0}; int main() {     int n,m;     for(int i=3;i<1000003;i++){             int t=i;         while(t){             if(t%10==4||t%100==62)                 s[i]=1;                 t/=10;             }     }     while(scanf("%d%d",&n,&m)!=EOF){         if(n==0&&m==0)             break;             int sum=0;         for(int i=n;i<=m;i++)             sum+=s[i];             int result=m-n+1-sum;         cout<<result<<endl;     }     return 0; }
二,数位dp
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; int dp[10][10];//dp[i][j]代表满足条件的数字个数有多少个,i代表这个数的位数,j代表以j开头数。比如:dp[2][3]代表满足条件3开头的两位数,如30,31,32,33,35,36,37,38,39.以此类推
////////////////////////////重点搞懂这个哦 void init(){//数据处理函数     memset(dp,0,sizeof(dp));     dp[0][0]=1;     for(int i=1;i<8;i++)         for(int j=0;j<=9;j++)             for(int k=0;k<=9;k++){                 if(j!=4&&!(j==6&&k==2))                                     dp[i][j]+=dp[i-1][k];                             //cout<<dp[i][j];             } }
////////////////////////////////////////
int solve(int n){     init();//调用     int digit[10]={0};     int len=0;//数据位数     int ans=0;//结果     while(n){//将数据每一位存到digit数组里         digit[++len]=n%10;         n/=10;     }     digit[len+1]=0;
//求出[0,n)的结果     for(int i=len;i;i--){         for(int j=0;j<digit[i];j++){
if(j!=4&&!(digit[i+1]==6&&j==2))               ans+=dp[i][j];         }         //cout<<ans<<endl;         if(digit[i]==4||(digit[i]==2&&digit[i+1]==6))             break;     }     return ans; } int main(){     int n,m;     while(scanf("%d%d",&n,&m)!=EOF){         if(n+m==0)break;        int res= solve(m+1)-solve(n);//n到m[n,m]之间的个数结果简单变形一下就是[0,m+1]-[0,n],solve(m)未包含m,
       cout<<res<<endl;     }     return 0; }
关于数位dp做这个题之前我并不懂,感觉这个名字听起来就比较高大上了,但小伙伴们不要被这个噱头给吓到了,认为这个很难,其实很简单的,这还是数位dp入门题。。。。。。。。搞懂上面我画的重点就轻松多了。
搞笑的是下面我的机器的数据处理也是。。。。
### 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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