Hope and Possibilities

本文探讨了两位毕业生关于大学毕业后职业选择、面临的挑战及如何实现个人目标的对话。他们讨论了兴趣导向的职业追求与家人期望之间的冲突,以及在追求职业成功时保持现实的重要性。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A: I am looking forward to relaxing this coming weekend.

B: I hope that I can finally find some free time too. I have been so busy at work recently.

A: How might you spend your weekend?

B: I hope to do a little gardening. I find it very relaxing.

A: I might do that too. I hope that the weather is nice. I could go and play some golf.

B: I heard that the weather should to be good. There’s a possibility of a shower, but it is not very likely.

A: Hopefully, we both have a relaxing weekends.

B: Of course, Something could come out and stop that wish coming true.

A: Unfortunately, there’s always the possibility of some urgent work requiring our attention.

 

A: What do you want to do after graduate university?

B: I’d like to go into management. I’ve applied for several jobs already and I’m hopeful that I will get some job offers. How about you?

A: After I graduate, I have to do some studies to pass exams to become a layer. I think I’ve got a good chance of passing. There’s a possibility of getting a job with a law firm in London, provide that I do well.

B: We both have to overcome some obstacles if we are to achieve our ambitions.

A: If life were easy, then we’d achieve our ambition quickly, and then get bored.

B: Unfortunately, it is inevitable that some people are going to work hard yet not succeed.

A: That’s why ambition needs to be realistic. You can’t achieve something that’s totally unrealistic.

B: As long as you plan carefully, most thing are possible. It is always good to have a backup plan in case things go wrong.

A: I think it’s important to be successful in a field you truly interested in, not something that other people force you to be interested it.

B: My father wanted me to become a doctor, but I knew it is impossible for me to be successful in that field.

A: I hope that my parents don’t try to interfere in my choice of career.

This is an interesting problem! To solve it, you can use dynamic programming. Let dp[i][j] be the minimum cost to sort the first i characters of the pattern such that the i-th character is j (0 or 1). The base case is dp[0][0]=dp[0][1]=0, since an empty string is already sorted. Then, to compute dp[i][j], you can consider the last character of the pattern, which can be either 0, 1, or ?. If it's 0 or 1, you can simply match it with the i-th character of the binary string, and update dp[i][j] accordingly. If it's ?, you need to try both possibilities (0 or 1) and take the minimum cost. The final answer is the minimum of dp[n][0] and dp[n][1], where n is the length of the pattern. Here's the Python code: ``` s = input().strip() n = len(s) dp = [[float('inf')]*2 for _ in range(n+1)] dp[0][0] = dp[0][1] = 0 for i in range(1, n+1): if s[i-1] == '0': dp[i][0] = min(dp[i-1][0], dp[i-1][1]+1) elif s[i-1] == '1': dp[i][1] = min(dp[i-1][1], dp[i-1][0]+1) else: dp[i][0] = min(dp[i-1][0], dp[i-1][1]+1) dp[i][1] = min(dp[i-1][1], dp[i-1][0]+1) ans = [''] * n if dp[n][0] <= dp[n][1]: i, j = n, 0 while i > 0: if s[i-1] == '0': ans[i-1] = '0' i, j = i-1, 0 elif s[i-1] == '1': ans[i-1] = '1' i, j = i-1, 1 else: if dp[i][j] == dp[i-1][0]: ans[i-1] = '0' i, j = i-1, 0 else: ans[i-1] = '1' i, j = i-1, 1 else: i, j = n, 1 while i > 0: if s[i-1] == '0': ans[i-1] = '0' i, j = i-1, 0 elif s[i-1] == '1': ans[i-1] = '1' i, j = i-1, 1 else: if dp[i][j] == dp[i-1][1]: ans[i-1] = '1' i, j = i-1, 1 else: ans[i-1] = '0' i, j = i-1, 0 print(''.join(ans)) ``` Hope this helps!
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值