codeforces893D Credit Card

本文介绍了一种称为“区间贪心”的算法,用于解决信用卡账户管理问题。通过将连续的收支记录视为区间,实现了对账户余额的有效管理。文章详细解释了算法流程,包括如何处理不同类型的收支变化,以及如何最小化存款次数。

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对着AC代码苦思冥想两个小时终于顿悟…真是神奇的做法。
大概算奥妙重重的贪心?
我给它起名叫:区间贪心!(雾)


惯例题意。一个人在银行里有一张信用卡。
晚上会有钱进来,不过:
1)如果送进来正数元,那就很棒。但是如果超过某个上限k,这个人就会被双规。
2)如果送进来0元,那就是在传递信号,第二天会有人查账户,要求账户不透支。
3)如果送进来负的,那就很惨。而且,信用卡是有可能透支的。
为了拯救自己的账户,这个人借了很多钱,并且可以在任意一天存入。这里的问题是,他最少存多少次,就可以应付过每一次检查?
如果会被双规,输出-1


不妨设表示晚上送钱的数组为a
首先,我们会发现,在ai=0时处理即可。这是因为,在每一天的处理都可以等价到那一天。
考虑一下,可以将两次检查之间的结果作为一个区间。
然后不妨思考一下,假设现在刚刚检查过一次,存钱之后的剩余钱范围设为[low,top],显然初始时为[0,k],因为可以存0~k元。然后,我们对这个区间进行这些操作(ai0):
每扫到一个数先变成[low+ai,top+ai]。如果新的top>k,由于存钱不可能超过k,那么最大值不变;而如果low>k,表示最少需要存的钱也会比k大,GG。
如果遇到了一次新的检查,有两种情况:一种是上次最多存的钱已经被挥霍完了,那么又该存新的了;一种是还没有被挥霍完,可以到下一个区间去挥霍。对于前者,直接++ans然后初始化一下lowtop;对于后者,如果low<0,就让low=0,因为不能存0元以下。
大概可以看做下图:

这一贪心思想比较奇特,是对区间贪心的。至于题解给的…恕在下没有看懂…


#include <iostream>
#include <cstdio>
using namespace std;
int low, top, n, k, ans;
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0); 
    cin>>n>>k;
    low = 0; top = 0;
    int tmp;
    for(int i = 1; i <= n; ++i) {
        cin>>tmp;
        if(tmp) {
            low += tmp, top += tmp;
            if(low > k) return puts("-1"), 0;
            if(top > k) top = k;
        } else {
            if(top < 0){++ans, top = k, low = 0;}
            else low = max(low, 0);
        }
    }
    cout<<ans<<endl;
    return 0;
}
### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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