CodeForces 438D 浅谈区间取模线段树

博客探讨了CodeForces 438D问题中如何处理区间取模的线段树问题。文章指出,当遇到区间操作时,应考虑是否能进行区间合并。对于取模操作,如果模数大于区间内所有数,操作无效。分析表明,取模次数最多为nlogn,因此暴力取模在复杂度上是可行的。解决方案是在线段树中维护区间最大值,若取模数小于最大值则执行取模操作。

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这里写图片描述
世界真的很大
很多正解其实本来都是暴力,但是莫名其妙地过了,然后分析一波复杂度貌似没有问题,然后就是正解了
线段树的运用的却有很多,其实只要要处理的问题满足区间合并的性质就行了
在遇见新的需要处理的问题时,要优先考虑他的信息对于询问而言能不能区间合并
如果可以那就可以区间合并了
如果不行的话就只能想其他办法了,或者暴力上。
有些时候暴力并不一定不是正解,要好好分析题目性质,计算一波复杂度
看题先:
description:

长度为n的非负整数数列,3种操作
1. 求[L,R]所有数的和。
2. 将[L,R]中所有数都mod x。
3. 将a[i]修改为v。
n,m≤100000

input

第一行两个整数n,m,表示数列元素个数和操作数
接下来n个数,表示序列
接下来m行,每行开头一个整数表示操作

output

对于每一个询问操作,输出一个整数表示答案

如果没有2操作的话,还是很简单的线段树裸题,单点修改,区间查询
现在考虑怎么处理这个取模的问题
对于一段区间,如果取模的数比这段区间所有的数都大,那取模就是没有意义的,就是说,如果取模的数比区间最大的数还大,那么就不用取模了,所以我们在线段树里再记录一个区间最大值
考虑每次取模,对于每一个数x,取模y,x mod y的值必然比y小,如果y小于x/2,那x就变得小于x/2,如果y大于x/2,x剩下的部分也比x/2少,x也会变得比x/2小
那么就是说x每次取模都会变得比x/2小,就是说,对于一个数x,有效的取mod最多进行logx次,一共只会进行nlogn次取模,那么就算对所有的数取模,这个时间复杂度都是可以接受的
分析一波之后,发现,直接暴力去取模是可行的。
那么我们对于每个区间记录一个最大值,如果取模的数大于最大值,就不管,如果小于最大值,就暴力取模
完整代码:

#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long dnt;

struc
### 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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