Codeforces 817D Imbalanced Array

本文介绍了一种计算数组中所有子区间最大值与最小值差的总和的方法,并提供了详细的AC代码实现。通过分别计算每个元素作为最大值和最小值出现的次数,有效地解决了问题。

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You are given an array a consisting of n elements. The imbalance value of some subsegment of this array is the difference between the maximum and minimum element from this segment. The imbalance value of the array is the sum of imbalance values of all subsegments of this array.

For example, the imbalance value of array [1, 4, 1] is 9, because there are 6 different subsegments of this array:

  • [1] (from index 1 to index 1), imbalance value is 0;
  • [1, 4] (from index 1 to index 2), imbalance value is 3;
  • [1, 4, 1] (from index 1 to index 3), imbalance value is 3;
  • [4] (from index 2 to index 2), imbalance value is 0;
  • [4, 1] (from index 2 to index 3), imbalance value is 3;
  • [1] (from index 3 to index 3), imbalance value is 0;

You have to determine the imbalance value of the array a.

Input

The first line contains one integer n (1 ≤ n ≤ 106) — size of the array a.

The second line contains n integers a1, a2... an (1 ≤ ai ≤ 106) — elements of the array.

Output

Print one integer — the imbalance value of a.

Example
Input
3
1 4 1
Output
9
这个题,想了挺长时间的,就是不出啊,脑袋都想爆炸了,最后还是看的别人的代码,我就是水个积分,希望他不会怪罪啦(都是一个学校的,希望不要打我啊)如果想看最最最牛逼的代码与讲解请点击这里,想听我讲的也可以继续。。。。。
 
    
题目大意:计算所有子区间里最大值和最小值差的和
题解:我们可以分开来计算,计算每一个位置能做最大值的次数和能做最小值的次数,然后用做最大值得到的价值的和减去做最小值得到的价值的和,就是这个位置所做的贡献,然后求下和就好啦。
AC代码
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e6 + 7;
int a[N];
int l[N];
int r[N];

int32_t main()
{
    int ans = 0, n, i;
    for(i = 1, scanf("%I64d", &n); i <= n; scanf("%I64d", &a[i++]));
    for(i = 1; i <= n; l[i] = r[i] = i, i++);
    for(i = 2; i <= n; i++)
    {
        int num = i;
        while(num > 1 && a[num - 1] <= a[i]) num = l[num - 1];
        l[i] = num;
    }
    for(i = n - 1; i >= 1; i--){
        int num = i;
        while(num < n && a[num + 1] < a[i]) num = r[num + 1];
        r[i] = num;
    }
    for(i = 1; i <= n; ans += (i - l[i] + 1)*(r[i] - i + 1)* a[i], i++);
    for(i = 1; i <= n; l[i] = r[i] = i, i++);
    for(i = 2; i <= n; i++){
        int num = i;
        while(num > 1 && a[num - 1] >= a[i]) num = l[num - 1];
        l[i] = num;
    }
    for(i = n - 1; i >= 1; i--){
        int num = i;
        while(num < n && a[num + 1] > a[i]) num = r[num + 1];
        r[i] = num;
    }
    for(i = 1; i <= n; i++){
        ans -= (i - l[i] + 1)*(r[i] - i + 1) * a[i];
    }
    printf("%I64d\n", ans);
}

### 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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