Codeforces 817D Imbalanced Array

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最新推荐文章于 2021-10-21 15:32:45 发布

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分类专栏： Codeforces 数学文章标签： codeforces

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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... a_n (1 ≤ a_i ≤ 10⁶) — elements of the array.

Output

Print one integer — the imbalance value of a.

Example

Input

3
1 4 1

Output

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);
}