<codeforces>Little Elephant and Sorting

小象排序挑战

最新推荐文章于 2022-05-30 01:20:02 发布

原创最新推荐文章于 2022-05-30 01:20:02 发布 · 796 阅读

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文章标签：

#sorting #output #input #integer #numbers #pair

本文介绍了一个有趣的排序问题：如何使用最少的操作次数将整数数组变为非递减序列。通过简单的算法，文章提供了一种有效的方法来解决这个问题。

The Little Elephant loves sortings.

He has an array a consisting of n integers. Let's number the array elements from 1 to n, then the i-th element will be denoted as a_i. The Little Elephant can make one move to choose an arbitrary pair of integers l and r (1 ≤ l ≤ r ≤ n) and increase a_i by 1 for all i such that l ≤ i ≤ r.

Help the Little Elephant find the minimum number of moves he needs to convert array a to an arbitrary array sorted in the non-decreasing order. Array a, consisting of n elements, is sorted in the non-decreasing order if for any i (1 ≤ i < n) a_i ≤ a_i + 1 holds.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the size of array a. The next line contains nintegers, separated by single spaces — array a (1 ≤ a_i ≤ 10⁹). The array elements are listed in the line in the order of their index's increasing.

Output

In a single line print a single integer — the answer to the problem.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use thecin, cout streams or the %I64d specifier.

Sample test(s)

input
3
1 2 3

output
0

input
3
3 2 1

output
2

input
4
7 4 1 47

output
6

Note

In the first sample the array is already sorted in the non-decreasing order, so the answer is 0.

In the second sample you need to perform two operations: first increase numbers from second to third (after that the array will be: [3, 3, 2]), and second increase only the last element (the array will be:[3, 3, 3]).

In the third sample you should make at least 6 steps. The possible sequence of the operations is: (2; 3), (2; 3), (2; 3), (3; 3), (3; 3), (3; 3). After that the array converts to [7, 7, 7, 47].

#include <iostream>
using namespace std;
int main()
{
    int n,a,b;
    long long int add;
    while(cin>>n>>a)
    {
        add=0;
        while(--n)
        {
            cin>>b;
            if(a>b)
            {
                add+=a-b;
            }
            a=b;
        }
        cout<<add<<endl;
    }
    return 0;
}