C - Little Elephant and Sorting CodeForces - 205B

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 n integers, 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 the cin, cout streams or the %I64d specifier.

Examples
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].

题意：
可以选一段连续数字加一，要使得序列单调不减，求最少操作次数。

思路：
每次考虑相邻两个数字，如果后面数字小于前面数字，则必须进行两数差值次操作。当然还需要考虑这些操作是否能延续到后面或者能否收到前面延续来的操作。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include <map>
#include <unordered_map>

using namespace std;
typedef unsigned long long ll;

const int maxn = 2e5 + 7;

ll a[maxn];

int main() {
    
    int n;scanf("%d",&n);
    ll ans = 0,now = 0;
    for(int i = 1;i <= n;i++) scanf("%lld",&a[i]);
    for(int i = 2;i <= n;i++) {
        if(a[i] <= a[i - 1]) {
            if(a[i] + now <= a[i - 1]) {
                a[i] += now;
                now += a[i - 1] - a[i];
                ans += a[i - 1] - a[i];
                a[i] = a[i - 1];
            } else {
                now = a[i - 1] - a[i];
                a[i] = a[i - 1];
            }
        } else {
            now = 0;
        }
    }
    printf("%lld\n",ans);
    return 0;
}