【ECJTU_ACM 11级队员2012年暑假训练赛(7) - E - Little Elephant and Sorting】

E - Little Elephant and Sorting
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 ai. The Little Elephant can make one move to choose an arbitrary pair of integers l and r (1 ≤ l ≤ r ≤ n) and increase ai by 1 for all isuch 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) ai ≤ ai + 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 ≤ ai ≤ 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 cincout streams or the %I64dspecifier.

Sample Input

Input
3
1 2 3
Output
0
Input
3
3 2 1
Output
2
Input
4
7 4 1 47
Output
6

Hint

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].

 

 

 

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 
 5 typedef unsigned long long int longint;
 6 
 7 #define MAXN 100100
 8 
 9 longint iMap[MAXN];
10 longint n;
11 
12 void iInit()
13 {
14     for (longint i = 0; i < n; i++)
15     {
16         cin >> iMap[i];
17     }
18 }
19 
20 void iCalAnswer()
21 {
22     longint iCount = 0;
23     for (longint i = 1; i < n; i++)
24     {
25         if (iMap[i] < iMap[i-1])
26         {
27             iCount += (iMap[i-1] - iMap[i]);
28         }
29     }
30     cout << iCount << endl;
31 }
32 
33 int main()
34 {
35     //freopen("in.dat", "r", stdin);
36     while (cin >> n)
37     {
38         iInit();
39         iCalAnswer();
40     }
41     return 0;
42 }
43 
44 // end 
45 // ism 

 

posted @ 2012-08-05 23:48  ismdeep  阅读(196)  评论(0编辑  收藏  举报