【ECJTU_ACM 11级队员2012年暑假训练赛(7) - E - Little Elephant and Sorting】
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 cin, cout streams or the %I64dspecifier.
Sample Input
3
1 2 3
0
3
3 2 1
2
4
7 4 1 47
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