题干：
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 aiby 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 nelements, is sorted in the non-decreasing order if for any i (1 ≤ i < n) ai ≤ ai + 1holds.
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 %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].
题目大意：
给你n个数，定义一次操作：任选一个区间+1，目的是让后一个数不能小于前一个数（不下降子序列），然后最小操作数是多少？
解题报告：
脑补一下过程就好了。就好了。。注意longlong。（好像好多人int交是错的）
首先肯定是从前往后扫，因为这个问题如果倒着找的话，你不知道应该加到多少（因为前面的数字还没有定下来）。从前往后找但是相对差值是不变的，因为每次加的时候肯定要连带着后面的数字一块加，这样是最方便的。想到这里代码就出来了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll a[MAX];
int main()
{
int n;
ll ans = 0,cur = 0;
cin>>n;
for(int i = 1; i<=n; i++) scanf("%lld",a+i);
cur = a[1];
for(int i = 1; i<=n; i++) {
if(a[i] < cur) {
ans += (cur-a[i]);
}
cur = a[i];
}
printf("%lld\n",ans);
return 0 ;
}
错误代码（找到的一个错误代码）：总之这个代码错误很多，比如应该是if(arr[i]<base)，再比如else中也应该更新base，，，但是有一个值得注意的点就是那个else中，不能更新minn！！
#include <iostream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
int main(){
int n,i;
ll base=0;;
cin>>n;
ll arr[100010]={0},count=0,mint=0;
for(i=0;i<n;i++){
cin>>arr[i];
if(arr[i]>=base){
count+=base-mint;
base=arr[i];
mint=arr[i];
}
else{
mint=min(mint,arr[i]);
}
}
cout<<count+(base-mint)<<endl;
return 0;
} 总结：
其实整个题的思路是，看样例猜算法，猜出个大概的，（或者猜出好多个算法），然后证明一下看哪个是正确的，或者帮助我们排除掉一些错误的。
思维过程是，这题首先想到是不能模拟的啊，1e5的数据量，500ms的时间，只能O(n)，所以肯定不能模拟整个过程，然后开始找规律，找到，写代码，提交，AC。
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