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  • Codeforces Round #129 (Div. 2) B

    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 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 < nai ≤ 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.

    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:可以把区间的数增加相同的值(值的范围是1-n),让数列成为非递减

    解法:当然让不符合要求的数字增加到最近的最大值,比如 7 4 1变成7 7 7 ,我们只需要计算7 4,4 1之间的差值就行(增加同一个数,两个数的差值不变的)

    #include<bits/stdc++.h>
    using namespace std;
    long long a[100005];
    int pos;
    int n;
    int d;
    int main()
    {
        long long sum=0;
        int pos=0;
        int flag=0;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        d=a[1];
        for(int i=2;i<=n;i++)
        {
           if(a[i]<a[i-1])
           {
               sum+=(a[i-1]-a[i]);
             //  pos=(d-a[i]);
           }
         //  cout<<a[i]<<"A"<<endl;
        }
        cout<<sum<<endl;
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/5721280.html
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