Codeforces Round #331 (Div. 2) B. Wilbur and Array

B. Wilbur and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Wilbur the pig is tinkering with arrays again. He has the array a₁, a₂, ..., a_n initially consisting of n zeros. At one step, he can choose any index i and either add 1 to all elements a_i, a_i + 1, ... , a_n or subtract 1 from all elements a_i, a_i + 1, ..., a_n. His goal is to end up with the array b₁, b₂, ..., b_n.

Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the length of the array a_i. Initially a_i = 0 for every position i, so this array is not given in the input.

The second line of the input contains n integers b₁, b₂, ..., b_n ( - 10⁹ ≤ b_i ≤ 10⁹).

Output

Print the minimum number of steps that Wilbur needs to make in order to achieve a_i = b_i for all i.

Sample test(s)

Input

5 
1 2 3 4 5

Output

5

Input

4 
1 2 2 1

Output

3

Note

In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.

In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1.

题意：输入n 接下来 输入n个数

  n个数初始都为零 现在可以执行两种操作 增加1或者减少1 例如 i个数增加1时 第i+1，i+2..到n 个数 都增加1

执行一次 算一次操作  问最少经过多少次操作 使得这n个数的值为 输入的排列

解答： for循坏遍历一遍  就可以保证操作数最小

       比如第i个位置操作几次 只与第i-1位置上的数有关 因为题目规定的操作只对之后的有影响 （注意理解）！！

注意 ：当处理第一位的时候 默认之前一位为0

      __int64

      别乱用abs

      做的一手死！！！

#include<bits/stdc++.h>
using namespace std;
__int64 next,a;
__int64 re;
__int64 n;
__int64 ans;
int main()
{
    re=0;
    scanf("%I64d",&n);
    //scanf("%I64d",&a);
    re=0;
    next=0;
    for(int i=0;i<n;i++)
    {
        scanf("%I64d",&a);
        ans=a-next;
        next=a;
        if(ans<0)
            ans=-ans;
        re+=ans;

    }
    printf("%I64d
",re);
    return 0;
}