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, ..., an initially consisting of n zeros. At one step, he can choose any index i and either add 1 to all elements ai, ai + 1, ... , an or subtract 1 from all elements ai, ai + 1, ..., an. His goal is to end up with the array b1, b2, ..., bn.

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 ai. Initially ai = 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, ..., bn ( - 109 ≤ bi ≤ 109).

Output

Print the minimum number of steps that Wilbur needs to make in order to achieve ai = bi 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\n",re);
return 0;
}

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