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

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解决一个数组操作问题，目标是通过最少步数使数组元素与指定目标数组相等，每步可以对数组区间内的元素加减一。

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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 subtract1.

题意：给你n个数，每次操作可以把一个区间内的数同时+1或者-1，求操作的最少次数

思路：直接贪心就行了，相邻数相差的数之和

总结：刚开始没想巧法，暴力模拟，然后WA掉，然后就重新列数据，重新写，哎。。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1000100
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int a[MAXN];
int main()
{
	int n;
	int i,j;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1;i<=n;i++)
		scanf("%d",&a[i]);
		ll ans=0;
		for(i=1;i<=n;i++)
		ans+=fabs(a[i]-a[i-1]);
		printf("%I64d\n",ans);
	}
	return 0;
}