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

B. Wilbur and Array

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/596/problem/B

Description

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 Input

5
1 2 3 4 5

Sample Output

5

HINT

题意

a[i]数组一开始都是0

然后你可以使得i-n都加1，也可以使得i-n都减一

然后问你最少操作数

题解：

扫一遍就好了，直接暴力，注意会爆int

代码

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define maxn 200005
long long b[maxn];
int main()
{
    int n;cin>>n;
    for(int i=;i<n;i++)
        scanf("%lld",&b[i]);
    long long now = ;
    long long ans = ;
    for(int i=;i<n;i++)
    {
        if(now!=b[i])
        {
            ans+= abs(b[i]-now);
            now = b[i];
        }
    }
    printf("%lld\n",ans);
}