CF--B. Wilbur and Array_wilbur the pig is tinkering with arrays again. he -优快云博客

本文链接：https://blog.youkuaiyun.com/Inite/article/details/49864753

这是一道关于数组操作的算法题，题目要求从全零数组通过加减1的操作变成目标数组，求最小操作步数。解题思路是遍历数组，累加每个位置上的元素差值，注意数据范围可能导致的溢出问题。

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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

Input

4
1 2 2 1

Output

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：原串(长度为 n)是全 0，给出一个目标串，要求从原串变成目标串的最少步数。每一步可以任选一个原串下标 i , 然后进行加 1 或减 1 的操作(对数组下标 i 到 n的数字都操作)。

思路：因为操作是*左影响了右*，所以想到从左往右遍历，依次比较原串与目标串的对位数字之差，再用变量 sum 累加改变量，用临时变量 tmp 存某一次原串与目标串对位数字之差的绝对值，每次加上此绝对值即得答案 ans。

坑点：一定要用 long long，不然会超 int 范围导致WA！！！

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f
typedef long long ll;
ll src[200020], des[200020];
using namespace std;
int main()
{
#ifdef OFFLINE
	freopen("t.txt","r",stdin);
#endif
	ll i, j, n, ans=0;
	scanf("%lld", &n);
	for(i=1;i<=n;i++){
		scanf("%lld", &des[i]);
	}
	memset(src, 0, sizeof(src));
	ll tmp=0, sum=0;
	for(i=1;i<=n;i++){//从左往右(左影响右)
		if(des[i]==src[i]+sum) continue;//对位相等则不用改变
		tmp=des[i]-(src[i]+sum);//原串与目的串对位数字之差
		sum+=tmp;//累加改变量的变量sum
		ans+=abs(tmp);//加每位对位之差的绝对值
	}
	printf("%lld\n", ans);
	return 0;
}