Sicily 2013 Pay Back

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

"Never a borrower nor a lender be." O how Bessie wishes she had taken that advice! She has borrowed from or lent money to each of N (1 ≤ N ≤ 100,000) friends, conveniently labeled 1..N.

Payback day has finally come. She knows she is owed more money than she owes to the other cows. They have all lined up in a straight line, cow i standing i meters from the barn. Bessie is going to traverse the line collecting money from those who owe her and reimbursing money to those she owes.

As she moves down the line, she can request any cow who owes her money to give her the money. When she has enough money to pay off any or all of her debts, she can pay the (recently collected) money to those she owes. Cow i owes Bessie D_i money (-1,000 ≤ D_i ≤ 1,000; D_i != 0). A negative debt means that Bessie owes money to the cow instead of vice-versa.

Bessie starts at the barn, location 0. What is the minimum distance she must travel to collect her money and pay all those she owes? She must end her travels at the end of the line.

 

Input

Line 1: A single integer: N
Lines 2..N+1: Line i+1 contains a single integer: D_i

Output

Line 1: A single integer that is the total metric distance Bessie must travel in order to collect or pay each cow.

Sample Input

5100-200250-200200

Sample Output

9

Problem Source

每周一赛：2010中山大学新手赛

Solution

题意是给出一个序列，分别是欠钱和还钱的数额，在同一条路上，距离是i，问最短把全部债务解决的路的长度。

贪心就好啦，记录下最远一个要还钱的i，一旦有钱（手上总额大于等于0）就去还就ok啦。

#include <cstdio>int d[100005];int main(){  int i, n, last, sum = 0, ans = 0;  scanf("%d", &n);  for (i = 1; i <= n; ++i)  {    ++ans;    scanf("%d", &d[i]);    if (sum >= 0 && sum + d[i] < 0) last = i;    if (sum < 0 && sum + d[i] >= 0) ans += 2 * (i - last);    sum += d[i];  }  printf("%d\n", ans);  return 0;}