TOJ 3203.Pay Back

题目链接：http://acm.tju.edu.cn/toj/showp3203.html

3203.   Pay Back

---

Time Limit: 2.0 Seconds   Memory Limit: 65536K
Total Runs: 595   Accepted Runs: 260    Multiple test files

---

"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

Source: USACO 2009 March Competition

---

Submit   List    Runs   Forum   Statistics

水题，按照题意，一直模拟着进行就可：

#include <stdio.h>int debt[100001];int main(){int n;while(~scanf("%d",&n)){for(int i=0;i<n;i++)scanf("%d",&debt[i]);int sum=0,step=0,position=0;for(int i=0;i<n;i++){sum+=debt[i];if(sum<0&&sum>debt[i])position=i;else if(sum>=0&&sum<debt[i])step+=2*(i-position);}printf("%d\n",step+n);}}