Wine trading in Gergovia

As you may know from the comic “Asterix and the Chieftain’s Shield”, Gergovia consists of one street, and every inhabitant of the city is a wine salesman. You wonder how this economy works? Simple enough: everyone buys wine from other inhabitants of the city. Every day each inhabitant decides how much wine he wants to buy or sell. Interestingly, demand and supply is always the same, so that each inhabitant gets what he wants.

There is one problem, however: Transporting wine from one house to another results in work. Since all wines are equally good, the inhabitants of Gergovia don’t care which persons they are doing trade with, they are only interested in selling or buying a specific amount of wine. They are clever enough to figure out a way of trading so that the overall amount of work needed for transports is minimized.

In this problem you are asked to reconstruct the trading during one day in Gergovia. For simplicity we will assume that the houses are built along a straight line with equal distance between adjacent houses. Transporting one bottle of wine from one house to an adjacent house results in one unit of work.

Input

The input consists of several test cases. Each test case starts with the number of inhabitants n(2≤n≤100000)n(2 \le n \le 100000). The following line contains nn integers ai(−1000≤ai≤1000)a_i(-1000 \le a_i \le 1000). If ai≥0a_i \ge 0, it means that the inhabitant living in the ii-th house wants to buy aia_i bottles of wine, otherwise if ai<0a_i, he wants to sell aia_i bottles of wine. You may assume that the numbers aia_i sum up to 00.

The last test case is followed by a line containing ‘0’.

Output

For each test case print the minimum amount of work units needed so that every inhabitant has his demand fulfilled. You may assume that this number fits into a signed 64-bit integer (in C/C++ you can use the data type “long long”, in JAVA the data type “long”).

Sample Input

5
5 -4 1 -3 1
6
-1000 -1000 -1000 1000 1000 1000
0

Sample Output

9
9000

直线上有n(2≤n≤100000)n(2≤n≤100000)个等距的村庄，每个村庄要么买酒，要么卖酒。设第ii个村庄对酒的需求为ai(−1000≤ai≤1000)a_i(-1000≤a_i≤1000)，其中ai>0a_i>0表示买酒，ai<0a_i表示卖酒。所有村庄供需平衡，即所有aia_i之和等于00。

思路非常简单，就是从左到右，将酒累加到下一个上去，总移动单位则为每次累加的绝对值之和。

其证明过程为，nn个村庄，无论是买酒还是卖酒，最左边的a1a_1都需要从a2a_2获取到∣∣a1∣∣\left| a_1 \right|个移动单位，这时a2=a1+a2a_2 = a_1 + a_2，这样就等价于n−1n-1个村庄的问题了。

#include <iostream>
#include <cmath>
using namespace std;
typedef long long LL;
int main() {int n;while(cin >> n && n) {LL sum = 0;LL ans = 0;LL num;for(int i = 0; i < n; i++) {cin >> num;sum += num;ans += abs(sum);}cout << ans << endl;}return 0;
}

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