UVA 11054 Wine trading in Gergovia(贪心）

2006/2007 ACM International Collegiate Programming Contest 

University of Ulm Local Contest

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 Specification

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

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

Output Specification

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

题意:n个人，每个人都要买酒或者卖酒，正数代表买酒负数卖酒。运费等于和两人的距离。。需求和供应总是平衡的。要求出买卖最少运费。。
思路：贪心。。有个注意点：比如第1个人要卖100瓶酒给第3个人。。运费等同于先把酒卖给第2个人，再由第2个人卖给第三个人。。如果这样考虑。每次就只要从左边开始不断和自己右边的人交易即可。。运费肯定都是一样的。。

代码：

#include <stdio.h>
#include <stdlib.h>

int n, num, sb;
long long sum;

int main() {
while (~scanf("%d", &n) && n) {
sum = num = 0;
for (int i = 0; i < n ; i ++) {
scanf("%d", &sb);
num += sb;
sum += abs(num);
}
printf("%lld\n", sum);
}
return 0;
}