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Wine trading in Gergovia
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描述
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.
输入
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.
输出
示例输入
5 5 -4 1 -3 1 6 -1000 -1000 -1000 1000 1000 1000 0
示例输出
9 9000
分析：
一个供求平衡的城市，各个商家一字摆开，且单位路程交易产生单位工作量，贪心求最小工作量。
从第一家开始向后每个商家都跟邻居交易，另辟数组储存每次交易工作量，加和即可
#include<stdio.h>
#include<math.h>
int s[100010],t[100010];
int main()
{int n,i;long long sum;while(scanf("%d",&n),n!=0){for(i=1;i<=n;i++){scanf("%d",&s[i]);}sum=0;t[0]=0;for(i=1;i<=n;i++){t[i]=t[i-1]+s[i];sum+=fabs(t[i]);}printf("%lld\n",sum);}return 0;
}
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