UVA 11054 Wine trading in Gergovia （Gergovia 的酒交易）（贪心+模拟）

题意：直线上有n(2<=n<=100000)个等距的村庄，每个村庄要么买酒，要么卖酒。设第i个村庄对酒的需求为ai(-1000<=ai<=1000)，其中ai>0表示买酒，ai<0表示卖酒。所有村庄供需平衡，即所有ai之和等于0。把k个单位的酒从一个村庄运到相邻村庄需要k个单位的劳动力。计算最少需要多少劳动力可以满足所有村庄的需求。

分析：从最左面的村庄考虑，不管他是买酒还是卖酒，相对于他的相邻村庄都会有a0的运输量，所以运输量不断累加或抵消，一直算到最右边村庄即可。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int main(){
    int n;
    while(scanf("%d", &n) == 1){
        if(!n) return 0;
        LL x, ans = 0, last = 0;
        for(int i = 0; i < n; ++i){
            scanf("%lld", &x);
            ans += abs(last);
            last += x;
        }
        printf("%lld\n", ans);
    }
    return 0;
}