uva 11054 wine trading in gergovia （归纳【好吧这是我自己起的名字】）——yhx

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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 specic amount of wine. They are clever enough
to gure 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). The following line contains n integers ai (??1000  ai  1000). If ai  0, it
means that the inhabitant living in the i-th 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
For each test case print the minimum amount of work units needed so that every inhabitant has his
demand fullled. You may assume that this number ts into a signed 64-bit integer (in C/C++ you
can use the data type \long long", in JAVA the data type \long").

 1 #include<cstdio>
 2 #include<cstring>
 3 long long a[100010];
 4 long long abs(long long x)
 5 {
 6     return x>=0?x:-x; 
 7 }
 8 int main()
 9 {
10     int i,j,k,m,n,p,q;
11     long long x,y,z,ans;
12     while (scanf("%d",&n)&&n)
13     {
14         for (i=1;i<=n;i++)
15           scanf("%lld",&a[i]);
16         ans=0;
17         for (i=1;i<n;i++)
18         {
19             ans+=abs(a[i]);
20             a[i+1]+=a[i];
21         }
22         printf("%lld\n",ans);
23     }
24 }

用归纳【我以前见过的一个词，不知道恰不恰当】法分析如下。

1.想让1号点满足要求，一定来自2号点（不管是2号点本身还是从更靠后的点来），于是1号点的需求得到满足，其转移到2号点。

2.若n-1号点及以左已满足要求，则对于n号点来说，他的需求必须由n+1号点满足（不管是n+1本身还是从更后的来）。于是n号点得到满足，转移到n+1号点。

由以上两点都成立可知应从左往右扫描，把需求转移到下一个点。

uva 11054 wine trading in gergovia （归纳【好吧这是我自己起的名字】）——yhx

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原文地址：http://www.cnblogs.com/AwesomeOrion/p/5402272.html