UVa11300 - Spreading the Wealth （分金币）

最新推荐文章于 2018-04-25 20:48:34 发布

转载最新推荐文章于 2018-04-25 20:48:34 发布 · 131 阅读

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原文链接：http://www.cnblogs.com/nigel-jw/archive/2013/05/24/3097599.html

本文介绍了一个关于共产主义制度下财富平均分配的问题。通过将个人财产转化为等值货币，并采用特定算法来计算最少的货币转移数量，使得每个人最终拥有相等的货币数。文章提供了完整的C++代码实现。

F. Spreading the Wealth

Problem

A Communist regime is trying to redistribute wealth in a village. They have have decided to sit everyone around a circular table. First, everyone has converted all of their properties to coins of equal value, such that the total number of coins is divisible by the number of people in the village. Finally, each person gives a number of coins to the person on his right and a number coins to the person on his left, such that in the end, everyone has the same number of coins. Given the number of coins of each person, compute the minimum number of coins that must be transferred using this method so that everyone has the same number of coins.

The Input

There is a number of inputs. Each input begins with n(n<1000001), the number of people in the village. nlines follow, giving the number of coins of each person in the village, in counterclockwise order around the table. The total number of coins will fit inside an unsigned 64 bit integer.

The Output

For each input, output the minimum number of coins that must be transferred on a single line.

Sample Input

Sample Output

0
4

 1 #include<cstdio>
 2 #include<math.h>
 3 #include<algorithm>
 4 #define maxn 1000010
 5 using namespace std;
 6 long long A[maxn] , C[maxn];
 7 int main()
 8 {
 9     int n;
10     int i;
11     long long total , M , ans;
12     while(scanf("%d" , &n)!=EOF)
13     {
14         total = 0;
15         for(i = 0; i < n; i++)
16         {
17             scanf("%lld" , &A[i]);
18             total += A[i];
19         }
20         M = total / n;
21         C[0] = 0;
22         for(i = 1; i < n; i++)
23         {
24             C[i] = C[i-1] + A[i] - M;
25         }
26         sort(C , C+n);
27         long long x  = C[n / 2];
28         ans = 0;
29         for(i = 0; i < n; i++)
30         {
31             ans += abs(x - C[i]);
32         }
33         printf("%lld\n" , ans);
34     }
35     return 0;
36 }

View Code

转载于:https://www.cnblogs.com/nigel-jw/archive/2013/05/24/3097599.html