UVA 11300 - Spreading the Wealth(数学推导+中位数)

最新推荐文章于 2020-01-09 10:51:30 发布

lab104_yifan

最新推荐文章于 2020-01-09 10:51:30 发布

阅读量1.6k

点赞数 1

CC 4.0 BY-SA版权

分类专栏：算法竞赛入门经典训练指南-第一章

本文链接：https://blog.youkuaiyun.com/accelerator_/article/details/17185853

算法竞赛入门经典训练指南-第一章专栏收录该内容

181 篇文章

订阅专栏

本文介绍了一个共产主义村落通过相邻人员间货币转移实现财富平均分配的算法。重点在于如何计算最小货币转移数量，确保最终每个人拥有相同的货币量。文章提供了一种有效的方法，并给出了具体实现的代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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. n lines 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

题意:n个人坐成一圈，每个人有一些钱，现在要平分这些钱，每个人只能把钱给周围的人，问最少要转移多少钱才能平分。

思路：推导，每个人最终得钱数可以算出为M，对于第i个人来说，Xi为他给上一个人的钱，如此一来 X(i+1) = M - Ai + Xi；

X2 = M - A1 + X1 = X1 - C1.依次类推，Xi + 1 = X1 - Ci. Ci数组是可以递推出来的，然后答案就是X1 + |X1 - C1| + |X1 - C2| 。。。。 + |X1 - Cn -1|。中位数为最佳答案。

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 1000005;

int n;
long long A[N], C[N], sum, M, ans;

int main() {
    while (~scanf("%d", &n)) {
	sum = ans = 0;
	for (int i = 1; i <= n; i ++) {
	    scanf("%lld", &A[i]);
	    sum += A[i];
	}
	M = sum / n;
	for (int i = 1; i < n ; i ++)
	    C[i] = C[i - 1] + A[i] - M;
	sort(C, C + n);
	int x = C[n / 2];
	for (int i = 0; i < n; i ++)
	    ans += abs(C[i] - x);
	printf("%lld\n", ans);
    }
    return 0;
}