Uva 11300.Spreading the Wealth

261

题目



Description  


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.


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.


Output  


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


Sample Input  


3
100
100
100
4
1
2
5
4


Sample Output  


0
4



题解


n 个人循环给金币,使大家最后的金币数一样

直接模拟非常麻烦
根据数学推导
假设第 i 个人原有 Ai 个金币,他会给第 i+1 个人(最后一个人给第 1 个) xi 个金币
金币平均数为 M

  • 对于第 1 个: M = A1 - x1 + xn
  • 对于第 2 个: M = A2 - x2 + x1
    可化为: x2 = A2 - M + x1
  • 对于第 3 个: M = A3 - x3 + x2
    可化为: x3 = A3 - M + x2 = A2 + A3 - 2M + x1
  • ……
  • 对于第 n 个: xn = A2 + …… + An - (n-1)M + x1

要求的就是 |x1| + |x2| + |x3| + …… + |xn| 的最小值
也即 |x1| + |A2 - M + x1| + |A2 + A3 - 2M + x1| + …… + |A2 + …… + An - (n-1)M + x1
其中每部分非 x1 的部分全部都是常数,可化成 |x1 - c1| + |x1 - c2| + …… + |x1 - cn|
只需求出距离坐标轴上 c1 c2 c3 …… cn 距离和最短的点
显然最中间的点就是答案,可以取中间点(或中间两个点中间任意一个点),分别求出其到各点的和就是答案

需要使用 long long

状态不佳,简直药丸

代码


/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

const int maxn = 1000005;

long long a[maxn];

bool Do() {
    int n;
    if(!(cin >> n))
        return false;

    long long sum = 0;
    for(int i = 0;i < n;i++) {
        long long t;
        cin >> t;
        sum += t;
        if(i == 0)
            a[0] = t;
        else
            a[i] = sum - a[0];
    }
    sum /= n;
    for(int i = 0;i < n;i++) {
        if(i == 0)
            a[0] = 0;
        else
            a[i] -= i*sum;
    }
    sort(a,a + n);

    sum = 0;
    for(int i = 0;i < n;i++)
        sum += abs(a[i] - a[n / 2 - 1]);

    cout << sum << endl;
    return true;
}

int main() {
    while(Do());
    return 0;
}
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