uva 11300

这是个纯粹推理数学公式的题

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

大致意思就是分硬币 每个人有一个初始的硬币数量
然后人都坐了一圈
每个人可以向相邻的两个人给出硬币
最后保证硬币数量相同

求 最少移动的硬币数量

首先 这是一个纯粹的数学推理

首先设人为A1

A1给出的硬币数量为x1 A2给出的为x2。。。。。。。。。

以上就是推理过程

x1为什么是等于负的c[n-2]？
为了最小

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <queue>#include <cmath>using namespace std;long long c[10000005],b[10000005],j[100005];int main(){    int n;    while(cin>>n)    {        int i;        long long x=0;        for(i=1;i<=n;i++)        {            cin>>b[i];            x+=b[i];        }        x=x/n;        c[0]=0;        for(i=1;i<n;i++)        {            c[i]=c[i-1]+x-b[i];        }        sort(c,c+n); //以此确定点的顺序        x=-c[n/2];  //然后找到中间点        long long y=0;        for(i=0;i<n;i++)        {            y=y+abs(c[i]+x);        }        cout<<y<<endl;    }    return 0;}