UVA11300 Spreading the Wealth (数学推导+中位数)

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 t inside an unsigned 64 bit integer.
Output
For each input, output the minimum number of coins that must be transferred on a single line.
SampleInput
3
100
100
100
4
1
2
5
4
SampleOutput
0
4

这道题数学味道太浓了。。

我们假设原来金币数为a1,a2,““`an.
最终的金币为这些数的平均值，设为A
xi表示i给i+1的金币个数
a1+xn-x1=A(每个人可以理解为被给了多少金币，给出了多少金币使得所有人的金币数平衡)
a2+x1-x2=A
a3+x2-x3=A
a4+x3-x4=A
……
an+x(n-1)-xn=A

用x1表示 x2,x3,…xn
x2=x1-(A-a2)
x3=x1-(2*A-a2-a3);

结果就是求|x1|+|x1-b1|+|x1-b2|+…+|x1-b[n-1]|的最小值，取中位数
这里的bi=A-ai，这里的A指的是平均数，ai表示第i个输入的数字。

#include<cstdio>#include<iostream>using namespace std;#include<cstring>#include<algorithm>typedef  long long LL;const int maxn=1e6+5;int a[maxn];LL C[maxn];int main() {    int n;    while(~scanf("%d",&n)) {        LL fl=0;        for(int i=0; i<n; ++i) {            scanf("%d",&a[i]),fl+=a[i];        }        fl/=n;        LL ans=0;        C[0]=0LL;        for(int i=1; i<n; ++i) {            C[i]=C[i-1]+a[i]-fl;        }        sort(C,C+n);        LL ave=C[n>>1];        for(int i=0;i<n;++i){            ans+=abs(ave-C[i]);        }        printf("%lld\n",ans);    }    return 0;}