Uva 11300 Spreading the Wealth

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.

Solution

we assume that M is the final number of coins. And xi means the number of coins that ith people gives to the (i-1)th people and x1 means the number of coins that 1st people gives to the nth people. Why? if people 1 gives 2 coins to people 2 and people 2 gives 4 coins to people 1, we can regard people 2 gives 2 coins to people as this and xi could be negative.

Every people will have M coins in the last. Let Ai be the coins people i have at the beginning time.

As for people 1, we have A1 + x2 - x1 = M ==> x2 = x1 - (A1 - M)
As for people 2, we have
A2 + x3 - x2 = M ==> x3 = x2 - (A2 - M) ==> x3 = x1 - (A2 + A1 - 2M)
and so on
xi = x1 - (A(n-1) + A(n-2) + …. + A1 - (n-1)M )

let Ci = Ai+..+A1 - iM

Now we want |x1| + |x2| + … + |xn| be minimum. In another word is |x1| + |x1-C1| + |x2-C2| + … + |x1-Cn-1| be minimum.

That is to say, we need to find a point in number axis so that the sum of the distances between the point and other points is minimum. And we can find the Median of the C[]. Let x1 be the Median, and the answer is clear.

code:

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1000010;typedef long long ll;ll a[maxn],c[maxn];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        ll sum = 0;        for(int i = 0;i < n;++ i)        {            scanf("%lld",&a[i]);            sum += a[i];        }        ll M = sum/n;        c[0] = 0;        for(int i = 1;i < n;++ i)            c[i] = c[i-1] + a[i-1] - M;        sort(c,c+n);        ll avg = c[(n-1)/2];        ll ans = 0;        for(int i = 0;i < n;++ i)            ans += fabs(avg-c[i]);        printf("%lld\n",ans);    }    return 0;}