uva11300Spreading the Wealth<数学>

链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2275

思路

原来金币数为a1,a2,`````an;

最终的金币为这些数的平均值，设为M

xi表示i给i+1的金币个数

a1+xn-x1=M
a2+x1-x2=M
a3+x2-x3=M
a4+x3-x4=M
.......
an+x(n-1)-xn=M
利用后面n-1个等式，用x1表示 x2,x3,...xn
x2=x1-(M-a2)
x3=x1-(2*M-a2-a3);
....

结果就是求|x1|+|x1-b1|+|x1-b2|+...+|x1-b[n-1]|的最小值，取中位数;

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
int N;
LL c[1000005], sum, M, a;
LL Labs(  LL a )
{
    return a>0?a:-a;
}
int main( )
{
    while( scanf("%d", &N )!= EOF ){
        sum=0, c[0]=0;
        for( int i=1; i<=N; ++ i ){
            scanf( "%lld", &a );
            sum+=a;    
            c[i]=sum;
        }
        M=sum/N;
        sum=0;
        for( int i=1; i<=N; ++ i ){
            c[i]-=i*M;
        }
        sort( c, c+N );
        M=c[N/2];
        for( int i=0; i<N; ++ i ){
            c[i]-=M;
            sum+=Labs(c[i]);
        }
        printf( "%lld\n", sum );
    }
    return 0;
}