厦门大学 ACM 1465 连续数列 三分

http://acm.xmu.edu.cn/JudgeOnline/problem.php?id=1465

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;

double arr[112345];
int N;
inline double Abs( double x ){ return x>0?x:-x;}
inline double Min( double a,double b ){return a>b?b:a;}

double work( double ans )
{
    double res = 0;
    for( int i = 1; i <= N; i++ )
        res += Abs( arr[i] - ans  );
    return res;
}
int main( )
{
    while( scanf("%d",&N) != EOF )
    {
        for( int i = 1; i <= N; i++ )
            scanf("%lf",&arr[i]);
        sort( &arr[1],&arr[1]+N );
        for( int i = 1; i <= N; i++ )
        arr[i] -= (i-1);
        sort( &arr[1],&arr[1]+N );
        double lt = arr[1],rt = arr[N];
        while( rt - lt >= 3  )
        {
            double ans1 = (rt+lt+lt)/3.;
            double ans2 = (rt+rt+lt)/3.;
            double res1 = work(ans1);
            double res2 = work(ans2);
            if( res1 < res2 ) rt = ans2;
            else              lt = ans1;
        }
        lt = int (lt-1); rt = int( rt+1 );
        double res = work(lt*1.);
        for( int i = lt; i <= rt; i++ )
            res = Min( res,work(i*1.) );
        printf("%.lf\n",res);
    }
    return 0;
}

　另一种方法

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;

long long arr[112345];
int main( )
{
    long long N;
    while( scanf("%lld",&N) != EOF )
    {
        for( long long i = 1; i <= N; i++ )
        scanf("%lld",&arr[i]);
        sort( &arr[1],&arr[1]+N );
        for( long long i = 1; i <= N; i++ )
        arr[i] -= (i-1);
        sort( &arr[1],&arr[1]+N );
        if( N%2 )
        {
            long long tab = N/2+1,res = 0;
            for( long long i = 1; i <= N; i++ )
               res += abs( arr[i] - arr[tab] );
            printf("%lld\n",res);
        }
        else
        {
            long long tab1 = N/2;
            long long tab2 = N/2+1;
            long long res = 0,ans = 0;
            for( long long i = 1; i <= N; i++ )
               ans += abs( arr[i] - arr[tab1] );
            for( long long i = 1; i <= N; i++ )
               res += abs( arr[i] - arr[tab2] );
            printf("%lld\n",min( res,ans ) );
        }
    }
    return 0;
}