CF-Educational Codeforces Round 44 (Rated for Div. 2)-A-Chess Placing_educational codeforces round 46

ACM模版

描述

题解

凑等差数列，一种是 $a_1 = 1$，一种是 $a_1 = 2$，取凑成这两种等差数列的代价中较小的，注意序列并没有保证有序，所以需要先排序。

代码

#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int MAXN = 111;

int n;
int p[MAXN];

int main(int argc, const char * argv[])
{
    while (cin >> n)
    {
        int ans0 = 0, ans1 = 0, pos0 = 1, pos1 = 2;
        n /= 2;
        for (int i = 1; i <= n; i++)
        {
            cin >> p[i];
        }

        sort(p + 1, p + n + 1);

        for (int i = 1; i <= n; i++)
        {
            if (p[i] != pos0)
            {
                ans0 += abs(p[i] - pos0);
            }
            if (p[i] != pos1)
            {
                ans1 += abs(p[i] - pos1);
            }
            pos0 += 2;
            pos1 += 2;
        }

        cout << min(ans0, ans1) << '\n';
    }

    return 0;
}
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