题目传送门

算法

(暴力枚举) $O(n)$

算出总和之后枚举$p_2$的所有可能位置

C++ 代码

#include <iostream>

using namespace std;
using LL = long long; //一定要开long long

const int N = 100010;

int n;
int a[N];
int m, p1, s1, s2, p2;

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++ i )
        scanf("%d", &a[i]);
    scanf("%d%d%d%d", &m, &p1, &s1, &s2);

    LL d = 0, t = 0;
    for (int i = 1; i < m; ++ i )
        d += a[i] * (m - i);
    for (int i = m + 1; i <= n; ++ i )
        t += a[i] * (i - m);

    if (p1 < m)
        d += s1 * (m - p1);
    else if (p1 > m)
        t += s1 * (p1 - m);

    LL mi = 0x7fffffffffffffff;
    for (int i = 1; i <= n; ++ i )
    {
        int l = abs(i - m);
        if (i < m)
        {
            if (abs(d + s2 * l - t) < mi)
            {
                mi = abs(d + s2 * l - t);
                p2 = i;
            }
        }
        else if (i > m)
        {
            if (abs(t + s2 * l - d) < mi)
            {
                mi = abs(t + s2 * l - d);
                p2 = i;
            }
        }
        else
        {
            if (abs(t - d) < mi)
            {
                mi = abs(t - d);
                p2 = i;
            }
        }
    }

    printf("%d", p2);
    return 0;
}