CodeForces 1084A The Fair Nut and Elevator

题意：

电梯要停在x层，每天早上要从x层去a层接人送到1层再回到x层的等待，每天晚上要从x层下到1层接人送到a层，再回到x层（对每层的乘客都是如此），电梯每走一层耗电1，问电梯停在那一层，每天的耗电最少。（电梯每次运载1人）

思路：

此题不难，就是读题太费劲，数据量不大，暴力去找每一层即可，对于a层的乘客，每天的耗电为：

a[i] * (fabs(i - index) + i + index - 2) * 2

Code：

#include <iostream>
#include <cmath>#define INF 0x3f3f3f3f
#define max(x, y) (x >= y ? x : y)
#define min(x, y) (x < y ? x : y)using namespace std;const int maxn = 105;int a[maxn];
int n;int solve(int index)
{int sum = 0;for (int i = 1; i <= n; i++){sum += a[i] * (fabs(i - index) + i + index - 2) * 2;}return sum;
}int main()
{cin >> n;for (int i = 1; i <= n; i++) cin >> a[i];int ans = INF;for (int i = 1; i <= n; i++){int val = solve(i);ans = min(ans, val);}cout << ans << endl;return 0;
}

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