【CodeForces】610B - Vika and Squares（模拟）

Vika and Squares

Description

Vika has n jars with paints of distinct colors. All the jars are numbered from 1 to n and the i-th jar contains a_i liters of paint of color i.

Vika also has an infinitely long rectangular piece of paper of width 1, consisting of squares of size 1 × 1. Squares are numbered 1, 2, 3and so on. Vika decided that she will start painting squares one by one from left to right, starting from the square number 1 and some arbitrary color. If the square was painted in color x, then the next square will be painted in color x + 1. In case of x = n, next square is painted in color 1. If there is no more paint of the color Vika wants to use now, then she stops.

Square is always painted in only one color, and it takes exactly 1 liter of paint. Your task is to calculate the maximum number of squares that might be painted, if Vika chooses right color to paint the first square.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of jars with colors Vika has.

The second line of the input contains a sequence of integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), where a_i is equal to the number of liters of paint in the i-th jar, i.e. the number of liters of color i that Vika has.

Output

The only line of the output should contain a single integer — the maximum number of squares that Vika can paint if she follows the rules described above.

Sample Input

Input

52 4 2 3 3

Output

12

Input

35 5 5

Output

15

Input

610 10 10 1 10 10

Output

11

Hint

In the first sample the best strategy is to start painting using color 4. Then the squares will be painted in the following colors (from left to right): 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.

In the second sample Vika can start to paint using any color.

In the third sample Vika should start painting using color number 5.

先找出最小的数，然后找所有最小数的距离，取一个最大距离，然后根据代码上的公式解就好了。

代码如下：

#include <cstdio>__int64 a[200022];//一怒之下全换成longlong，居然过了 __int64 b[200022];int main(){__int64 n;__int64 m;__int64 min;__int64 ans;while (~scanf ("%I64d",&n)){min = 1e9+1;for (__int64 i = 1 ; i <= n ; i++){scanf ("%I64d",&a[i]);if (a[i] < min)min = a[i];}m = 1;for (__int64 i = 1 ; i <= n ; i++){if (a[i] == min)b[m++] = i;}m--;if (m == 1){ans = n * (min + 1) -1;printf ("%I64d\n",ans);}else{__int64 dis = b[1] + n - b[m];for (__int64 i = 2 ; i <= m ; i++){if (dis < b[i] - b[i-1])//什么时候脑子抽抽在这打了个分号 = = ,每次比赛都要出现这样的意外，唉 dis = b[i] - b[i-1];}ans = n * min + dis - 1;printf ("%I64d\n",ans);}}return 0;}