Codeforce - 610 - B. Vika and Squares

B. Vika and Squares
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vika has n jars with paints of distinct colors. All the jars are numbered from 1 to n and the i-th jar contains ai 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, 3 and 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, …, an (1 ≤ ai ≤ 109), where ai 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.

Examples
input
5
2 4 2 3 3
output
12
input
3
5 5 5
output
15
input
6
10 10 10 1 10 10
output
11
Note
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.

题意：有 n 个数围成圈，要求选取一个起点，每走一步，所到的数值 -1 ，直到遇到遇到 0 时停止。求所能走的最长的步数。

AC代码思路： n 个数中的最小值，是所要走的圈数，所有数减去最小值之后， 圈中两个 0 之间的个数的最大值，是多走的步数。

AC代码：

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

int main()
{
    long long n;
    cin>>n;
    long long a[n+5];
    for (int i=0;i<n;i++)
        cin>>a[i];
    long long mi= *min_element(a,a+n);
    long long cnt=0,s=0;
    for (int i=1;i<=2*n;i++)
    {
        if (a[i%n]==mi)
        {
            s=max(s,cnt);
            cnt=0;
        }
        else cnt++;
    }
    long long ans=s+n*mi;
    cout<<ans;
}