CodeForces - 610B Vika and Squares (模拟)

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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, 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, ..., 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

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.

//题意：

小明有有n罐油漆，并且每罐剩余的油漆量不同，现在他想把这些油漆刷在（1*无穷大）的墙上，每个1*1的正方形要消耗一升油漆，并且刷油漆的顺序是固定的(x--n--x--n...)，也就是说你可以选择起始点（第几罐），但选择以后就得按照给定的油漆顺序进行循环，直到某一罐油漆耗完为止，才结束，问小明怎样选择起始点，才能使他刷的矩形最长。

//思路，就是模拟，找到最少的油漆m的位置，可能有多个，将这多个位置记录下来，先用n*m求出不用考虑刷完的情况的长度，然后再考虑刷完的情况，这就得遍历先前记录的最少油漆的位置了，找出相邻两个位置之间的最长长度即可。

//Hait：数很大，输入输出要注意；

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 200010
#define M 1000000007
using namespace std;
IN a[N];
int b[N];
int main()
{
	int n,mk;
	int i;
	IN k,m;
	while(scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
		m=INF;k=0;
		for(i=0;i<n;i++)
		{
			scanf("%I64d",&a[i]);
			if(a[i]<m)
				m=a[i];
		}
		k+=m*n;
		for(i=0;i<n;i++)
			a[i]-=m;
		int kk=0;
		for(i=0;i<n;i++)
		{
			if(!a[i])
				b[kk++]=i;
		}
		int mm=n-b[kk-1]+b[0]-1;
		for(i=1;i<kk;i++)
		{
			mk=b[i]-b[i-1]-1;
			mm=max(mm,mk);
		}
		k+=mm;
		printf("%I64d\n",k);
	}
	return 0;
}