Codeforces Round #337 (Div. 2) 610B Vika and Squares(脑洞)

B. Vika and Squares

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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, 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, ..., 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.

Sample test(s)

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.

题目链接:点击打开链接

给出n种油漆及每种油漆数量, 要求在长方形纸上挨个涂色, 从任意颜色开始涂, 问最多可以涂多少.

找到数量最小的, 求出数量最少的个数, 若数量为n, 则全部可以涂, 若数量为1, 则从下一个开始涂, 其他情况则求出最长的间隔.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 5;
ll n, num, len, tmp, cnt, a[MAXN];
int main(int argc, char const *argv[])
{
	scanf("%lld", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%lld", &a[i]);
	ll mi = INF;
	for(int i = 1; i <= n; ++i)
		mi = min(mi, a[i]);
	for(int i = 1; i <= n; ++i)
		if(a[i] == mi) num++;
	if(num == n) printf("%lld\n", n * a[1]);
	else if(num == 1) printf("%lld\n", n * mi + n - 1);
	else {
		for(int i = 1; i <= n; ++i) {
			if(a[i] == mi) {
				if(tmp > len) len = tmp;
				tmp = 0;
			}
			else tmp++;
		}
		for(int i = 1; i <= n; ++i)
			if(a[i] == mi) break;
			else cnt++;
		len = max(cnt + tmp, len);
		printf("%lld\n", n * mi + len);
	}
	return 0;
}