B. Vika and Squares
题目连接：
http://www.codeforces.com/contest/610/problem/B
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 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.
Sample Input
5
2 4 2 3 3
Sample Output
12
Hint
题意
你有n种颜色，然后每种颜色有ai个，你需要依次涂色。
比如第一个你涂x，那么下一个就得涂x+1,然后x+2.....
问你最多能涂多少个格子
题解：
贪心一下，我们肯定从最小的后面一格开始走。
这样可以得到答案，是 n * min + tmp
tmp是多少呢？tmp是最小值的两个之间的差值中最大的那个
证明略 hhh
代码
#include<bits/stdc++.h>
using namespace std;
int Min = 1e9+5,ans=0;
int a[200005];
vector<int>P;
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
{
if(Min>=a[i])
{
Min = a[i];
ans = i;
}
}
int first = 0;
for(int i=1;i<=n;i++)
{
if(a[i]==Min)
P.push_back(i);
}
P.push_back(P[0]+n);
long long tmp = 0;
for(int i=1;i<P.size();i++)
tmp = max(tmp,1LL*P[i]-1LL*P[i-1]-1LL);
printf("%lld
",1LL*n*Min+1LL*tmp);
}