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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.
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.
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.
5
2 4 2 3 3
12
3
5 5 5
15
6
10 10 10 1 10 10
11
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：给你 了个序列 ，你可以从任意一个起点开始，从做走到右，一直循环走，每离开一个点，权值-1，问你 最多走几步
题解：扫一遍找最小的值，对于 找最后一段路，我们可以中间一段值，或者 两端的值，遍历选一个最大的就好了
//meek///#include<bits/stdc++.h> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include<iostream> #include<bitset> #include<vector> #include <queue> #include <map> #include <set> #include <stack> using namespace std ; #define mem(a) memset(a,0,sizeof(a)) #define pb push_back #define fi first #define se second #define MP make_pair typedef long long ll; const int N = 201000; const int M = 1000001; const int inf = 0x3f3f3f3f; const int MOD = 1000000007; const double eps = 0.000001; ll a[N],n; int main() { scanf("%I64d",&n); ll ans= inf; for(int i=1;i<=n;i++) { scanf("%I64d",&a[i]); ans = min(ans,a[i]); } for(int i=1;i<=n;i++) a[i] -= ans; ans*=n; ll mm=0,mn=0; for(int i=1;i<=n;i++) { if(a[i]) mm++; else { mn=max(mm,mn); mm=0; } mn=max(mm,mn); } mn=max(mm,mn); mm=0; for(int i=1;i<=n;i++) if(a[i]) mm++; else break; for(int i=n;i>=1;i--) if(a[i]) mm++; else break; mn=max(mm,mn); cout<<ans+mn<<endl; return 0; }
Codeforces Round #337 (Div. 2) B. Vika and Squares 水题
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原文地址：http://www.cnblogs.com/zxhl/p/5081050.html
