CodeForces 610 B - Harmony Analysis_codeforces610b-优快云博客

本文链接：https://blog.youkuaiyun.com/AleiChen/article/details/50676630

本文介绍了一个有趣的编程挑战，任务是计算使用特定颜色组合和有限数量的油漆可以绘制的最大连续方格数。通过分析输入数据，利用循环和条件判断来找出最优解。

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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

Input

5
2 4 2 3 3

Output

Input

3
5 5 5

Output

Input

6
10 10 10 1 10 10

Output

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>

using namespace std;
//#define LOCAL
#define MAX_N 200005
#define INF 0x3f3f3f3f
int num[MAX_N];
int main()
{
	#ifdef LOCAL
		freopen("b:\\data.in.txt", "r", stdin);
	#endif
	int n;
	while(~scanf("%d", &n))
    {
        int ma, mi;
        ma = -1;
        mi = INF;
        for(int i=0; i<n; i++){
            scanf("%d", &num[i]);
            ma = max(ma,num[i]);
            mi = min(mi,num[i]);
        }
//        cout<<ma<<" "<<mi<<endl;
        long long ans = ((long long)mi)*n;//基础的解
        int ed, fir;
        for(int i=n-1; i>=0; i--){
            if(num[i] == mi){
                ed = i;
                break;
            }
        }
        for(int i=0; i<n; i++){
            if(num[i] == mi){
                fir = i;
                break;
            }
        }
//        cout<<fir<<" "<<ed<<endl;
        int ans2 = n-1-ed + fir;
        int ma2 = 0;
        int ma3 = 0;
        for(int i=fir+1; i<=ed; i++){
            if(num[i] > mi){
                ma2++;
            }
        else
            ma2 = 0;
            ma3 = max(ma3, ma2);
        }
//        cout<<ans2<<" "<<ma2<<endl;
        ans = ans + max(ans2, ma3);
        cout<<ans<<endl;
    }
    return 0;
}

总结：本来考虑到了有数据溢出的情况，然后定义了long long 类型的ans, 没有想到的是为什么是在赋值之前就溢出了，赋值的值本身就是溢出后的值，肯定错啊。一定不放过A不过的题目。