zoukankan      html  css  js  c++  java
  • CodeForces 1095B Array Stabilization

    You are given an array aa consisting of nn integer numbers.

    Let instability of the array be the following value: maxi=1naimini=1naimaxi=1nai−mini=1nai .

    You have to remove exactly one element from this array to minimize instability of the resulting (n1)(n−1) -elements array. Your task is to calculate the minimum possible instability.

    Input

    The first line of the input contains one integer nn (2n1052≤n≤105 ) — the number of elements in the array aa .

    The second line of the input contains nn integers a1,a2,,ana1,a2,…,an (1ai1051≤ai≤105 ) — elements of the array aa .

    Output

    Print one integer — the minimum possible instability of the array if you have to remove exactly one element from the array aa .

    Examples

    Input
    4
    1 3 3 7
    
    Output
    2
    
    Input
    2
    1 100000
    
    Output
    0
    

    Note

    In the first example you can remove 77 then instability of the remaining array will be 31=23−1=2 .

    In the second example you can remove either 11 or 100000100000 then instability of the remaining array will be 100000100000=0100000−100000=0 and 11=01−1=0 correspondingly.

    #include <cstdio>
    #include <iostream>
    #include <cmath>
    #include <algorithm>
    #include <cstring>
    #include <string>
    
    using namespace std;
    
    int n, a[100000+8], sum[100000+8], r = 0, number[100000+8], g = 0;
    int main()
    {
        scanf("%d", &n);
        for(int i = 0; i<n; i++)
        {
            scanf("%d", &a[i]);
        }
        sort(a, a+n);
        if(n == 2)printf("0
    ");
        else printf("%d
    ", min(a[n-2]-a[0], a[n-1]-a[1]));
        return 0;
    }
  • 相关阅读:
    sae-xhprof调试性能
    VC++常用函数
    C++资料收集&整理
    C++宽窄字符串转换
    #MySQL for Python(MySQLdb) Note
    VC++ ADO相关
    MFC各种控件的常见操作(逐步添加中......)
    关于Sql注入的那些事
    Python中socket经ssl加密后server开多线程
    网络编程
  • 原文地址:https://www.cnblogs.com/RootVount/p/10345302.html
Copyright © 2011-2022 走看看