Codeforces Beta Round #97 (Div. 2)

最新推荐文章于 2024-09-25 11:53:33 发布

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最新推荐文章于 2024-09-25 11:53:33 发布

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分类专栏： Codeforces 文章标签： numbers output sorting arrays input each

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本文介绍了一个算法问题，即如何通过一次元素替换并排序后得到数组每个位置上的最小可能值。输入包括数组长度及元素，输出为替换排序后的最小值数组。

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Little Petya very much likes arrays consisting of n integers, where each of them is in the range from 1 to 10⁹, inclusive. Recently he has received one such array as a gift from his mother. Petya didn't like it at once. He decided to choose exactly one element from the array and replace it with another integer that also lies in the range from 1 to 10⁹, inclusive. It is not allowed to replace a number with itself or to change no number at all.

After the replacement Petya sorted the array by the numbers' non-decreasing. Now he wants to know for each position: what minimum number could occupy it after the replacement and the sorting.

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Input

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The first line contains a single integer n (1 ≤ n ≤ 10⁵), which represents how many numbers the array has. The next line contains n space-separated integers — the array's description. All elements of the array lie in the range from 1 to 10⁹, inclusive.

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Output

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Print n space-separated integers — the minimum possible values of each array element after one replacement and the sorting are performed.

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Sample test(s)

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input

5
1 2 3 4 5

output

1 1 2 3 4

input

5
2 3 4 5 6

output

1 2 3 4 5

input

3
2 2 2

output

1 2 2

#include<cstdio>
#include<algorithm>
using std::sort;
using std::min;

int main()
{
    int n, m, mm = 0;
    int array[199999];
    scanf("%d",&n);
    for ( int i = 1; i <= n; i++ )
    {
        scanf("%d",&array[i]);
        if ( array[i] > mm ) mm = array[i], m = i;
    }
    array[m] = ( mm == 1 ? 2 : 1 );
    sort(array+1,array+1+n);
    for ( int i = 1; i <= n; i++ )
        printf("%d ",array[i]);
    printf("\n");
    return 0;
}