Codeforces 135A-Replacement（思维）

A. Replacement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

Output

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

Sample test(s)

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

题意：给出一个长度为n的数组，要求替换掉数组中的一个数。（替换的含义是要用另外一个与次数不同的数来替换）全部的数的范围在[1,10^9],然后排序后的要求是使每一个位置上的数尽可能小，一种比較巧的做法是对于这个数组先排序（升序） 然后将最大的那个数换成1 假设最大的为1 则换成2（想想 为什么 ） 

然后在排序输出就能够了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
#define LL long long
int a[100050];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<n;i++)
			scanf("%d",a+i);
		sort(a,a+n);
		a[n-1]=a[n-1]==1?

2:1;
		sort(a,a+n);
		for(int i=0;i<n;i++)
			if(i!=n-1)
			printf("%d ",a[i]);
		    else
			printf("%d\n",a[i]);
	}
	return 0;
}

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