HDU5166Missing number

Missing number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 936    Accepted Submission(s): 471

Problem Description

There is a permutation without two numbers in it, and now you know what numbers the permutation has. Please find the two numbers it lose.

 

Input

There is a number T shows there are T test cases below. (T≤10)
For each test case , the first line contains a integers n , which means the number of numbers the permutation has. In following a line , there are n distinct postive integers.(1≤n≤1,000)

 

Output

For each case output two numbers , small number first.

 

Sample Input

2
3
3 4 5
1
1

 

Sample Output

1 2
2 3

 

题目比较水，不懂题意的去best code里面看看中文版的：http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=564&pid=1001

不过这里还是给不了解题意的小伙伴们说下：给定一个排列，小yb不小心弄丢了这个排列中的两个数字，输入数据中给出的是小明现在手上有的序列，要求得丢失的那两个数列，并且按从小到大的顺序输出就好了；从给出的数据我们很容易就可以推断，这个所给的数列就是 1至n+2 ，不然题目没法下笔，可以抱着试试的心态去写下，提交AC了证明是对的，所以就不要纠结于推断这个排列为什么会是这样的了；

给出代码:

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
	int T, n;
	int num[1000];
	cin >> T;
	while (T--)
	{
		cin >> n;
		for (int i = 1; i <= n; i++)
			cin >> num[i];
		sort(num+1, num+n+1);
		int count=1;//记录输出了几个数；
		int j;
		for (int i = 1,j=1; i <= n + 2; i++)
		{
			if (count < 2)
			{
				if (i == num[j])
				{
					j++;
					continue;
				}
				else if (count == 1)
				{
					count++;
					cout << i;
				}
				else if (count==2)
				{
					cout << " " << i << endl;
					count++;
				}
			}
			else break;
		}
	}
	return 0;
}