题目大意：
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 
shows there are 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<=1000)
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
思路分析：
总共有n+2个数，丢失了两个数。我们把输入的n个数用数组a[maxn]存起来，用b[maxn]来模拟一个排序从1~n+2,用a与数组b做比较，如果比较得出了不相同，那就把数组b上这个位置上的数变为0，最后只要用一个循环找出前两个不为0的数即可。
源代码：
1 #include<iostream>
2 #define maxn 1010
3 using namespace std;
4 int main()
5 {
6 int t;
7 cin >> t;
8 while (t--)
9 {
10 int n, a[maxn], b[maxn], i;
11 int count = 0;
12 cin >> n
13 for (i = 0; i < n; i++)
14 cin >> a[i];
15 for (i = 1; i <= n + 2; i++)
16 {
17 b[i] = i; //建立一个1~n+2的排序数组
18
19 for (int j = 0; j<n; j++)
20 if (b[i] == a[j]) //两个数组作比较，找出相同的标记为0
21 b[i] = 0;
22 }
23 for (i = 1; i <= n + 2; i++)
24 {
25 if (b[i] != 0)
26 {
27
28 count++; //找出不为0的数
29 if (count == 2)
30 {
31 cout << b[i] << endl; //最后一个数输出时不用空格
32 break;
33 }
34 cout << b[i] << " ";
35 }
36
37
38 }
39
40
41 }
42 return 0;
43
44 }
心得：
这样的思路以后应该还会用到，好好积累。刷题也可以带来乐趣（=_=关键是要会），好好加油吧！
跟这么多同届的同学，还有学姐学长一起学习，感觉很充实，但是自己平时问的题目还太少了，跟别人交流也太少了，要改改，因为这样才能学到更多的东西。进步才会更快~~~~~