Bestcoder BestCoder Round #28 A Missing number(查找缺失的合法数字)

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<=10T≤10)
For each test case , the first line contains a integers  nn , which means the number of numbers the permutation has. In following a line ,

there are  nnN distinct postive integers.(1<=n<=10001≤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

Source

BestCoder Round #28

 

代码：

 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <queue>
 #define FOR(i, a, b) for(int i=a; i<b; i++)
 #define FOR_e(i, a, b) for(int i=a; i<=b; i++)

 using namespace std;
 int f[];
 int main()
 {
     int t, dd;
     int ma, mi;
     scanf("%d", &t);
     int i, n;
     while(t--)
     {
         scanf("%d", &n);
         memset(f, , sizeof(f));
         ma=-; mi=;
         for(i=; i<n; i++)
         {
             scanf("%d", &dd);
             f[dd]=;
             if(dd>ma) ma=dd;
             if(dd<mi) mi=dd;
         }
         queue<int>q;
         while(!q.empty()) q.pop();
         for(i=mi; i<=ma; i++)
             if(f[i]==)
             {
                 //printf("%d--", i);
                 q.push(i);
             }
         if(q.size()>=)
         {
             dd=q.front(); q.pop(); printf("%d ", dd);
             dd=q.front(); q.pop(); printf("%d\n", dd);
         }
         else if(q.size()==)
         {
             dd=q.front(); q.pop(); printf("%d ", dd);
             if(mi==)
             {
                 printf("%d\n", ma+);
             }
             else if(mi>)
             {
                 printf("%d\n", mi-);
             }
         }
         else if(q.size()==)
         {
             if(mi>)
             {
                 printf("%d %d\n", mi-, mi-);
             }
             else if(mi==)
             {
                 printf("%d %d\n", mi-, ma+);
             }
             else
             {
                 printf("%d %d\n", ma+, ma+);
             }
         }
     }
     return ;
 }