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：

 1 #include <stdio.h>

 2 #include <string.h>

 3 #include <math.h>

 4 #include <queue>

 5 #define FOR(i, a, b) for(int i=a; i<b; i++)

 6 #define FOR_e(i, a, b) for(int i=a; i<=b; i++)

 7 

 8 using namespace std;

 9 int f[1001];

10 int main()

11 {

12     int t, dd;

13     int ma, mi;

14     scanf("%d", &t);

15     int i, n;

16     while(t--)

17     {

18         scanf("%d", &n);

19         memset(f, 0, sizeof(f));

20         ma=-1; mi=2000;

21         for(i=0; i<n; i++)

22         {

23             scanf("%d", &dd);

24             f[dd]=1;

25             if(dd>ma) ma=dd;

26             if(dd<mi) mi=dd;

27         }

28         queue<int>q;

29         while(!q.empty()) q.pop();

30         for(i=mi; i<=ma; i++)

31             if(f[i]==0)

32             {

33                 //printf("%d--", i);

34                 q.push(i);

35             }

36         if(q.size()>=2)

37         {

38             dd=q.front(); q.pop(); printf("%d ", dd);

39             dd=q.front(); q.pop(); printf("%d\n", dd);

40         }

41         else if(q.size()==1)

42         {

43             dd=q.front(); q.pop(); printf("%d ", dd);

44             if(mi==1)

45             {

46                 printf("%d\n", ma+1);

47             }

48             else if(mi>1)

49             {

50                 printf("%d\n", mi-1);

51             }

52         }

53         else if(q.size()==0)

54         {

55             if(mi>2)

56             {

57                 printf("%d %d\n", mi-2, mi-1);

58             }

59             else if(mi==2)

60             {

61                 printf("%d %d\n", mi-1, ma+1);

62             }

63             else

64             {

65                 printf("%d %d\n", ma+1, ma+2);

66             }

67         }

68     }

69     return 0;

70 }

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