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 }
View Code

 

posted @ 2015-02-07 17:10  我喜欢旅行  阅读(163)  评论(0)  编辑  收藏  举报