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[1001];
int main()
{
    int t, dd;
    int ma, mi;
    scanf("%d", &t);
    int i, n;
    while(t--)
    {
        scanf("%d", &n);
        memset(f, 0, sizeof(f));
        ma=-1; mi=2000;
        for(i=0; i<n; i++)
        {
            scanf("%d", &dd);
            f[dd]=1;
            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]==0)
            {
                //printf("%d--", i);
                q.push(i);
            }
        if(q.size()>=2)
        {
            dd=q.front(); q.pop(); printf("%d ", dd);
            dd=q.front(); q.pop(); printf("%d
", dd);
        }
        else if(q.size()==1)
        {
            dd=q.front(); q.pop(); printf("%d ", dd);
            if(mi==1)
            {
                printf("%d
", ma+1);
            }
            else if(mi>1)
            {
                printf("%d
", mi-1);
            }
        }
        else if(q.size()==0)
        {
            if(mi>2)
            {
                printf("%d %d
", mi-2, mi-1);
            }
            else if(mi==2)
            {
                printf("%d %d
", mi-1, ma+1);
            }
            else
            {
                printf("%d %d
", ma+1, ma+2);
            }
        }
    }
    return 0;
}