2015 HUAS Provincial Select Contest #2 A

题目：

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  shows there are  test cases below. () 
For each test case , the first line contains a integers  , which means the number of numbers the permutation has. In following a line , there are  distinct postive integers.()

 

Output

For each case output two numbers , small number first.

 

题目大意：输入N连续个数，求它的最小的两个数。

解题思路：用一个数组存储这N个数，再用另个数组存储1—N+2的连续数，再两个数组比较，输出缺了的两项。

代码：

#include<iostream>
#include<cstdio>
#define maxn 1010
using namespace std;
int main()
{
    int T;
    cin >> T;
    if(T<=10)
    {
        while (T--)
        {
            int n, a[maxn],b[maxn] ,i,d;
            cin >> n;
            if(n>=1&&n<=1000)
            {
                d=0;
                for (i = 0; i <n; i++)
                    cin >> a[i];
                for(i=0;i<n+2;i++)
                    b[i]=i+1;
                
                for(i=0;i<n;i++)
                {
                    for(int j=0;j<n+2;j++)
                    {
                        if(a[i]==b[j])
                            b[j]=0;
                    }
                }
                for(i=0;i<n+2;i++)
                {
                    if(b[i]!=0)
                    {    
                        d++;
                        if(d==2)
                            printf("%d
",b[i]);
                        else printf("%d ",b[i]);
                        
                    }
                }
            }        
        }
    }    
    return 0;    
}