HDU 5166

Time Limit:1000MS  Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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. (0<=T <=10) 
For each test case , the first line contains a integers \(n\) , which means the
number of numbers the permutation has. In following a line , there are $n$
distinct postive integers.(1 <=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

题解：找出一个数列中缺的两个数（这两个数要为最小的两个数）。

注意：利用bool函数把数组中所有数全部返回0

#include<stdio.h>

  int main()

  {

      int T,i;

     scanf("%d",&T);

     while(T--)

      {

          int count,t=,j[];;

          scanf("%d",&count);

         bool flag[]={  false  };

        for( i=;i<=count;i++)

         {

             int tem;

             scanf("%d",&tem);

             flag[tem]=true;//每输入一个数s，便给第s个赋值为1

         }

         for(i=;i<=count+;i++)

         {

             if(flag[i]==)//如果有为0的出现，则为缺失的数

             {

                 j[t++]=i;

             }

        }

         printf("%d %d\n",j[],j[]);

     }

     return ;

 }

多多交流~