HDU5166Missing number

Missing number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 936    Accepted Submission(s): 471

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≤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

233 4 511

 

Sample Output

1 22 3

 

题目比较水，不懂题意的去best code里面看看中文版的：http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=564&pid=1001

不过这里还是给不了解题意的小伙伴们说下：给定一个排列，小yb不小心弄丢了这个排列中的两个数字，输入数据中给出的是小明现在手上有的序列，要求得丢失的那两个数列，并且按从小到大的顺序输出就好了；从给出的数据我们很容易就可以推断，这个所给的数列就是 1至n+2 ，不然题目没法下笔，可以抱着试试的心态去写下，提交AC了证明是对的，所以就不要纠结于推断这个排列为什么会是这样的了；

给出代码:

#include<iostream>#include<cstdio>#include<string>#include<algorithm>using namespace std;int main(){int T, n;int num[1000];cin >> T;while (T--){cin >> n;for (int i = 1; i <= n; i++)cin >> num[i];sort(num+1, num+n+1);int count=1;//记录输出了几个数；int j;for (int i = 1,j=1; i <= n + 2; i++){if (count < 2){if (i == num[j]){j++;continue;}else if (count == 1){count++;cout << i;}else if (count==2){cout << " " << i << endl;count++;}}else break;}}return 0;}