Given N integers, you are supposed to find the smallest positive integer that is NOT in the given list.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10​5​​). Then N integers are given in the next line, separated by spaces. All the numbers are in the range of int.

Output Specification:

Print in a line the smallest positive integer that is missing from the input list.

Sample Input:

105 -25 9 6 1 3 4 2 5 17

Sample Output:

7

题目大意：给定n个数，找出这n个数里丢失的最小的正整数

注意细节：

（1）可能n个数全是负数，输出1

（2）正整数要从1开始记录，数中可能存在重复的数

（3）可能n个数都是连续的正整数，且从1开始，那么输出最大的数+1

代码如下：

#include
    
     using namespace std;int a[1000002];int main(){	int n;	while(cin >> n)	{		for(int i = 0;i < n;i ++)		{			cin >> a[i];		}		sort(a,a + n);		int i,fl = 0;		for(i = 0;i < n - 1;i ++)		{			if(a[i] > 0)			{				if(a[i] == 1)				fl = 1;				if(fl)				{					if(a[i] == a[i + 1] || a[i + 1] == a[i] + 1)					continue;					else					{						cout << a[i] + 1 << endl;						break;					}				}				else				{					cout << fl + 1 << endl;					break;				}			}		}		if(i == n - 1)		{			if(a[n - 1] > 0)			cout << a[n - 1] + 1 << endl;			else			cout << 1 << endl;		}	}	return 0;}