Missing number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1178 Accepted Submission(s): 608
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
2
3
3 4 5
1
1
Sample Output
1 2
2 3
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
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[1005];
int main()
{
int t;
cin >> t;
while (t--)
{
memset(a, 0, sizeof(a));
int n;
cin >> n;
for (int i = 0; i < n; i++)
{
int x;
cin >> x;
a[x]++;
}
int ans[2], s = 0;
for (int i = 1; i < 1000 + 5; i++)
{
if (a[i] == 0)
{
ans[s] = i;
s++;
if (s == 2)
{
break;
}
}
}
cout << ans[0] << ' ' << ans[1] << endl;
}
return 0;
}
