hdu 5166 Missing number

2021-11-03 03:11:19

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5166

Missing number

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

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<vector>

#include<map>

using std::map;

using std::min;

using std::find;

using std::pair;

using std::vector;

using std::multimap;

#define pb(e) push_back(e)

#define sz(c) (int)(c).size()

#define mp(a, b) make_pair(a, b)

#define all(c) (c).begin(), (c).end()

#define iter(c) __typeof((c).begin())

#define cls(arr, val) memset(arr, val, sizeof(arr))

#define cpresent(c, e) (find(all(c), (e)) != (c).end())

#define rep(i, n) for(int i = 0; i < (int)n; i++)

#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)

const int N = 1010;

const int INF = 0x3f3f3f3f;

bool vis[N];

void solve(int n) {

    int v;

    vector<int> ans;

    cls(vis, false);

    rep(i, n) {

        scanf("%d", &v);

        vis[v] = true;

    }

    rep(i, n + 2) {

        if(!vis[i + 1]) ans.pb(i + 1);

    }

    printf("%d %d\n", ans[0], ans[1]);

}

int main() {

#ifdef LOCAL

    freopen("in.txt", "r", stdin);

    freopen("out.txt", "w+", stdout);

#endif

    int t, n;

    scanf("%d", &t);

    while(t--) {

        scanf("%d", &n);

        solve(n);

    }

    return 0;

}

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