Codeforces Round #623 (Div. 2) Restoring Permutation

解題報告：

思路：看懂題意後，模擬就行，時間很充裕。

代碼：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 100+10;
ll a[N<<1], b[N], visit[N<<2];
void solve(){
    memset(a, 0, sizeof(a));
    memset(visit, 0, sizeof(visit));
    ll n;
    scanf("%lld", &n);
    for(ll i=0; i<n; ++i){
        scanf("%lld", b+i);
        a[i<<1] = b[i], visit[b[i]] = 1;
    }
    for(ll i=0; i<n; ++i){
        ll flag = 0;
        for(ll j=1; j<=2*n; ++j){
            if(!visit[j] && j > a[i<<1]){
                visit[j] = 1, a[i<<1|1] = j;
                flag = 1;
                break;
            }
        }
        if(!flag){
            printf("-1\n");
            return ;
        }
    }
    for(ll i=0; i<2*n; ++i){
        printf("%lld%s", a[i], i == 2*n-1 ? "\n" : " ");
    }
}
int main(){
    ll t;
    scanf("%lld", &t);
    while(t--)solve();
    return 0;
}