2021-8-16 Mocha and Hiking

Posted 2021-08-17 洛水天iriya

难度 

题目 Codeforces:

Mocha and Hiking

The city where Mocha lives in is called Zhijiang. There are n+1n+1 villages and 2n−12n−1 directed roads in this city.

There are two kinds of roads:

• n−1n−1 roads are from village ii to village i+1i+1, for all 1≤i≤n−11≤i≤n−1.
• nn roads can be described by a sequence a1,…,ana1,…,an. If ai=0ai=0, the ii-th of these roads goes from village ii to village n+1n+1, otherwise it goes from village n+1n+1 to village ii, for all 1≤i≤n1≤i≤n.

Mocha plans to go hiking with Taki this weekend. To avoid the trip being boring, they plan to go through every village exactly once. They can start and finish at any villages. Can you help them to draw up a plan?

Input

Each test contains multiple test cases.

The first line contains a single integer tt (1≤t≤201≤t≤20) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains a single integer nn (1≤n≤1041≤n≤104) — indicates that the number of villages is n+1n+1.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤10≤ai≤1). If ai=0ai=0, it means that there is a road from village ii to village n+1n+1. If ai=1ai=1, it means that there is a road from village n+1n+1 to village ii.

It is guaranteed that the sum of nn over all test cases does not exceed 104104.

Output

For each test case, print a line with n+1n+1 integers, where the ii-th number is the ii-th village they will go through. If the answer doesn\'t exist, print −1−1.

If there are multiple correct answers, you can print any one of them.

 

题目解析

我吐了这道题，一直wa就是因为我想的太多，各种计算条件，反正是真的够惨的，本来上大分的，这题卡了老久了，不想解析了，看代码吧

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
typedef long long ll;
int x[10005];
int main()
{//1是从临时点出来
    int t; cin >> t;
    while (t--)
    {
        ll n; cin >> n;

        for (int i = 0; i < n; i++)cin >> x[i];
        if (x[n - 1] == 0)
        {
            for (int i = 1; i <= n; i++)cout << i << " ";
            cout << n + 1 << "\\n";
        }
        else if (x[0] == 1)
        {
            cout << n + 1;
            for (int i = 1; i <= n; i++)cout << " " << i;
            cout << "\\n";
        }
        else
        {
            bool tf = 1;
            for (int i = 1; i <= n; i++)
            {
                if (tf && i < n && x[i - 1] == 0 && x[i] == 1)
                {
                    cout << i << " " << n + 1 << " ";
                    tf = 0;
                }
                else if (i != n) cout << i << " ";
                else if (i == n) cout << i;
            }
            cout << "\\n";
        }
    }
    return 0;
}