Codeforces Round #686 (Div. 3) A. Special Permutation

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AlberTeslaWizard

最新推荐文章于 2024-10-25 20:01:42 发布

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A. Special Permutation
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given one integer n (n>1).

Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation of length 5, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Your task is to find a permutation p of length n that there is no index i (1≤i≤n) such that pi=i (so, for all i from 1 to n the condition pi≠i should be satisfied).

You have to answer t independent test cases.

If there are several answers, you can print any. It can be proven that the answer exists for each n>1.

Input
The first line of the input contains one integer t (1≤t≤100) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2≤n≤100) — the length of the permutation you have to find.

Output
For each test case, print n distinct integers p1,p2,…,pn — a permutation that there is no index i (1≤i≤n) such that pi=i (so, for all i from 1 to n the condition pi≠i should be satisfied).

If there are several answers, you can print any. It can be proven that the answer exists for each n>1.

Example
input
2
2
5
output
2 1
2 1 5 3 4

My Answer Code:

/*
	Author:Albert Tesla Wizard
	Time:2020/11/27 17:50
*/
#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    vector<int>a(t);
    for(int i=0;i<t;i++)cin>>a[i];
    for(int i=0;i<t;i++)
    {
        int record=a[i];
        vector<int>b(record);
        for(int j=0;j<record;j++)b[j]=j+1;
        if(record%2==0)
        {
            for(int k=record-1;k>=0;k--)cout<<b[k]<<" ";
            cout<<'\n';
        }
        else
        {
            int temp;
            temp=b[0];
            b[0]=b[record/2];
            b[record/2]=temp;
            for(int m=record-1;m>=0;m--)cout<<b[m]<<" ";
            cout<<'\n';
        }
    }
}