2020牛客暑期多校训练营（第十场）A.Permutation

2020牛客暑期多校训练营（第十场）A.Permutation

题目链接

题目描述

You are given a prime number p, you want to find a permutation of numbers $1,2,\dots ,p-1$, denoted as $x_1,x_2,\dots ,x_{p-1}$, such that for all $i(1\le i\le p-2)$, $x_{i+1}\equiv 2x_i(\mathrm{m}\mathrm{o}\mathrm{d}p)$x or $x_{i+1}\equiv 3x_i(\mathrm{m}\mathrm{o}\mathrm{d}p)$.

输入描述:

The first line contains an integer $T(1\le T\le 100)$ indicating the number of test cases.
For each test case, there is one prime number $p(2\le p\le 10^6)$ in the first line.
It’s guaranteed that $\sum p\le 10^6$.

输出描述:

If there is no solution for this prime, print ‘-1’. Otherwise print p-1 integers $x_1,x_2,\dots ,x_{p-1}$.

示例1

输入

2
3
5

输出

1 2
1 2 4 3

感觉是很捞的一道题，貌似只要暴力就过了，不难发现，从 $1$ 开始选，能选 $2$ 的倍数就选 $2\ast x\%n$，能选 $3$ 的倍数就选 $3\ast x\%n$，不能选就输出 $-1$ 就行，AC代码如下:

#include
using namespace std;
typedef long long ll;
const int N=1e6+5;
int t,n,vis[N];
int main()
{
    scanf("%d",&t);
    while(t--){
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        vector<int>ans;
        ans.push_back(1);
        vis[1]=1;
        int cnt=1,flag=1;
        while(1){
            int a=ans.back(),ok=0;
            if(!vis[a*2%n]&&a*2%n) {
                vis[a*2%n]=1;
                ans.push_back(a*2%n);
                cnt++;
                ok=1;
            }else if(!vis[a*3%n]&&a*3%n){
                vis[a*3%n]=1;
                ans.push_back(a * 3 % n);
                cnt++;
                ok=1;
            }
            if(cnt==n-1) break;
            if(ok==0){
                flag=0;
                break;
            }
        }
        if(flag==0) printf("-1");
        else for(auto i:ans) printf("%d ",i);
        printf("\n");
    }
	return 0;
}