Codeforces Round #836 (Div. 2）C
1.AtCoder Beginner Contest 293(C,D ,E,F)2.Educational Codeforces Round 115 (Rated for Div. 2)（D,E)3.AtCoder Beginner Contest 294(E,F,G)4.AtCoder Beginner Contest 2465.中国石油大学（北京）第三届“骏码杯”程序设计竞赛（同步赛）(D,E,F)6.Codeforces Global Round 16(D,E,F)7.AtCoder Beginner Contest 209（D，E）8.Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）（E，F）9.AtCoder Beginner Contest 285（B,D,E,F)10.AtCoder Beginner Contest 242（D，E)11.AtCoder Beginner Contest 223（D,E,F)12.AtCoder Beginner Contest 207（D,E)13.AtCoder Beginner Contest 247（E,F)14.AtCoder Beginner Contest 226(E,F,G)15.AtCoder Beginner Contest 229(F,G)16.AtCoder Beginner Contest 273(E)17.AtCoder Beginner Contest 286(G)18.AtCoder Beginner Contest 287(C,D,E,F)19.AtCoder Beginner Contest 288(D,E,F)20.AtCoder Beginner Contest 289(E,F)21.AtCoder Beginner Contest 290(D,E)22.AtCoder Beginner Contest 292(E,F,G)23.AtCoder Beginner Contest 298(D,F)24.AtCoder Beginner Contest 299(E,F)25.AtCoder Beginner Contest 300(E,F)26.AtCoder Beginner Contest 302（E，F，G）27.AtCoder Beginner Contest 253(E,F)28.AtCoder Beginner Contest 245(D，E，F)29.AtCoder Beginner Contest 248(D，E，F) 30.AtCoder Beginner Contest 206（Sponsored by Panasonic）（E，F）31.Codeforces Round 875 (Div. 2)(D)32.AtCoder Beginner Contest 178(E,F)33.AtCoder Beginner Contest 307（E,F,G)34.CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C35.Educational Codeforces Round 151 (Rated for Div. 2)(C,D)36.AtCoder Beginner Contest 212(E,F)37.牛客小白月赛5138.2022年浙大城市学院新生程序设计竞赛（同步赛）（补题）39.Codeforces Round #770 (Div. 2)B，C40.Codeforces Global Round 24(B,C)
41.Codeforces Round #836 (Div. 2）C
42.Codeforces Round #840 (Div. 2) C43.Good Bye 2022: 2023 is NEAR C44.Codeforces Round #765 (Div. 2)A，B，C45.Codeforces Round #766 (Div. 2)C，D46.Codeforces Round #841 (Div. 2) and Divide by Zero 202247.Codeforces Round #767 (Div. 2)C ,D 48.Codeforces Round #768 (Div. 2)C ,D49. Codeforces Round #769 (Div. 2) B，C50.Educational Codeforces Round 122 (Rated for Div. 2)，C，D51.Educational Codeforces Round 119 (Rated for Div. 2)52.AtCoder Beginner Contest 25853.Codeforces Round #763 (Div. 2)C54.Codeforces Round #843 (Div. 2)（B，C，D，E）55.Educational Codeforces Round 141 (Rated for Div. 2)（B，C，D）56.AtCoder Beginner Contest 275（B，C，D，E，F）57.Codeforces Round #842 (Div. 2)（B，D，E）58.AtCoder Beginner Contest 284（D，E，F）59.The 14th Jilin Provincial Collegiate Programming Contest（补题）60.牛客小白月赛65（C，D，E，F）61.AtCoder Beginner Contest 281（D，E，F）62.Good Bye 2021: 2022 is NEAR D63.Hello 202364.The 15th Jilin Provincial Collegiate Programming Contest(补题)65.Codeforces Round #781 (Div. 2)C 66.Hello 2022（B，D）67.AtCoder Beginner Contest 272（D,E)68.Codeforces Round 751 (Div. 2)（D）69.Codeforces Round 856 (Div. 2)(C,D)70.Codeforces Round 752 (Div. 2)（C，D，E）71.Codeforces Round 855 (Div. 3)（E，F）72.AtCoder Regular Contest 131（A,B,C）73.Educational Codeforces Round 144 (Rated for Div. 2)（A，B，C，D）74.Codeforces Round 853 (Div. 2)（C,D)75.牛客练习赛109（C,D)76.AtCoder Beginner Contest 291（Sponsored by TOYOTA SYSTEMS）（D，E，F）77.Educational Codeforces Round 143 (Rated for Div. 2)（A，C,D)78.Codeforces Round #852 (Div. 2)（C,D)79.Educational Codeforces Round 118 (Rated for Div. 2)（D，E）80.AtCoder Beginner Contest 236（D，E，F）81.Codeforces Round #850 (Div. 2, based on VK Cup 2022 - Final Round)(B,D)82.Codeforces Round #848 (Div. 2)（B,C,D）83.TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) (B,C,D) 84.Codeforces Round #846 (Div. 2)（B，E） 85.Educational Codeforces Round 142 (Rated for Div. 2)（C，D）86.2023牛客寒假算法基础集训营687.Codeforces Round #845 (Div. 2) and ByteRace 2023（A，B，C）88.2023牛客寒假算法基础集训营5 89.2023牛客寒假算法基础集训营3 90.2023牛客寒假算法基础集训营291.2023牛客寒假算法基础集训营192.Educational Codeforces Round 120 (Rated for Div. 2) C，D93.AtCoder Beginner Contest 254（C，D，E，F) 这一次呢，题目要求让a[1]=x,a[n]=1,我们需要找到一个最小的序列使得每一个a[i]都可以被它的下标整除，没找到就是输出-1
我第一次是想着先让a[1]=x,a[n]=1,然后在中间一个一个的找它的下标的倍数，结果wa了
最后还是没有做出来！ε=(´ο｀*)))唉
看来题解之后，才知道他们是这么做的（好想知道他们聪明的脑袋瓜子是怎么想出来的）
第一步是先判断n%x==0，如果不满足，那么就一定是输出-1
然后，找出所有n的因数（除了a[1]=x,a[n]=1,a[x]=n,其他的都让a[i]=i)，
比如n=16,那么就有2,4,8，（不包括1和本身），x=2,那么我们可以把a[2]=4,a[4]=8,a[8]=16,a[16]=1(a[n]=1),一个这样的交换，这也是为什么我们要在前面判断n%x==0这一条件，这样的情况，刚好就是字典序最小的情况。
而且记得在交换的时候（swap(a[i],a[now],这个now是上一次的下标）也要记得判断是否满足a[i]%now等于0这一条件，因为我们也要保证交换了之后a[i]%i==0呀
比如n=36时，x=6，可以交换的就只有6,12,18，而不是2,3,4,6,9,12,18，如果是2,3,4,6,9,12,18，那么a[2]=3就这一个个就不符合条件了，所以还要判断上一次交换的下标和这一次的值是否可以整除
接下来就是看代码了
#include <iostream>
#include <stdlib.h>
#include <algorithm>
using namespace std;
const int maxn=2e5+10;
int t,n,x,ans[maxn];
void solve()
{
cin>>n>>x;
ans[1]=x,ans[n]=1;
for (int i=2;i<n;i++)
{
if (i==x) ans[i]=n;
else ans[i]=i;
}
if (n%x)
{
cout<<-1<<'\n';
return ;
}
if (n==x)
{
for (int i=1;i<=n;i++)
{
cout<<ans[i]<<" ";
}
cout<<'\n';
return ;
}
int now=x;
for (int i=x;i<n;i++)
{
if (n%i==0&&ans[i]%now==0)
{
swap(ans[i],ans[now]);
now=i;
}
}
for (int i=1;i<=n;i++)
{
cout<<ans[i]<<" ";
}
cout<<'\n';
return ;
}
int main ()
{
cin>>t;
while (t--)
{
solve();
}
system ("pause");
return 0;
}
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