意甲冠军:
对于数字n, 他询问是否有1~n置换 这种布置能够在产品上模每个前缀n 有可能0~n-1
解析:
通过观察1肯定要在首位,n一定要在最后
除4意外的合数都没有解
其它质数构造 a[i]=i*inv[i-1] , 这样用逆元把前面每一个数的影响都消除掉
Consider a sequence [a1, a2, ... , an]. Define its prefix product sequence 
.
Now given n, find a permutation of [1, 2, ..., n], such that its prefix product sequence is a permutation of [0, 1, ..., n - 1].
The only input line contains an integer n (1 ≤ n ≤ 105).
In the first output line, print "YES" if such sequence exists, or print "NO" if no such sequence exists.
If any solution exists, you should output n more lines. i-th line contains only an integer ai. The elements of the sequence should be different positive integers no larger than n.
If there are multiple solutions, you are allowed to print any of them.
7
YES 1 4 3 6 5 2 7
6
NO
For the second sample, there are no valid sequences.
/* ***********************************************
Author :CKboss
Created Time :2015年03月12日 星期四 19时58分14秒
File Name :CF487C.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long int LL;
const int maxn=100100;
int n;
LL a[maxn],inv[maxn];
bool isprime(int x)
{
if(x==2||x==1) return true;
if(x%2==0) return false;
for(int i=3;i*i<=x;i+=2) if(x%i==0) return false;
return true;
}
void solve()
{
inv[1]=1LL;
for(int i=2;i<=n;i++) inv[i]=inv[n%i]*(n-n/i)%n;
a[1]=1LL; a[n]=n;
for(int i=2;i<n;i++) a[i]=(i*inv[i-1])%n;
for(int i=1;i<=n;i++) printf("%I64d\n",a[i]);
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
scanf("%d",&n);
if(n==4)
{
puts("YES"); puts("1"); puts("3"); puts("2"); puts("4");
return 0;
}
if(isprime(n)==false) puts("NO");
else
{
puts("YES");
solve();
}
return 0;
}
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