首先第一个必须放1(不然放在中间会有两个重复的数) 最后一个必须放n (n! equiv 0 pmod n)
[ ext{中间的数放}frac{2}{1},frac{3}{2},...
]
这样第 (i) 个数就是 $$
prod_{j=1}^i{a_j}=1 imes frac{2}{1} imes frac{3}{2} imes ...=i
[
现在前缀积各不相同了，如何证明每个元素不同呢？发现中间的数减去１都是１／ｘ的形式，因为逆元互不相同(百度百科:群G中任意一个元素a，都在G中有唯一的逆元a‘,具有性质aa'=a'a=e，其中e为群的单位元。)，所以每个元素也是不同的
```cpp
namespace QvvQ {
int mod;
struct Mod {
int a;
Mod(int _a) : a(_a) {}
Mod(ll _a) : a(_a % mod) {}
Mod() {}
int inv() {return Pow(a, mod - 2, mod);}
Mod &operator += (const int b) {a += b; if (a < 0) a += mod; else if (a >= mod) a -= mod; return *this;}
Mod &operator -= (const int b) {a -= b; if (a < 0) a += mod; else if (a >= mod) a -= mod; return *this;}
Mod &operator *= (const int b) {a = a * 1ll * b % mod; if (a < 0) a += mod; return *this;}
Mod &operator /= (const int b) {a = a * 1ll * Pow(b, mod - 2, mod) % mod; if (a < 0) a += mod; return *this;}
Mod &operator += (const Mod rhs) {return *this += rhs.a;}
Mod &operator -= (const Mod rhs) {return *this -= rhs.a;}
Mod &operator *= (const Mod rhs) {return *this *= rhs.a;}
Mod &operator /= (const Mod rhs) {return *this /= rhs.a;}
friend Mod operator + (Mod c, const Mod rhs) {return c += rhs.a;}
friend Mod operator - (Mod c, const Mod rhs) {return c -= rhs.a;}
friend Mod operator * (Mod c, const Mod rhs) {return c *= rhs.a;}
friend Mod operator / (Mod c, const Mod rhs) {return c /= rhs.a;}
Mod &operator = (const int x) {a = x; return *this;}
Mod &operator = (const ll x) {a = x % mod; return *this;}
Mod &operator = (const Mod rhs) {a = rhs.a; return *this;}
} inv[N];
void init() {
}
void solve() {
int n = in;
mod = n;
if (n == 1) {
out, "YES
1";
return ;
}
if (n == 4) {
out, "YES
1
3
2
4";
return ;
}
for (int i = 2; i * i <= n; ++i) if (n % i == 0) return puts("NO"), void();
puts("YES");
inv[0] = inv[1] = 1;
lo(i, 2, n) inv[i] = (n - n / i) * inv[n % i];
lo1(i, n-1) out, (inv[i - 1] * i).a, '
';
out, n;
}
}
```]