Codeforces 487C Prefix Product Sequence[数论+构造]

转载于 2019-03-03 00:25:00 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/storz/p/10463740.html

本文探讨了一种构造特定序列的方法，确保序列中元素各不相同且满足特定数学条件。通过将序列的第i个元素定义为从1到i的乘积，利用逆元的唯一性证明了每个元素的独特性。

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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 \(n! \equiv 0 \pmod n\)

\[ \text{中间的数放}\frac{2}{1},\frac{3}{2},... \]

这样第 \(i\) 个数就是 \[ \prod_{j=1}^i{a_j}=1\times \frac{2}{1}\times \frac{3}{2}\times ...=i \]

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

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\n1";
    return ;
  }
  if (n == 4) {
    out, "YES\n1\n3\n2\n4";
    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, '\n';
  out, n;
}

}

转载于:https://www.cnblogs.com/storz/p/10463740.html