HDU 2619 - Love you Ten thousand years (原根)

Description：

Love you Ten thousand years------Earth’s rotation is a day that is the representative of a day I love you. True love, there is no limit and no defects. Earth’s revolution once a year, it is on behalf of my love you more than a year. Permanent horizon, and my heart will never change ……

We say that integer $x, 0 < x < n,$( $n$ is a odd prime number) is a LovePoint-based-on n if and only if the set ${ (x i mod n) | 1 <= i <= n-1 }$ is equal to ${ 1, ..., n-1 }$. For example, the powers of $3$ modulo $7$ are $3, 2, 6, 4, 5, 1$, and thus $3$ is a LovePoint-based-on $7$.
Now give you a integer $n >= 3$( $n$ will not exceed $2 31$).
We say the number of LovePoint-based-on n is the number of days the earth rotating.
Your task is to calculate the number of days someone loved you.

Input

Each line of the input contains an integer $n$. Input is terminated by the end-of-file.

Output

For each $n$, print a single number that gives the number of days someone loved you.

Sample Input

5

Sample Output

2

求有多少 $i(<=n-1)$，使 $x^i \mod n$ 的值为 $[1,n-1]$，其实也就是满足完全剩余类的原根数量。

设 $m > 1$ 且 $gcd(a, m) = 1$, 则使得 $a^t ≡ 1(mod m)$ 成立的最小的正整数 $t$ 称为 $a$ 对模 $m$的阶, 记为 $δm(a)$。

如果 $a$的阶 $(mod m)$为 $ϕ(m)$, 则称 $a$ 为 $m$ 的一个原根。 即若 $δm(a)=ϕ(m)$, 则称 $a$为 $m$ 的一个原根。

定理1：若 $g$ 是 $m$ 的一个原根，则 $g,g^2,⋯,g^ϕ(m)$

各数对模 $m$ 的最小剩余，恰是小于 $m$ 且与 $m$ 互素的 $ϕ(m)$ 个正整数的一个排列。

定理2：每一个素数 $p$ 都有 $ϕ(p−1)$ 个原根。 $ϕ(m)$ 为小于 $m$ 的素数的个数，事实上, 每一个数 $m$ 都有 $ϕ(ϕ(m))$ 个原根(如果有的话)。即 若 $n$ 为素数 $ϕ(p−1) = ϕ(ϕ(p))$

这个题中 $n$ 为奇素数，故形成的为完全剩余系。求 $n$ 的原根数量即可。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a)
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(int m, int k, int mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

ll phi(ll n)
{
    ll ans = n;
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            ans -= ans / i;//素数倍数的个数减去
            while (n % i == 0)
                n /= i;//除去素数倍数
        }
    }
    if (n > 1)//对素数特判
        ans -= ans / n;
    return ans;
}//互素数的个数

ll n, m;
int main()
{
    while (~sld(n))
    {
        pld(phi(phi(n)));
    }
    return 0;
}