這道題不知道這個定理很難做出來。

除非暴力找規律。

我原本找的時候出了問題

暴力找出的從13及以上的答案就有問題了

因為13的12次方會溢位

那麼該怎麼做？

快速冪派上用場。

把前幾個素數的答案找出來。

然後因為原根的一個條件是與p互質，所以可以把尤拉函式的值求出來嘗試一下

打印出來，就可以發現規律

答案就是phi(p-1)

暴力找答案程式碼

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cctype>
#define REP(i, a, b) for(int i = (a); i < (b); i++) 
#define _for(i, a, b) for(int i = (a); i <= (b); i++) 
using namespace std;

typedef long long ll;
const int MAXN = 21234567;
bool vis[MAXN];

void read(ll& x)
{
	int f = 1; x = 0; char ch = getchar();
	while(!isdigit(ch)) { if(ch == '-') f = -1; ch = getchar(); }
	while(isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
	x *= f;
}

ll cal(ll a, ll b, ll p)
{
	ll ret = 1 % p; a %= p;
	while(b)
	{
		if(b & 1) ret = (ret * a) % p;
		b >>= 1;
		a = (a * a) % p;
	}
	return ret;
}

int main()
{
	while(1)
	{
		ll p;
		read(p);
		int ans = 0;
		_for(x, 1, p - 1)
		{
			memset(vis, 0, sizeof(vis));
			_for(i, 1, p - 1)
			{
				ll t = cal(x, i, p);
				if(vis[t] || t == 0) break;
				vis[t] = 1;
				if(i == p - 1) ans++;
			}	
		}
		printf("p: %d ans: %d\n", p, ans);
	}
	return 0;
}
 

求尤拉函式值程式碼

#include<cstdio>
#include<cctype>
#define REP(i, a, b) for(int i = (a); i < (b); i++) 
#define _for(i, a, b) for(int i = (a); i <= (b); i++) 
using namespace std;

typedef long long ll;
const int MAXN = 21234567;

ll euler(ll x)
{
	ll ret = x;
	for(int i = 2; i * i <= x; i++)
		if(x % i == 0)
		{
			ret = ret / i * (i - 1);
			while(x % i == 0) x /= i;
			if(x == 1) break;
		}
	if(x > 1) ret = ret / x * (x - 1);
	return ret;
}

void read(ll& x)
{
	int f = 1; x = 0; char ch = getchar();
	while(!isdigit(ch)) { if(ch == '-') f = -1; ch = getchar(); }
	while(isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
	x *= f;
}

int main()
{
	_for(i, 1, 30) printf("i: %d euler[i]: %lld\n", i, euler(i));
	return 0;
}
 

AC程式碼

#include<cstdio>
#include<cctype>
#define REP(i, a, b) for(int i = (a); i < (b); i++) 
#define _for(i, a, b) for(int i = (a); i <= (b); i++) 
using namespace std;

typedef long long ll;
const int MAXN = 21234567;

ll euler(ll x)
{
	ll ret = x;
	for(int i = 2; i * i <= x; i++)
		if(x % i == 0)
		{
			ret = ret / i * (i - 1);
			while(x % i == 0) x /= i;
			if(x == 1) break;
		}
	if(x > 1) ret = ret / x * (x - 1);
	return ret;
}

void read(ll& x)
{
	int f = 1; x = 0; char ch = getchar();
	while(!isdigit(ch)) { if(ch == '-') f = -1; ch = getchar(); }
	while(isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
	x *= f;
}

int main()
{
	ll n, x; read(n);
	while(n--)
	{
		read(x);
		printf("%lld\n", euler(x-1));
	}
	return 0;
}