原根就是一種數論的定義,詳情請見ssy的部落格.我這裡主要講的是poj的1284.

poj這道題要求的是一個數的原根有多少個,我一開始自己瞎做把所有原根都求出來了,程式碼如下:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(register int i = a;i <= n;i++)
#define
 
 lv(i,a,n) for(register int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        
 
if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}
int
 
 p;
int used[100005],pri[50004],top = 0;
int st[70000],tp = 0;
void pri_table()
{
    for(int i = 2;i <= 1000;i++)
    {
        if(used[i] == 0)
        {
            for(int j = 2;j <= 1000 / i;j++)
            {
                used[i * j] = 1;
            }
            pri[++top] = i;
        }
    }
}
void doit()
{
    clean(st);tp = 0;
    int now = p - 1;
    duke(i,1,top)
    {
        if(now == 1) break;
        if(pri[i] > ceil(sqrt(p)))
        {
            break;
        }
        if(now != 0 && now % pri[i] == 0)
        st[++tp] = pri[i];//cout<<pri[i]<<" ";
        while(now != 0 && now % pri[i] == 0)
        {
            now /= pri[i];
        }
    }
    if(now != 1 && now !=  0)
    {
        st[++tp] = now;
        //cout<<now<<endl;
    }
    /*puts("");*/
}
int qpow(int a,int b)
{
    int tot = 1;
    //cout<<a<<" "<<b<<endl;
    while(b)
    {
        if(b % 2 == 1)
        {
            tot *= a;
            tot %= p;
        }
        a *= a;
        a %= p;
        b >>= 1;
    }
    //cout<<tot<<endl;
    return tot;
}
void done()
{
    int num = 0;
    int now = p - 1;
    /*duke(i,1,tp)
    printf("%d ",st[i]);
    puts("");*/
    duke(i,2,now)
    {
        int flag = 0;
        //cout<<i<<endl;
        //cout<<tp<<endl;
        duke(j,1,tp)
        {
            if(qpow(i,now / st[j]) == 1)
            {
                flag = 1;
                break;
            }
        }
        if(flag == 0)
        {
            num++;
        }
    }
    printf("%d\n",num);
}
int main()
{
    pri_table();
    /*duke(i,1,10)
    printf("%d ",pri[i]);
    puts("");*/
    while(scanf("%d",&p) != EOF)
    {
        tp = 0;
        doit();
        done();
    }
    return 0;
}

但是這樣會T,百度了一下,發現一個性質,就是一個數的原根個數就是phi[phi[n]],而此題n是素數,phi[n]=n - 1;

所以直接輸出phi[n - 1]就行了.

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(register int i = a;i <= n;i++)
#define lv(i,a,n) for(register int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}
int eul[66666];
void phi()
{
    duke(i,1,66666)
    {
        eul[i] = i;
    }
    for(int i = 2;i <= 66666;i++)
    {
        if(eul[i] == i)
        {
            for(int j = i;j <= 66666;j += i)
            {
                eul[j] = eul[j] / i * (i - 1);
            }
        }
    }
    return;
}
int main()
{
    int n;
    phi();
    while(~scanf("%d",&n))
    {
        printf("%d\n",eul[n - 1]);
    }
    return 0;
}

妙極.