POJ 1284 Primitive Roots(原根)

2014-11-24 02:42:10 · 作者: · 浏览: 2

素数的原根个数:p是素数,则p有φ(p-1)个原根(其中φ(p)表示p的欧拉函数)

代码:

1:


[cpp]
#include
#include
#include
using namespace std;
bool boo[50000];
long long p[20000];
void prim()//线性筛素数,在main函数里面先调用函数prim
{
memset(boo, 0, sizeof (boo));
boo[0]=boo[1]=1;
int k=0;
for (int i=2; i<50000; i++)
{
if (!boo[i]) p[k++]=i;
for (int j=0; j {
boo[i*p[j]]=1;
if (!(i%p[j])) break;
}
}
}
long long phi(long long n)
{
long long rea = n;
for (int i=0; p[i]*p[i]<=n; i++)//对于一些不是素数的可以不用遍历
{
if (n%p[i]==0)
{
rea-=rea/p[i];
do
n/=p[i];
while (n%p[i]==0);
}
}
if (n>1)
rea-=rea/n;
return rea;
}
int main()
{
int n;
prim();
while (cin>>n)
{
cout< }
return 0;
}

#include
#include
#include
using namespace std;
bool boo[50000];
long long p[20000];
void prim()//线性筛素数,在main函数里面先调用函数prim
{
memset(boo, 0, sizeof (boo));
boo[0]=boo[1]=1;
int k=0;
for (int i=2; i<50000; i++)
{
if (!boo[i]) p[k++]=i;
for (int j=0; j {
boo[i*p[j]]=1;
if (!(i%p[j])) break;
}
}
}
long long phi(long long n)
{
long long rea = n;
for (int i=0; p[i]*p[i]<=n; i++)//对于一些不是素数的可以不用遍历
{
if (n%p[i]==0)
{
rea-=rea/p[i];
do
n/=p[i];
while (n%p[i]==0);
}
}
if (n>1)
rea-=rea/n;
return rea;
}
int main()
{
int n;
prim();
while (cin>>n)
{
cout< }
return 0;
}2:


[cpp]
#include
#include
using namespace std;
long long phi(long long n)
{
long long rea = n;
for (int i=2; i*i<=n; i++)
{
if (n%i==0)
{
rea-=rea/i;
do
n/=i;
while (n%i==0);
}
}
if (n>1)
rea-=rea/n;
return rea;
}
int main()
{
int n;
while (cin>>n)
cout< return 0;
}