数论学习总结 | 各种模板

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 欧拉筛法(过程中求欧拉函数)+求一个数的欧拉函数

bool isPrime[N];
ll prime[N],Eu[N];
void euler(ll Range)
{
    ll cntPrime=0;
    memset(isPrime,1,sizeof(isPrime));
    isPrime[1]=0;
    for (ll i=2;i<=Range;i++)
    {
    if (isPrime[i]) prime[++cntPrime]=i,Eu[i]=i-1;
    for (int j=1;j<=cntPrime && prime[j]*i<=Range;j++)
    {
        isPrime[prime[j]*i]=0;
        if (i%prime[j]==0)
        {
        Eu[i*prime[j]]=Eu[i]*prime[j];
        break;
        }
        Eu[i*prime[j]]=Eu[i]*(prime[j]-1);
    }
    }
}
int oula(int n)
{
    int ans=n,a=n;
    for(int i=2;i*i<=n;i++)
    {
    if(a%i==0)
    {
        ans-=ans/i;
        while(a%i==0)
        a/=i;
    }
    }
    if(a>1) ans-=ans/a;
    return ans;
}

---

EXGCD

ll exGcd(ll a,ll b,ll &x,ll &y)
{
    if (b==0) return x=1,y=0,a;
    ll r=exGcd(b,a%b,y,x);
    y-=(a/b)*x;
    return r;
}

因为ax+by=1 → ax+by+ab-ab=1 → a(x+b)+b(y+a)=1

所以在求出一个可行解x之后,不断的把x+-b仍是该方程的一个解,利用这个性质可以求类似最小解的问题

---

给出A,B,C 求最小的x满足A^x=B (mod C)

BSGS问题

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
typedef long long ll;
using namespace std;
ll x,z,k;
ll Gcd(ll x,ll y)
{
    return y==0?x:Gcd(y,x%y);
}
ll exGcd(ll a,ll b,ll &x,ll &y)
{
    if (b==0) return x=1,y=0,a;
    ll r=exGcd(b,a%b,y,x);
    y-=a/b*x;
    return r;
}
ll inv(ll a,ll m)
{
    ll x,y;
    exGcd(a,m,x,y);
    return (x%m+m)%m;
}
namespace Hash
{
    const ll N=50000;
    const ll H=999979;
    struct adj
    {
    ll nxt,v,num,val;
    }e[N];
    ll  head[H],ecnt=0;
    void init()
    {
    ecnt=0;
    memset(head,0,sizeof(head));
    }
    void insert(ll x,ll val)
    {
    ll org=x;
    x%=H;
    for (int i=head[x];i;i=e[i].nxt)
    {
        if (e[i].num==org)
        {
        e[i].val=val;
        return ;
        }
    }
    e[++ecnt].num=org;
    e[ecnt].val=val;
    e[ecnt].nxt=head[x];
    head[x]=ecnt;
    }
    ll query(ll x)
    {
    ll org=x;
    x%=H;
    for (int i=head[x];i;i=e[i].nxt)
        if (e[i].num==org) return e[i].val;
    return -1;
    }
}
ll BSGS(ll a,ll b,ll c)
{
    ll cnt=0,G,d=1;
    while ((G=Gcd(a,c))!=1)
    {
    if (b%G!=0) return -1;
    cnt++,b/=G,c/=G;
    d=d*(a/G)%c;
    }
    b=b*inv(d,c)%c;
    Hash::init();
    ll s=sqrt(c*1.0);
    ll p=1;
    for (int i=0;i<s;i++)
    {
    if (p==b) return i+cnt;
    Hash::insert(p*b%c,i);
    p=p*a%c;
    }
    ll q=p,t;
    for (int i=s;i-s+1<=c-1;i+=s)
    {
    t=Hash::query(q);
    if (t!=-1) return i-t+cnt;
    q=q*p%c;
    }
    return -1;
}
int check()
{
    for (ll i=0,j=1;i<=10;i++)
    {
    if (j==k)
    {
        printf("%lld
",i);
        return 1;
    }
    j=j*x%z;
    }
    if (x==0)
    {
    puts("No Solution");
    return 1;
    }
    return 0;
}
int main()
{
    while (scanf("%lld%lld%lld",&x,&z,&k) && x+z+k>0)
    {
    x%=z,k%=z;
    if (check()) continue;
    ll ans=BSGS(x,k,z);
    if (ans==-1) puts("No Solution");
    else printf("%lld
",ans);
    }
    return 0;
}