BZOJ 2242 [SDOI2011]计算器 BSGS+快速幂+EXGCD

题意:链接

方法: BSGS+快速幂+EXGCD

解析:

BSGS…

题解同上..

代码:

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MOD 140345
using namespace std;
typedef long long ll;
ll t,k,ans;
ll y,z,p;
int head[MOD+10],cnt;
struct node
{
    ll from,to,val,next;
}edge[MOD+10];
void init()
{
    memset(head,-1,sizeof(head));
    cnt=1;
}
void edgeadd(ll from,ll to,ll val)
{
    edge[cnt].from=from,edge[cnt].to=to,edge[cnt].val=val;
    edge[cnt].next=head[from];
    head[from]=cnt++;
}
ll quick_my(ll x,ll y)
{
    ll ret=1;
    while(y)
    {
        if(y&1)ret=(ret*x)%p;
        x=(x*x)%p;
        y>>=1;
    }
    return ret;
}
void exgcd(ll a,ll b,ll &x,ll &y,ll &gcd)
{
    if(!b)
    {
        x=1,y=0,gcd=a;
        return;
    }
    exgcd(b,a%b,y,x,gcd);
    y=y-a/b*x;
}
void BSGS(ll A,ll B,ll C)
{
    //A^x=B(mod C)
    ll m=(int)ceil(sqrt(C));
    ll k=1;
    for(int i=0;i<m;i++)
    {
        int flag=1;
        for(int j=head[k%MOD];j!=-1;j=edge[j].next)
        {
            if(edge[j].val==k){flag=0;continue;}
        }
        if(flag)edgeadd(k%MOD,i,k);
        k=k*A%C;
    }
    ll X,Y,GCD;
    exgcd(k,C,X,Y,GCD);
    ll invk=(X%C+C)%C;
    ll D=1,invD=1;
    for(int i=0;i<=m;i++)
    {
        ll tmpB=B*invD%C;
        for(int j=head[tmpB%MOD];j!=-1;j=edge[j].next)
        {
            if(edge[j].val==tmpB){ans=i*m+edge[j].to;return;}
        }
        D=D*k%C;
        invD=invD*invk%C;
    }
}
int main()
{
    scanf("%lld%lld",&t,&k);
    switch(k)
    {
        case 1:
            while(t--)
            {
                scanf("%lld%lld%lld",&y,&z,&p);
                printf("%lld\n",quick_my(y,z));
            }
            break;
        case 2:
            while(t--)
            {
                ll x,tmp,gcd;
                scanf("%lld%lld%lld",&y,&z,&p);
                exgcd(y,p,x,tmp,gcd);
                if(z%gcd!=0)puts("Orz, I cannot find x!");
                else
                {
                    ll mod=p/gcd;
                    printf("%lld\n",((x*z/gcd)%mod+mod)%mod);
                }
            }
            break;
        case 3:
            while(t--)
            {
                init();
                scanf("%lld%lld%lld",&y,&z,&p);
                ans=-1;
                if(y%p==0&&z!=0){puts("Orz, I cannot find x!");continue;}
                BSGS(y,z,p);
                if(ans==-1)puts("Orz, I cannot find x!");
                else printf("%lld\n",ans);
            }
    }
}