【hdu2815-Mod Tree】高次同余方程-拓展BadyStepGaintStep

http://acm.hdu.edu.cn/showproblem.php?pid=2815

题意：裸题。。。

关于拓展BSGS的详细解释我写了一篇博文：http://www.cnblogs.com/KonjakJuruo/p/5178600.html

题解：又有一个坑，就是N>=P的时候输出无解。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long LL;
const int SIZE=40000;
LL K,P,N;
int bl;
struct node{
    LL d,id;
}bit[SIZE];

bool cmp(node x,node y){
    if(x.d==y.d) return x.id<y.id;
    return x.d<y.d;
}

LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0) {x=1,y=0;return a;}
    LL tx,ty;
    LL d=exgcd(b,a%b,tx,ty);
    x=ty;y=tx-(a/b)*ty;
    return d;
}

LL find(LL x)
{
    int l=0,r=bl;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(bit[mid].d==x) return bit[mid].id;
        if(bit[mid].d<x) l=mid+1;
        if(bit[mid].d>x) r=mid-1;
    }
    return -1;
}

LL exBSGS()
{
    if(N>=P) return -1;
    LL t,m,a,b,c,g,k,x,y,add,pm,am;
    k=1;a=K;b=N;c=P;add=0;t=1;
    for(int i=0;i<=100;i++)//在约分前
    {
        if(t%c==b) return i;
        t=t*a%c;
    }
    while((g=exgcd(K,c,x,y)) != 1)
    {
        k=(k*K/g)%c;
        c/=g;//不是mod
        if(b%g) return -1;
        b/=g;add++;
    }
    m=(LL)(ceil((double)sqrt((double)c)));
    pm=1%c;bit[0].d=k%c;bit[0].id=0;
    for(int i=1;i<=m;i++)
    {
        bit[i].d=bit[i-1].d*a%c;
        bit[i].id=i;
        pm=pm*a%c;
    }
    sort(bit,bit+1+m,cmp);
    bl=0;
    for(int i=1;i<=m;i++)
    {
        if(bit[i].d!=bit[bl].d) bit[++bl]=bit[i];
    }
    exgcd(pm,c,x,y);
    am=x%c+c;//x
    t=b%c;
    for(int i=0;i<=m;i++)
    {
        x=find(t);
        if(x!=-1) return i*m+x+add;
        t=t*am%c;
    }
    return -1;
}

int main()
{
    // freopen("a.in","r",stdin);
    // freopen("a.out","w",stdout);
    while(scanf("%I64d%I64d%I64d",&K,&P,&N)!=EOF)
    {
        LL ans=exBSGS();
        if(ans==-1) printf("Orz,I can’t find D!
");
        else printf("%I64d
",ans);
    }
    return 0;
}