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  • BZOJ 3239 Discrete Logging

    BSGS。细节蛮多的。。。。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #define maxn 100500
    using namespace std;
    struct hhh
    {
        long long id,num;
    }hash[maxn];
    long long b,n,p,lim;
    bool cmp(hhh x,hhh y)
    {
        if (x.num!=y.num) 
            return x.num<y.num;
        return x.id<y.id;
    }
    inline long long mul(long long a,long long b)
    {
        long long d=(long long)floor(a*(double)b/p+0.5);
        long long ret=a*b-d*p;
        if (ret<0) ret+=p;
        return ret;
    }
    inline long long mull(long long a,long long b)
    {
        long long ans=1,base=a;
        while (b)
        {
            if (b&1) ans=mul(ans,base)%p;
            base=mul(base,base)%p;
            b>>=1;
        }
        return ans;
    }
    long long found(long long now,long long x)
    {
        long long ans;
        while (hash[now].num%p==x%p)
        {
            ans=hash[now].id;
            now--;
        }
        return ans;
    }
    long long find(long long x)
    {
        long long left=0,right=lim-1;
        while (left<=right)
        {
            long long mid=(left+right)>>1;
            long long regis1=hash[mid].num%p,regis2=x%p;
            if (regis1==regis2) 
            {
                return found(mid,x);
            } 
            else if (regis1>regis2) right=mid-1;
            else left=mid+1;
        }
        return -1;
    }
    int main()
    {
        while (scanf("%lld%lld%lld",&p,&b,&n)!=EOF)
        {
            long long flag=0;
            long long inv=mull(b,p-2)%p,base=n%p;lim=ceil(sqrt(p));
            for (long long i=0;i<lim;i++)
            {
                hash[i].id=i;
                hash[i].num=base%p;
                base=mul(base,inv)%p;    
            }
            sort(hash,hash+lim,cmp);
            base=1;long long regis=mull(b,lim)%p;
            for (long long i=0;i<=lim;i++)
            {
                long long rr=find(base);
                if ((rr!=-1) && (flag==0))
                {
                    printf("%d
    ",mul(i,lim)+rr);
                    flag=1;
                }
                base=mul(base,regis)%p;
            }
            if (flag==0) printf("no solution
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ziliuziliu/p/5513295.html
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