CF995E Number Clicker

题目描述

Allen is playing Number Clicker on his phone.

He starts with an integer u u on the screen. Every second, he can press one of 3 buttons.

Turn $u \to u+1 \pmod{p}$.

Turn $u \to u+p-1 \pmod{p}$.

Turn $u \to u^{p-2} \pmod{p}$.

Allen wants to press at most 200 buttons and end up with v v on the screen. Help him!

输入输出格式

输入格式：

The first line of the input contains 3 positive integers: $u, v, p$( $0 \le u, v \le p-1$ , $3 \le p \le 10^9 + 9$ ). $p$ is guaranteed to be prime.

输出格式：

On the first line, print a single integer $\ell$ , the number of button presses.

On the second line, print integers $c_1, \dots, c_\ell$, the button presses.

For $1 \le i \le \ell$ , $1 \le c_i \le 3$.

We can show that the answer always exists.

长见识了。

发现这道题可以建成一个比较随机的图。

根据生日攻击，我们基本上只需要图的点数$p$开根号个$\sqrt p$，就可以找到答案了。

于是进行双向搜索，合并答案即可。

Code:

#include <cstdio>

#include <map>

#define ll long long

const int N=2e5;

ll u,v,p,q[N+10],l=1,r=0,s[N],tot;

std::map <ll,ll > pre,used,opt,pre0,opt0;

void in(ll now,ll to,ll op)

{

    if(!used[to])

    {

        used[to]=1;

        opt[to]=op;

        pre[to]=now;

        q[++r]=to;

    }

}

ll quickpow(ll d,ll k)

{

    ll f=1;

    while(k)

    {

        if(k&1) f=f*d%p;

        d=d*d%p;

        k>>=1;

    }

    return f;

}

int bfs0()

{

    q[++r]=u;

    while(l<=r&&r<=N)

    {

        ll now=q[l++];

        if(now==v)

        {

            while(now!=u) s[++tot]=opt[now],now=pre[now];

            printf("%lld\n",tot);

            for(int i=tot;i;i--)

                printf("%lld ",s[i]);

            return 1;

        }

        ll to=(now+1)%p;

        in(now,to,1);

        to=(now+p-1)%p;

        in(now,to,2);

        to=quickpow(now,p-2);

        in(now,to,3);

    }

    return 0;

}

void in0(ll now,ll to,ll op)

{

    if(!used[to])

    {

        used[to]=1;

        opt0[to]=op;

        pre0[to]=now;

        q[++r]=to;

    }

}

ll tmp;

void swap(ll &x,ll &y){tmp=x,x=y,y=tmp;}

void bfs1()

{

    used.clear();

    l=1,r=0;

    q[++r]=v;

    while(l<=r&&r<=N)

    {

        ll now=q[l++];

        if(pre.find(now)!=pre.end())

        {

            ll t=now;

            while(now!=u) s[++tot]=opt[now],now=pre[now];

            for(int i=1;i<=tot>>1;i++) swap(s[i],s[tot+1-i]);

            now=t;

            while(now!=v) s[++tot]=opt0[now],now=pre0[now];

            printf("%lld\n",tot);

            for(int i=1;i<=tot;i++)

                printf("%lld ",s[i]);

            return;

        }

        ll to=(now+1)%p;

        in0(now,to,2);

        to=(now+p-1)%p;

        in0(now,to,1);

        to=quickpow(now,p-2);

        in0(now,to,3);

    }

}

int main()

{

    //freopen("dew.out","w",stdout);

    scanf("%lld%lld%lld",&u,&v,&p);

    if(bfs0()) return 0;

    bfs1();

    return 0;

}

2018.10.9