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
原文地址：https://www.cnblogs.com/ppprseter/p/9762632.html