题目链接
思路：BFS
简单模拟两个函数，然后通过BFS不断将当前可能的两种操作放入队列，只要变成了B就说明是最小次数，代码如下：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int N=5005,mod=233;
int vis[300];
const ll INF=999999999999999999;
ll a,b;
struct A{
    ll now,num;
};
ll P(ll x){
    return ((x*x%mod*x)%mod+x*x%mod)%mod;
}

ll G(ll x){
    return ((x*x%mod*x)%mod-x*x%mod+mod)%mod;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        memset(vis,0,sizeof(vis));
        scanf("%lld%lld",&a,&b);
        if(a==b){printf("0\n");continue;}
        if(b>=mod){printf("-1\n");continue;}
        queue<A>qqq;
        if(a>=233){
            ll c=P(a);
            ll d=G(a);
            vis[c]=1;vis[d]=1;
            qqq.push(A{c,1});
            qqq.push(A{d,1});
        }
        else{
            vis[a]=1;
            qqq.push(A{a,0});
        }
        ll ans=INF;
        //把出现过的状态都标记一下，只入队没出现过的状态
        while(!qqq.empty()){//队列不为空
            A tmp=qqq.front();
            qqq.pop();
            if(tmp.now==b){
                ans=tmp.num;
                break;
            }
            else {
                ll t1=P(tmp.now),t2=G(tmp.now);
                if(!vis[t1])qqq.push(A{t1,tmp.num+1});
                if(!vis[t2])qqq.push(A{t2,tmp.num+1});
                vis[t1]=1;vis[t2]=1;
            }
        }
        if(ans==INF)printf("-1\n");
        else printf("%lld\n",ans);
    }
}