題目鏈接:https://ac.nowcoder.com/acm/contest/885/C
思路:感覺自己說不清楚,具體見文末參考鏈接博客。
AC代碼:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,x0,a,b,p;
ll q,v;
ll qpow(ll x,ll y,ll mo)
{
ll ans=1;
while(y){
if(y&1){
ans=ans*x%mo;
}
x=x*x%mo;
y>>=1;
}
return ans;
}
ll t1=1001;
ll t2;
ll maxn=1e6;
unordered_map < ll , ll > mp;//手寫hash更快,但是不會,要是超時就多交幾發
int main()
{
int t;
scanf("%d",&t);
ll ans;
while(t--){
mp.clear();
scanf("%lld%lld%lld%lld%lld",&n,&x0,&a,&b,&p);
ll t=qpow(a,t1,p);
ll tmp=1;
for(ll x=1;x<=maxn;x++){
tmp=tmp*t%p;
if(mp.count(tmp)==0){//因爲是求最小,不能覆蓋
mp[tmp]=x*t1;
}
}
scanf("%lld",&q);
ll fenmu=(x0*(a-1)%p+b+p)%p;
ll inv_fenmu=qpow(fenmu,p-2,p);
ll inv_b=qpow(b,p-2,p);
while(q--){
scanf("%lld",&v);
if(a==0){
if(v==x0){
printf("0\n");
}
else if(v==b){
printf("1\n");
}
else{
printf("-1\n");
}
continue;
}
else if(a==1){
ans=(v-x0+p)%p*inv_b%p;
if(ans<n){
printf("%lld\n",ans);
}
else{
printf("-1\n");
}
continue;
}
else{
ans=p+1;
v=(v*(a-1)%p+b)%p*inv_fenmu%p;
for(ll j=0;j<=t1;j++){
if(mp.count(v)){
ans=min(ans,mp[v]-j);
}
v=v*a%p;
}
if(ans==p+1||ans>=n){
ans=-1;
}
printf("%lld\n",ans);
}
}
}
return 0;
}