题意:给sum,m组询问,每组x,y求(x^t=ymod p,p|sum),p是素数,求最小的t
题解:先处理sum的所有质因子p,求出p的原根rt,(rt^a=xmod p,rt^b=ymod p),(rt^{a*t}=rt^bmod p),
(a*t=bmod p-1),先预处理bsgs的表,然后求出ab,再exgcd求t,找最小值
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <bits/extc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
//#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll mul(ll a,ll b,ll c){return (a*b-(ll)((ld)a*b/c)*c+c)%c;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=mul(ans,a,c);a=mul(a,a,c),b>>=1;}return ans;}
using namespace std;
//using namespace __gnu_pbds;
const ld pi=acos(-1);
const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=1000000+7,maxn=2000000+10,inf=0x3f3f3f3f;
struct prime{
int p,rt;
struct HashTable
{
int top, head[N];
struct Node
{
int x, y, next;//x是原根p的幂,y是幂次数,next是y+1次幂所在的head桶
}node[N];
void init()
{
top = 0;
memset(head, 0, sizeof(head));
}
void Insert(int x, int y)
{
node[top].x = x; node[top].y = y; node[top].next = head[x%N];
head[x%N] = top++;
}
int Find(int x)//求出x是原根proot的几次幂
{
for (int tx = head[x%N]; tx; tx = node[tx].next)
if (node[tx].x == x)return node[tx].y;
return -1;
}
}ma;
void init()
{
ma.init();
int a=rt;
int m=100000,now=1;
ma.Insert(now,0);
for(int i=1;i<=m;i++)
{
now=(1ll*now*a)%p;
ma.Insert(now,i);
}
}
int query(int b)
{
int m=100000;
int ans=-1,t=qp(rt,m,p),now=qp(b,p-2,p);
for(int i=1;i<=m;i++)
{
now=1ll*now*t%p;
int te=ma.Find(now);
if(~te)
{
ans=i*m-te;
break;
}
}
if(ans!=-1)ans=(ans%p+p)%p;
return ans;
}
}p[15];
vi v;
int findroot(int n)
{
int p=n-1;v.clear();
for(int i=2;i*i<=p;i++)if(p%i==0)
{
v.pb(i);
while(p%i==0)p/=i;
}
if(p!=1)v.pb(p);
for(int i=1;i<=n;i++)
{
bool ok=1;
for(int j=0;j<v.size();j++)
if(qp(i,(n-1)/v[j],n)==1)
ok=0;
if(ok)return i;
}
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
if(!b){x=1,y=0;return a;}
ll ans=exgcd(b,a%b,x,y);
ll t=x;x=y;y=t-a/b*y;
return ans;
}
int main()
{
int t,cas=0;scanf("%d",&t);
while(t--)
{
int cnt=0;
int s,m;scanf("%d%d",&s,&m);
for(int i=2;i*i<=s;i++)if(s%i==0)
{
p[++cnt].p=i,p[cnt].rt=findroot(i),p[cnt].init();
while(s%i==0)s/=i;
}
if(s!=1)p[++cnt].p=s,p[cnt].rt=findroot(s),p[cnt].init();
printf("Case #%d:
",++cas);
for(int o = 1; o <= m; o++)
{
int x,y,ans=inf;scanf("%d%d",&x,&y);
for(int i=1;i<=cnt;i++)
{
int a=p[i].query(x),b=p[i].query(y);
assert(a!=-1&&b!=-1);
if(b%gcd(a,p[i].p-1)!=0)continue;
else
{
ll c,d;
exgcd(a,p[i].p-1,c,d);
c=c*b/gcd(a,p[i].p-1);
ll te=(p[i].p-1)/gcd(a,p[i].p-1);
c=(c%te+te)%te;
ans=min(ans,(int)c);
}
}
printf("%d
",ans==inf?-1:ans);
}
}
return 0;
}
/********************
1
175 2
2 3
********************/