# bzoj2242(高次同余方程，线性同余方程）

#include<cstdio>
#include<map>
#include<algorithm>
#include<cmath>
using namespace std;

typedef long long ll;

ll power(ll a,ll b,ll mod){
ll ans=1%mod;
for (;b;b>>=1){
if(b&1) ans=ans*a%mod;
a=a*a%mod;
}
return ans%mod;
}

ll exgcd(ll a,ll b,ll &x,ll &y){
if(b==0){
x=1;
y=0;
return a;
}
ll d=exgcd(b,a%b,x,y);
ll z=x;
x=y;
y=z-(a/b)*y;
return d;
}

ll babe(ll a,ll b,ll p){
map<int,int>hashl;
hashl.clear();
b%=p;
ll t=(int)sqrt(p)+1;
for (int j=0;j<t;j++){
ll val=b*power(a,j,p)%p;
hashl[val]=j;
}
a=power(a,t,p);
if(a==0) return b==0?1:-1;
for (int i=0;i<=t;i++){
ll val=power(a,i,p);
ll j;
if(hashl.find(val)==hashl.end()) j=-1;
else j=hashl[val];
if(j>=0&&i*t-j>=0) return i*t-j;
}
return -1;
}

int main(){
int t,k;
scanf("%d%d",&t,&k);
ll y,z,p,x;
if(k==1){
for (int i=1;i<=t;i++)
{
scanf("%lld%lld%lld",&y,&z,&p);
printf("%lld\n",power(y,z,p));
}
}
if(k==2){
ll a,b;
for (int i=1;i<=t;i++){
scanf("%lld%lld%lld",&a,&b,&p);
int d=exgcd(a,p,x,y);
if(b%d) printf("Orz, I cannot find x!\n");
else {
ll s=p/d;
x=(b/d*x%s+s)%s;//怎样找最小解
printf("%lld\n",x);
}
}
}
if(k==3){
ll a,b;
for (int i=1;i<=t;i++){
scanf("%lld%lld%lld",&a,&b,&p);
int h=babe(a,b,p);
if(h==-1) printf("Orz, I cannot find x!\n");
else printf("%lld\n",h%p);记得mod p
}
}
return 0;
}

posted @ 2018-07-14 17:14  lmjer  阅读(164)  评论(0编辑  收藏  举报