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