【BZOJ】【3239】Discrete Logging

BSGS

　　BSGS裸题，嗯题目中也有提示：求a^m (mod p)的逆元可用快速幂，即 pow(a,P-m-1,P) * (a^m) = 1 (mod p)

　　

 /**************************************************************
     Problem: 3239
     User: Tunix
     Language: C++
     Result: Accepted
     Time:208 ms
     Memory:2872 kb
 ****************************************************************/

 //BZOJ 3239
 #include<cmath>
 #include<map>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;
 int getint(){
     int v=,sign=; char ch=getchar();
     while(!isdigit(ch)) {if(ch=='-') sign=-; ch=getchar();}
     while(isdigit(ch))  {v=v*+ch-''; ch=getchar();}
     return v*sign;
 }
 /*******************template********************/
 typedef long long LL;
 LL pow(LL a,LL b,LL P){
     LL r=,base=a%P;
     while(b){
         if (b&) r=r*base%P;
         base=base*base%P;
         b>>=;
     }
     return r;
 }
 LL log_mod(LL a,LL b,LL P){
     LL m,v,e=;
     m=ceil(sqrt(P+0.5));
     v=pow(a,P-m-,P);
     map<LL,LL>x;
     x[]=;
     for(int i=;i<m;++i){
         e=e*a%P;
         if(!x.count(e)) x[e]=i;
     }
     rep(i,m){
         if (x.count(b)) return (LL) i*m+x[b];
         b=b*v%P;
     }
     return -;
 }
 int main(){
     LL P,b,n;
     while(scanf("%lld%lld%lld",&P,&b,&n)!=EOF){
         b%=P; n%=P;
         if (!b && !n) {printf("1\n"); continue;}
         if (!b) {printf("no solution\n"); continue;}
         LL ans=log_mod(b,n,P);
         if (ans==-){printf("no solution\n");continue;}
         printf("%lld\n",ans);
     }
     return ;
 }