bzoj 3285 离散对数解指数方程

 /**************************************************************
     Problem: 3285
     User: idy002
     Language: C++
     Result: Accepted
     Time:756 ms
     Memory:32072 kb
 ****************************************************************/

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <cctype>
 #define N 1000010

 typedef long long dnt;

 const int Hmod = ;
 struct Hash {
     int head[N], key[N], val[N], next[N], etot;
     void init() {
         etot = ;
         memset( head, , sizeof(head) );
     }
     void insert( int k, int v ) {
         int kk = k%Hmod;
         etot++;
         key[etot] = k;
         val[etot] = v;
         next[etot] = head[kk];
         head[kk] = etot;
     }
     int query( int k ) {
         int kk = k%Hmod;
         for( int t=head[kk]; t; t=next[t] )
             if( key[t]==k ) return val[t];
         return -;
     }
 }hash;

 dnt mpow( dnt a, int b, int c ) {
     dnt rt;
     for( rt=; b; b>>=,a=(a*a)%c )
         if( b& ) rt=(rt*a)%c;
     return rt;
 }
 int findroot( int p ) {
     int phi = p-;
     int tmp = phi;
     int stk[], top;
     top = ;
     for( int i=; i<=(<<); i++ ) {
         if( tmp%i== ) {
             stk[++top] = i;
             do {
                 tmp/=i;
             }while( tmp%i== );
         }
     }
     if( tmp!= )
         stk[++top] = tmp;
     for( int r=; ; r++ ) {
         bool ok = true;
         for( int i=; i<=top; i++ ) {
             if( mpow(r,phi/stk[i],p)== ) {
                 ok=false;
                 break;
             }
         }
         if( ok ) return r;
     }
 }
 dnt ind( dnt r, int a, int p ) {    //  ind_r(a) mod p-1
     int m = ceil(sqrt(p-));
     hash.init();
     dnt cur = ;
     for( int i=; i<m; i++ ) {
         if( cur==a ) return i;
         hash.insert( cur, i );
         cur = (cur*r) % p;
     }
     dnt base;
     base = cur = mpow(cur,p-,p);
     for( int i=m; i<p; i+=m,cur=(cur*base)%p ) {
         int j = hash.query( a*cur%p );
         if( j!=- ) return i+j;
     }
     return -;  //  impossible
 }
 dnt gcd( dnt a, dnt b ) {
     return b ? gcd(b,a%b) : a;
 }
 void exgcd( dnt a, dnt b, dnt &d, dnt &x, dnt &y ) {
     if( b== ) {
         d=a, x=, y=;
     } else {
         exgcd(b,a%b,d,y,x);
         y-=a/b*x;
     }
 }
 dnt meq( dnt a, dnt b, dnt c ) {    //  ax=b mod c
     dnt d, dd, x, y;
     a = (a%c+c)%c;
     b = (b%c+c)%c;
     d = gcd(a,c);
     if( b%d!= ) return -;
     exgcd(a/d,c/d,dd,x,y);
     x = x*(b/d);
     x = (x%(c/d)+(c/d))%(c/d);
     if( x== ) x+=c/d;
     return x;
 }

 dnt a, b, c, g, p, r;
 int aa[N], bb[N], cc[N], gg[N];

 void read( int a[] ) {
     int i;
     char ch;
     for( i=; isdigit(ch=getchar()); i++ )
         a[i] = ch-'';
     a[i] = -;
 }
 dnt modulo( int a[], dnt mod ) {
     dnt rt = ;
     for( int i=; a[i]!=-; i++ )
         rt = (rt* + a[i]) % mod;
     return rt;
 }
 int main() {
     read(aa);
     read(bb);
     read(cc);
     read(gg);
     scanf( "%lld", &p );
     a = modulo(aa,p-);
     b = modulo(bb,p-),
     c = modulo(cc,p);
     g = modulo(gg,p);

     if( g%p== || c%p== ) {
         if( g%p== && c%p== )
             printf( "1\n" );
         else
             printf( "no solution\n" );
         return ;
     }
     r = findroot(p);
 //  fprintf( stderr, "%d\n", (int)r );
     dnt ans = meq( a*ind(r,g,p), ind(r,c,p)-b*ind(r,g,p), p- );
     if( ans< )
         printf( "no solution\n" );
     else
         printf( "%lld\n", ans );
 }