Discrete Logging (poj2417)

Discrete Logging

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5120		Accepted: 2319

Description

Given a prime P, 2 <= p < 231, an integer B, 2 <= B < P, and an integer n, 1 <= n < P, compute the DISCR Ete logarithm of N, base B, modulo P. That's, find an integer L such that

    B

L

= = N (mod P)

Input

Read several lines of input, each containing p,b,n separated by a space.

Output

For each line print the logarithm to a separate line. If There is several, print the smallest; If there is none, print "no solution".

Sample Input

5 2 15 2 25 2 35 2 45 3 15 3 25 3 35 3 45 4 15 4 25 4 35 4 412345701 2 11111111111111121 65537 1111111111

Sample Output

013203120no Solutionno solution19584351462803587

Idea: Baby_step,giant_step algorithm template problem

1#include <stdio.h>2#include <algorithm>3#include <iostream>4#include <stdlib.h>5#include <queue>6#include <vector>7#include <math.h>8#include <string.h>9#include <map>Ten#include <Set> One using namespacestd; AtypedefLong LongLL; -LL mod[65539]; - BOOLjudge (LL N) the { -LL Ac=sqrt (1.0*n); -     if(ac*ac==N) -         return true; +     Else return false; - } +Map<ll,ll>my; A Set<LL>que; at Set<LL>:: iterator it; - ll Quick (ll n,ll m,ll p); - ll gcd (ll n,ll m) - { -     if(m==0) -         returnN; in     Else returnGCD (m,n%m); - } toPair<ll,ll>Pc (LL n,ll m) + { -     if(m==0) the         returnMake_pair (1,0); *     Else $     {Panax NotoginsengPAIR&LT;LL,LL&GT;N=PC (m,n%m); -         returnMake_pair (n.second,n.first-(n/m) *n.second); the     } + } Atypedefstructpp the { + LL x; - LL ID; $ } SS; $SS table[655390]; -SS tt[655390]; - BOOLCMP (pp p,pp q) the { -     if(p.x==q.x)Wuyi         returnp.id<q.id; the     returnp.x<q.x; - } Wu intMainvoid) - { About LL p,b,n; $      while(SCANF ("%lld%lld%lld", &p,&b,&n)! =EOF) -     { -         BOOLA=judge (P); -LL Ask=sqrt (1.0*P); A         if(!a) +ask+=1; the         inti,j; -mod[0]=1; $         intvv=0; thetable[0].id=0; thetable[0].x=1; the         intcn=1; the          for(i=1; i<=ask; i++) -         { inmod[i]=mod[i-1]*b%P; theTable[i].id=i; thetable[i].x=Mod[i]; About         } theSort (table,table+ask+1, CMP); thett[0].id=table[0].id; thett[0].x=table[0].x; +LL yy=tt[0].x; -          for(i=1; i<=ask;i++) the         {Bayi             if(table[i].x!=yy) the             { theyy=table[i].x; -tt[cn].x=yy; -Tt[cn].id=table[i].id; thecn++; the             } the         } the         intFl=0; - LL ack; theLL Nn=quick (b,p-2, P); thenn=Quick (nn,ask,p); theLL ni=1;94          for(i=0; i<=ask; i++) the         { theLL ap=ni*n%P; theni%=P;98             intL=0; About             intr=cn-1; -LL ic=-1;101              while(l<=R)102             {103                 intMid= (L+R)/2;104                 if(tt[mid].x>=AP) the                 {106Ic=mid;107r=mid-1;108ack=tt[ic].id;109                 } the                 ElseL=mid+1;111             } the             if(ic!=-1&&tt[ic].x==AP)113             { the                 //printf ("%lld\n", table[ic].x);p rintf ("%d\n", I); theFl=1; the                  Break;117             }118ni= (NI*NN)%P;119         } -         if(!FL)121printf"No solution\n");122         Elseprintf"%lld\n", ack+ (LL) i*ask);123     }124     return 0; the }126 ll Quick (ll n,ll m,ll p)127 { -n%=p;129LL ans=1; the      while(m)131     { the         if(m&1)133         {134ans=ans*n%p;135         }136n=n*n%p;137M/=2;138     }139     returnans; $}

Discrete Logging (poj2417)