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 discrete logarithm of N, base B, modulo P. That is, 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 on a separate line. If there are several, print the smallest; if there is none, print "no solution".

Sample Input

5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111

Sample Output

0
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587

思路：baby_step,giant_step算法模板题

  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>
 10 #include<set>
 11 using namespace std;
 12 typedef long long LL;
 13 LL mod[65539];
 14 bool judge(LL n)
 15 {
 16     LL ac=sqrt(1.0*n);
 17     if(ac*ac==n)
 18         return true;
 19     else return false;
 20 }
 21 map<LL,LL>my;
 22 set<LL>que;
 23 set<LL>::iterator it;
 24 LL quick(LL n,LL m,LL p);
 25 LL gcd(LL n,LL m)
 26 {
 27     if(m==0)
 28         return n;
 29     else return gcd(m,n%m);
 30 }
 31 pair<LL,LL>Pc(LL n,LL m)
 32 {
 33     if(m==0)
 34         return make_pair(1,0);
 35     else
 36     {
 37         pair<LL,LL>N=Pc(m,n%m);
 38         return make_pair(N.second,N.first-(n/m)*N.second);
 39     }
 40 }
 41 typedef struct pp
 42 {
 43     LL x;
 44     LL id;
 45 } ss;
 46 ss table[655390];
 47 ss tt[655390];
 48 bool cmp(pp p,pp q)
 49 {
 50     if(p.x==q.x)
 51         return p.id<q.id;
 52     return p.x<q.x;
 53 }
 54 int main(void)
 55 {
 56     LL P,B,N;
 57     while(scanf("%lld %lld %lld",&P,&B,&N)!=EOF)
 58     {
 59         bool a=judge(P);
 60         LL ask=sqrt(1.0*P);
 61         if(!a)
 62             ask+=1;
 63         int i,j;
 64         mod[0]=1;
 65         int vv=0;
 66         table[0].id=0;
 67         table[0].x=1;
 68         int cn=1;
 69         for(i=1; i<=ask; i++)
 70         {
 71             mod[i]=mod[i-1]*B%P;
 72             table[i].id=i;
 73             table[i].x=mod[i];
 74         }
 75         sort(table,table+ask+1,cmp);
 76         tt[0].id=table[0].id;
 77         tt[0].x=table[0].x;
 78         LL yy=tt[0].x;
 79         for(i=1;i<=ask;i++)
 80         {
 81             if(table[i].x!=yy)
 82             {
 83                 yy=table[i].x;
 84                 tt[cn].x=yy;
 85                 tt[cn].id=table[i].id;
 86                 cn++;
 87             }
 88         }
 89         int fl=0;
 90         LL ack;
 91         LL nn=quick(B,P-2,P);
 92         nn=quick(nn,ask,P);
 93         LL ni=1;
 94         for(i=0; i<=ask; i++)
 95         {
 96             LL ap=ni*N%P;
 97             ni%=P;
 98             int l=0;
 99             int r=cn-1;
100             LL ic=-1;
101             while(l<=r)
102             {
103                 int mid=(l+r)/2;
104                 if(tt[mid].x>=ap)
105                 {
106                     ic=mid;
107                     r=mid-1;
108                     ack=tt[ic].id;
109                 }
110                 else l=mid+1;
111             }
112             if(ic!=-1&&tt[ic].x==ap)
113             {
114                 //printf("%lld\n",table[ic].x);printf("%d\n",i) ;
115                 fl=1;
116                 break;
117             }
118             ni=(ni*nn)%P;
119         }
120         if(!fl)
121             printf("no solution\n");
122         else printf("%lld\n",ack+(LL)i*ask);
123     }
124     return 0;
125 }
126 LL quick(LL n,LL m,LL p)
127 {
128     n%=p;
129     LL ans=1;
130     while(m)
131     {
132         if(m&1)
133         {
134             ans=ans*n%p;
135         }
136         n=n*n%p;
137         m/=2;
138     }
139     return ans;
140 }