Discrete Logging Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 5865 Accepted: 2618 Description Given a prime P, 2 <= P < 231, an integer ...
Discrete Logging
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 5865 | Accepted: 2618 |
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 thatBL
== 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
Hint
The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat's theorem that statesB(P-1)
== 1 (mod P)
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m
B(-m)
== B(P-1-m)
(mod P) .
Source
Waterloo Local 2002.01.26 BSGS模板題1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<map> 6 #define LL long long 7 using namespace std; 8 LL a,b,c; 9 map<LL,LL>mp; 10 LL fastpow(LL a,LL p,LL c) 11 { 12 LL base=a;LL ans=1; 13 while(p!=0) 14 { 15 if(p%2==1)ans=(ans*base)%c; 16 base=(base*base)%c; 17 p=p/2; 18 } 19 return ans; 20 } 21 int main() 22 { 23 // a^x = b (mod c) 24 while(scanf("%lld%lld%lld",&c,&a,&b)!=EOF) 25 { 26 LL m=ceil(sqrt(c));// 註意要向上取整 27 mp.clear(); 28 if(a%c==0) 29 { 30 printf("no solution\n"); 31 continue; 32 } 33 // 費馬小定理的有解條件 34 LL ans;//儲存每一次枚舉的結果 b* a^j 35 for(LL j=0;j<=m;j++) // a^(i*m) = b * a^j 36 { 37 if(j==0) 38 { 39 ans=b%c; 40 mp[ans]=j;// 處理 a^0 = 1 41 continue; 42 } 43 ans=(ans*a)%c;// a^j 44 mp[ans]=j;// 儲存每一次枚舉的結果 45 } 46 LL t=fastpow(a,m,c); 47 ans=1;//a ^(i*m) 48 LL flag=0; 49 for(LL i=1;i<=m;i++) 50 { 51 ans=(ans*t)%c; 52 if(mp[ans]) 53 { 54 LL out=i*m-mp[ans];// x= i*m-j 55 printf("%lld\n",(out%c+c)%c); 56 flag=1; 57 break; 58 } 59 60 } 61 if(!flag) 62 printf("no solution\n"); 63 } 64 65 return 0; 66 }
