POJ 2417 Discrete Logging
2016-12-05 15:18
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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
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
Sample Output
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 states
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
Source
Waterloo Local 2002.01.26
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
BSGS~
BSGS—Baby-Step-Giant-Step,用于解方程A^x≡B(mod C):
设m=ceil(sqrt(C))(ceil即上取整),x=i*m-j,则A^(i*m)≡B*A^j(mod C)。
我们从0~m枚举j,将求得的B*A^j存入一个hash表中(用map替代),然后1~m枚举i,将对应的A^(i*m)与hash匹配,第一个出现再map中的即为答案。
(更详细的证明:http://blog.csdn.net/clover_hxy/article/details/50683832~)
(POJ上用C++交这题会CE,要用G++交~)
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
BL == 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 states
B(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~
BSGS—Baby-Step-Giant-Step,用于解方程A^x≡B(mod C):
设m=ceil(sqrt(C))(ceil即上取整),x=i*m-j,则A^(i*m)≡B*A^j(mod C)。
我们从0~m枚举j,将求得的B*A^j存入一个hash表中(用map替代),然后1~m枚举i,将对应的A^(i*m)与hash匹配,第一个出现再map中的即为答案。
(更详细的证明:http://blog.csdn.net/clover_hxy/article/details/50683832~)
(POJ上用C++交这题会CE,要用G++交~)
#include<cstdio> #include<cstring> #include<iostream> #include<map> #include<cmath> using namespace std; #define ll long long ll a,b,c,ans,now; int m; bool tr; map<ll,int> ma; ll mi(ll u,ll v) { ll num=1; while(v) { if(v&1) num=(num*u)%c; v>>=1;u=u*u%c; } return num; } int main() { while(scanf("%d%lld%lld",&c,&a,&b)!=EOF) { ma.clear(); if(!(a%c)) { printf("no solution\n");continue; } m=sqrt(c);tr=0; if(c!=m*m) m++; ans=b%c;ma[ans]=0; for(int i=1;i<=m;i++) { ans=(ans*a)%c;ma[ans]=i; } ans=1;now=mi(a,m); for(int i=1;i<=m;i++) { ans=(ans*now)%c; if(ma[ans]) { now=i*m-ma[ans]; printf("%lld\n",(now%c+c)%c); tr=1;break; } } if(!tr) { printf("no solution\n");continue; } } return 0; }
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