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…

Discrete Logging Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 2819   Accepted: 1386 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, b…

d. 式子B^L=N(mod P),给出B.N.P,求最小的L. s.下面解法是设的im-j,而不是im+j. 设im+j的话,貌似要求逆元什么鬼 c. /* POJ 2417,3243 baby step giant step a^x=b(mod n) n是素数或不是素数都可以 求解上式 0<=x<n的解 */ #include<iostream> #include<stdio.h> #include<string.h> #include<math.…

[题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=3239 [题目大意] 计算满足 Y^x ≡ Z ( mod P) 的最小非负整数 [题解] BSGS裸题. [代码] #include <cstdio> #include <cmath> #include <map> #include <algorithm> #include <tr1/unordered_map> using name…

POJ 2417 Discrete Logging Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 4860   Accepted: 2211 Description Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarith…

Discrete Logging Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 2831   Accepted: 1391 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, b…

Discrete Logging Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 3696   Accepted: 1727 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, b…

http://poj.org/problem?id=2417 BSGS 大步小步法( baby step giant step ) sqrt( p )的复杂度求出 ( a^x ) % p = b % p中的x https://www.cnblogs.com/mjtcn/p/6879074.html 我的代码中预处理a==b和b==1的部分其实是不必要的,因为w=sqrt(p)(向上取整),大步小步法所找的x包含从0到w^2. #include<iostream> #include<cst…

http://poj.org/problem?id=2417 A^x = B(mod C),已知A,B.C.求x. 这里C是素数,能够用普通的baby_step. 在寻找最小的x的过程中,将x设为i*M+j.从而原始变为A^M^i * A^j = B(mod C),D = A^M,那么D^i * A^j = B(mod C ), 预先将A^j存入hash表中,然后枚举i(0~M-1),依据扩展欧几里得求出A^j.再去hash表中查找对应的j,那么x = i*M+j. 确定x是否有解,就是在循环i…

链接:http://poj.org/problem?id=2417 题意: 思路:求离散对数,Baby Step Giant Step算法基本应用. 下面转载自:AekdyCoin [普通Baby Step Giant Step] [问题模型] 求解 A^x = B (mod C) 中 0 <= x < C 的解,C 为素数 [思路] 我们能够做一个等价 x = i * m + j  ( 0 <= i < m, 0 <=j < m) m = Ceil ( sqrt( C…