给a^x == b (mod c)求满足的最小正整数x, 用BSGS求,令m=ceil(sqrt(m)),x=im-j,那么a^(im)=ba^j%p;, 我们先枚举j求出所有的ba^j%p,1<=j<m复杂度O(sqrt(c)),然后枚举1<=i<=m,求出a^(im)在ba^j找满足条件的答案,最后的答案就是第一个满足条件的i*m-j,复杂度O(sqrt(c)) //#pragma comment(linker, "/stack:200000000") //…

[BZOJ3239]Discrete Logging Description Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 2 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that BL== N (mod P) Inpu…

Discrete Logging Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5577   Accepted: 2494 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 Given a prime P, 2 <= P < 2 31, 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 line…

本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! Description Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the disc…

<题目链接> 题目大意: P是素数,然后分别给你P,B,N三个数,然你求出满足这个式子的L的最小值 : BL== N (mod P). 解题分析: 这题是bsgs算法的模板题. #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <string> #include <math.h> #include…

题目: 给出A,B,C 求最小的x使得Ax=B  (mod C) 题解: bsgs算法的模板题 bsgs 全称:Baby-step giant-step 把这种问题的规模降低到了sqrt(n)级别 首先B的种类数不超过C种,结合鸽巢原理,所以Ax具有的周期性显然不超过C 所以一般的枚举算法可以O(C)解决这个问题 但是可以考虑把长度为C的区间分为k块,每块长度为b 显然x满足x=bi-p的形式(1<=i<=k,0<=p<b),所以Ax=B  (mod C)移项之后得到Abi=Ap*…

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…

[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…

http://www.lydsy.com/JudgeOnline/problem.php?id=3239 题意:原题很清楚了= = #include <bits/stdc++.h> using namespace std; map<int, int> s; typedef long long ll; int mpow(int a, int b, int p) { a%=p; int r=1; while(b) { if(b&1) r=((ll)r*a)%p; a=((ll)…