本题是一道 \(BSGS\) 裸题,用于求解高次同余方程,形如 \(a^x\equiv b(\mod p)\),其中 \(a\),\(p\) 互质(不互质还有 \(EXBSGS\)). 建议多使用 \(HASH\) 表,不要懒省事使 \(map\),数据大时会 \(T\) 飞. \(AC \kern 0.4emCODE:\) #include<bits/stdc++.h> #define int long long #define ri register int #define p(i) ++…

3239: Discrete Logging Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 729  Solved: 485[Submit][Status][Discuss] Description Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 2 <= N < P, compute the discrete logar…

[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 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…

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…

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: 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…

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 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, b…

我先转为敬? orz% miskcoo 贴板子 BZOJ 3239: Discrete Logging//2480: Spoj3105 Mod(两道题输入不同,我这里只贴了3239的代码) CODE #include<bits/stdc++.h> using namespace std; typedef long long LL; int p, a, b; int gcd(int a, int b) { return b ? gcd(b, a%b) : a; } inline int qpow…

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…

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, b…

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…

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…

[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…

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 (题目链接) 题意 求解$${A^X≡B~(mod~P)}$$ Solution BSGS. 细节 map TLE飞,只好写了hash挂链..从TLE到110MS→_→ 代码 // poj2417 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #in…

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)…

http://www.cnblogs.com/jianglangcaijin/archive/2013/04/26/3045795.html 给p,a,b求a^n==b%p #include<algorithm> #include<cstdio> #include<cmath> #include<cstring> #include<iostream> #include<map> #define ll long long ll p; s…

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…

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…

裸题 求\(ind_{n,a}b\),也就是\(a^x \equiv b \pmod n\) 注意这里开根不能直接下取整 这个题少了一些特判也可以过... #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> using namespace std; typedef lo…

就是求Ax三B(mod C)当C为素数时 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; typedef long long LL; << ) - ) * + ; int A, B, C; struct Hashmap //哈希表代替map { , maxe = ; ], nxt[maxe + ], w[maxe…

<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…

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…

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") //…

BSGS BSGS裸题,嗯题目中也有提示:求a^m (mod p)的逆元可用快速幂,即 pow(a,P-m-1,P) * (a^m) = 1 (mod p) /************************************************************** Problem: 3239 User: Tunix Language: C++ Result: Accepted Time:208 ms Memory:2872 kb ***********************…

一道裸的BSGS题目(叫baby step, giant step) 从爱酱的blog里学来的,是一个很神的根号算法. 如果我们有hash的工具的话,就是O(sqrt(p))的,这里又用了一个map所以是O(sqrt(p) * log(sqrt(p))) /************************************************************** Problem: 3239 User: rausen Language: C++ Result: Accepted…

题目链接:http://poj.org/problem?id=2417 题目: 题意: 求一个最小的x满足a^x==b(mod p),p为质数. 思路: BSGS板子题,推荐一篇好的BSGS和扩展BSGS的讲解博客:http://blog.miskcoo.com/2015/05/discrete-logarithm-problem 代码实现如下: #include <set> #include <map> #include <queue> #include <st…

[吐槽] 这题和[bzoj]2480一毛一样. 就是输入顺序和输出变了一下. 传送门:http://www.cnblogs.com/chty/p/6043707.html…