题意:矩阵快速幂取模 #include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;long long mod;long long p[]={1,1,1,0};void matrix(long long a[],long long b[],long long c[]
Description The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1≡x (mod m). This is equivalent toax≡1 (mod m). Input There are multiple test case
不考虑取模的快速幂运算(迭代法) /* Iterative Function to calculate (x^y) in O(logy) */int power(int x, unsigned int y){int res = 1; // Initialize resultwhile (y > 0){// If y is odd, multiply x with resultif (
计算模乘逆元原理上有四种方法: 1.暴力算法 2.扩展欧几里得算法 3.费尔马小定理 4.欧拉定理 模乘逆元定义:满足 ab≡1(mod m),称b为a模乘逆元。以下是有关概念以及四种方法及程序。 文章出处:Modular Multiplicative Inverse The modular multiplicative inverse of an integer a modu
The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence: F0 = 0 F1 = 1 Fi = Fi−1 + Fi−2 for i > 1 Write a program which calculates Mn = Fn mod 2^
原题: The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …) are defined by the recurrence: F 0 = 0 F 1 =1 F i = F i−1 + F i−2 for i > 1 Write a program which calculates M n = F n mod 2 m f
原文地址:http://colah.github.io/posts/2014-07-Conv-Nets-Modular/ Conv Nets: A Modular Perspective Posted on July 8, 2014 neural networks, deep learning, convolutional neural networks, modular neural ne
E. Modular Sequence 题意 对于一个长度为 n n n 的数组 a a a,定义它是 g o o d good good 的当且仅当: a 1 = x a_1 = x a1=x a i = a i − 1 + y a_{i} = a_{i - 1} + y ai=ai−1+y 或 a i = a i − 1 m o d y i ≥ 2 a_{i} = a
模逆元 定义 整数 a a a的模逆元是满足 a ⋅ x a\cdot x a⋅x模一个模数 m m m等于1。也就是找到一个数 x x x: a ⋅ x ≡ 1 mod m. a \cdot x \equiv 1 \text{ ~~~~mod m.} a⋅x≡1 mod m. 也可以把 x x x表示为 a − 1 a^{-1} a−1 需要注意模逆并不是总是存在。例如, m =
1. 引言 关于有限域的基础知识,可参考: RISC Zero团队2022年11月视频 Intro to Finite Fields: RISC Zero Study Club 有限域几乎是密码学中所有数学的基础。 ZKP证明系统中的所有运算都是基于有限域的: 使用布尔运算的数字电路:如AND、OR、NOT。使用有限域运算的算术电路:如addition、multiplication、ne