转载自: http://blog.csdn.net/txwh0820/article/details/46392293 矩阵的迹求导法则 1. 复杂矩阵问题求导方法:可以从小到大,从scalar到vector再到matrix 2. x is a column vector, A is a matrix d(A∗x)/dx=A d(xT∗A)/dxT=A d(xT∗A)/dx=AT d(xT∗A∗x)/dx=xT(AT+A) 3. Practice: 4. 矩阵求导计算法则 求导公式(撇号为
cr:http://blog.csdn.net/txwh0820/article/details/46392293 一.矩阵的迹求导法则 1. 复杂矩阵问题求导方法:可以从小到大,从scalar到vector再到matrix 2. x is a column vector, A is a matrix d(A∗x)/dx=A d(xT∗A)/dxT=A d(xT∗A)/dx=AT d(xT∗A∗x)/dx=xT(AT+A) 3. Practice: 4. 矩阵求导计算法则 求导公式(撇号为
# coding:utf-8 from __future__ import absolute_import from __future__ import unicode_literals from __future__ import print_function from __future__ import division import tensorflow as tf x = tf.placeholder(dtype=tf.float32) y = tf.placeholder(dtype=
#欧几里得求最大公约数 #!/usr/bin/env python #coding -*- utf:8 -*- #iteration def gcd(a,b): if b==0: return a else: return gcd(b, remainder(a, b)) #此方法仅仅书用于a和b都为正数 def gcd_1(a,b): while(b>0): rem = remainder(a,b) a = b b = rem return a def remainder(x,y): retur
gcd(欧几里得算法辗转相除法): gcd ( a , b )= d : 即 d = gcd ( a , b ) = gcd ( b , a mod b ):以此式进行递归即可. 之前一直愚蠢地以为辗转相除法输进去时 a 要大于 b ,现在发现事实上如果 a 小于 b,那第一次就会先交换 a 与 b. #include<stdio.h> #define ll long long ll gcd(ll a,ll b){ ?a:gcd(b,a%b); } int main(){ ll a,b; wh
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1576 A/B Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4020 Accepted Submission(s): 3091 Problem Description 要求(A/B)%9973,但由于A很大,我们只给出n(n=A%99
https://vjudge.net/problem/UVA-11768 题意: 给定两个点A(x1,y1)和B(x2,y2),均为0.1的整数倍.统计选段AB穿过多少个整点. 思路: 做了这道题之后对于扩展欧几里得有了全面的了解. 根据两点式公式求出直线 ,那么ax+by=c 中的a.b.c都可以确定下来了. 接下来首先去计算出一组解(x0,y0),因为根据这一组解,你可以写出它的任意解,其中,K取任何整数. 需要注意的是,这个 a' 和 b' 是很重要的,比如说 b' ,它代表的是x每隔 b
欧几里得算法求最大公约数算法思想: 求p和q的最大公约数,如果q=0,最大公约数就是p:否则,p除以q余数为r,p和q的最大公约数即q和r的最大公约数. java实现代码: public class Demo0 { public static void main(String[] args) { System.out.println(gcd(24,120)); } public static int gcd(int p,int q){ if(q==0) return p; int r=p%q;
One Person Game Time Limit: 2 Seconds Memory Limit: 65536 KB There is an interesting and simple one person game. Suppose there is a number axis under your feet. You are at point A at first and your aim is point B. There are 6 kinds of operations
The modular modular multiplicative inverse of an integer a modulo m is an integer xsuch that a-1≡x (mod m). This is equivalent to ax≡1 (mod m). Input There are multiple test cases. The first line of input is an integer T ≈ 2000 indicating the number
