<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Gig_Face的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Gig_Face</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Gig_Face]]></copyright><item><title><![CDATA[pytorch中的batch_norm, instance_norm, layer_norm, group_norm]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/119644720</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/119644720</guid><author>Gig_Face</author><pubDate>Thu, 12 Aug 2021 16:46:28 +0800</pubDate><description><![CDATA[1 BatchNorm

1.1 实现

Batch Normalization: Accelerating Deep Network Training by Reducing Internal Covariate Shift

http://d2l.ai/chapter_convolutional-modern/batch-norm.html?highlight=batchnorm2d










    import torch
    import torch.nn.functi...]]></description><category></category></item><item><title><![CDATA[pytorch数据加载遇到的坑]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/114737311</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/114737311</guid><author>Gig_Face</author><pubDate>Sat, 13 Mar 2021 13:19:04 +0800</pubDate><description><![CDATA[问题

pytorch 版本1.6

最近在处理scRNA-seq数据，发现数据读取贼慢，dataset，__getitem__函数如下：


    def __getitem__(self, index):
        return {"x":self.adata.X[index],
                "count":self.adata.layers["count"][index],
                "label":self.adata.obs["label"][in]]></description><category></category></item><item><title><![CDATA[binary/categorical crossentropy实现]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/99468198</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/99468198</guid><author>Gig_Face</author><pubDate>Tue, 13 Aug 2019 20:23:11 +0800</pubDate><description><![CDATA[import tensorflow as tf
from tensorflow import keras

print(tf.__version__)
# 2.0.0-beta0

def cross_entropy(x, y):
    return tf.reduce_sum(-y*tf.math.log(x), axis=-1)

def softmax_cross_entropy_wit...]]></description><category></category></item><item><title><![CDATA[使用python实现ACO解决ATT48]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/89441234</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/89441234</guid><author>Gig_Face</author><pubDate>Sun, 21 Apr 2019 22:30:34 +0800</pubDate><description><![CDATA[关键词：python、ACO(蚁群算法)、TSP(旅行商问题)、ATT48

参考文献：M. Dorigo, V. Maniezzo, and A. Colorni. The Ant System: An autocatalytic optimizing process. Technical Report 91-016 Revised, Dipartimento di Elettronica,Po...]]></description><category></category></item><item><title><![CDATA[连续手语翻译（S2T模型部署）]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/89190203</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/89190203</guid><author>Gig_Face</author><pubDate>Wed, 10 Apr 2019 21:52:44 +0800</pubDate><description><![CDATA[作为连续手语翻译基准实验模型

Neural Sign Language Translation

Camgöz, Necati Cihan, Hadfield S , Koller O , et al. Neural Sign Language Translation[C]// IEEE Conference on Computer Vision &amp; Pattern Recognitio...]]></description><category></category></item><item><title><![CDATA[树的直径]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50352265</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50352265</guid><author>Gig_Face</author><pubDate>Fri, 18 Dec 2015 15:46:28 +0800</pubDate><description><![CDATA[http://www.cnblogs.com/wuyiqi/archive/2012/04/08/2437424.html]]></description><category></category></item><item><title><![CDATA[最小逆序数]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50352035</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50352035</guid><author>Gig_Face</author><pubDate>Fri, 18 Dec 2015 15:23:48 +0800</pubDate><description><![CDATA[问题描述：有一个长度为 n 个正整数序列{a0，a1，…，an-1}。定义一次操作为：交换序列中任意两个相邻的元素。求不超过 k 次操作后，这个序列的最小的逆序数。
#include
#include
#define minin(a,b)  a<b?a:b
#define less(a,b)	a<=b
#define item int
struct bag
{
	item*a;
	item*b;]]></description><category></category></item><item><title><![CDATA[单调队列]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50351856</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50351856</guid><author>Gig_Face</author><pubDate>Fri, 18 Dec 2015 15:03:54 +0800</pubDate><description><![CDATA[问题描述：给你一个大小为N的数组，再给你一个K值，要你计算[1,K]，[2,K+1],[3,K+2]，……，[N-K+1,N]这几段区间的最大值的和并输出。
#include
#include
struct queue
{
	int data[100005];
	int lp;
	int rp;
	queue(){memset(data,0,sizeof(data));lp=rp=0;}
	vo]]></description><category></category></item><item><title><![CDATA[冬天来了]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50351694</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50351694</guid><author>Gig_Face</author><pubDate>Fri, 18 Dec 2015 14:46:37 +0800</pubDate><description><![CDATA[问题描述：

某个小村庄里面有一个农民小明，他一开始一天可以生产L 个玉米，在接下来的每天里，他可以选择让自己的生产能力+1（当天不生产），或者生产L 个玉米，但是距离冬天到来只有N 天了，他必须在冬天到来之前储备尽可能多的玉米过冬，可是每天都会有地主来收玉米，如果不能交出足够的玉米，地主会没收小明的土地，于是小明请求你帮忙计算他最后最多可以留下多少玉米？如果在中途不能满足地主要求，就输出“D]]></description><category></category></item><item><title><![CDATA[后缀表达式]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50351482</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50351482</guid><author>Gig_Face</author><pubDate>Fri, 18 Dec 2015 14:19:28 +0800</pubDate><description><![CDATA[将一个只包含 +,  -,  *,  / ， %，（，）的非负整数计算表达式的 中缀表达式 转化为 后缀表达式
#include       
#include   
#include             
std::stack opr;      
int order(char c) {return (c == '+' || c == '-') ? 0 : 1;}    
int main]]></description><category></category></item><item><title><![CDATA[单调栈]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50351466</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50351466</guid><author>Gig_Face</author><pubDate>Fri, 18 Dec 2015 14:14:49 +0800</pubDate><description><![CDATA[题意：给你一段区间，需要你求出（在这段区间之类的最小值*这段区间所有元素之和）的最大值



#include
#define maxn 100010
struct node
{
	int sum,num; //sum为每个元素两边延伸的和，num为该元素的值 
}a[maxn];

__int64 MAX(__int64 a,__int64 b){return a>b?a:b;}

in]]></description><category></category></item><item><title><![CDATA[再见——电脑鼠 Atmega128]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50342095</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50342095</guid><author>Gig_Face</author><pubDate>Thu, 17 Dec 2015 15:35:05 +0800</pubDate><description><![CDATA[学了一年的电脑鼠基本收获不多，也不能说少，至少看了眼界。
挺羡慕之后参见的同学可以有只完好的Micro Mouse,而不用担心电机坏了，红外坏了。
电脑鼠报废主要在红外吧，只能直走，不能避障。庆幸结课时要求不高，能顺利转几圈。
总之，作为软件部分：
快速PWM基本会用；
UART正常；
PID算法，P算法凑数吧；
红外发射接受会写，但硬件不能用；
遗憾；


代码留作纪念；]]></description><category></category></item><item><title><![CDATA[链表一]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50221727</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50221727</guid><author>Gig_Face</author><pubDate>Tue, 08 Dec 2015 17:31:56 +0800</pubDate><description><![CDATA[问题：若干只猪排成一队，从2开始编号。每次排在最前面的为幸运猪（假设编号是i）,
幸运猪及它后面的第i只，2*只，3*i只...的小猪出列，然后重复上述过程。
把幸运猪的编号从小到大排序，现在问第n个幸运猪的编号是多少？ 0


#include
#include
#define N 33810
int next[N];
int a[3001];
int main()
{
	int i,]]></description><category></category></item><item><title><![CDATA[异或三（彼岸花）]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50221659</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50221659</guid><author>Gig_Face</author><pubDate>Tue, 08 Dec 2015 17:26:23 +0800</pubDate><description><![CDATA[问题：
有一排彼岸花，其特点是：
1.雄花会向相邻的花授粉,雄花被一朵花授粉则变为雌花，被两朵或不被授粉，则保持不变；
2.雌花被一朵花授粉则变为雄花，被两朵或不被授粉，则保持不变；
现要想知道X年后每朵花的性别。
（用1表示雄花，0表示雌花）
#include   
#include 
#define N 1005
int a[2*N];
int main()   
{   
   ]]></description><category></category></item><item><title><![CDATA[位运算]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50221589</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50221589</guid><author>Gig_Face</author><pubDate>Tue, 08 Dec 2015 17:19:27 +0800</pubDate><description><![CDATA[问题： 输入一个32位无符号整数，将该数对应的二进制位翻转，输出翻转后的二进制数对应的十进制值。
#include
unsigned int bit[32]; 
int main()
{
	unsigned int ans;int i,k,x,n;
	bit[0]=1;
	for(i=1;i<32;i++)
		bit[i]=bit[i-1]<<1;
	scanf("%d",&n);
	whil]]></description><category></category></item><item><title><![CDATA[异或二]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50221165</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50221165</guid><author>Gig_Face</author><pubDate>Tue, 08 Dec 2015 16:59:14 +0800</pubDate><description><![CDATA[问题：给定一组数组（数组元素范围为0
#include
int arra[3000010];
int  Findbit(int flag) 
{
	int key=1;
	while(flag&key==0)
		key<<=1;
	return key;
}
int main()  
{
	int N,flag1=0,key,i,num1=0,num2=0,index;
	scanf ("%d]]></description><category></category></item><item><title><![CDATA[矩阵快速幂]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50220977</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50220977</guid><author>Gig_Face</author><pubDate>Tue, 08 Dec 2015 16:51:36 +0800</pubDate><description><![CDATA[问 题 简 述:
求Fibonacci数列第n项 F[n] % 10000。其中: n ≤ 10086110    询问次数 Q ≤ 10  
Hint:
Fibonacci数列定义：
F[0] = 0
F[1] = 1
F[n] = F[n − 1] + F[n − 2] （ n ≥ 2）



#include 
int A[3], B[3], temp[3], n, q;]]></description><category></category></item><item><title><![CDATA[异或一]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/50220737</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/50220737</guid><author>Gig_Face</author><pubDate>Tue, 08 Dec 2015 16:39:29 +0800</pubDate><description><![CDATA[题目：给定一个数组，除了一个数只出现一次以外，其他数都出现了两次。
  目的就是找出只出现一次的数。#include
int main() 
{
    int n,c,i,c1=0;
    scanf( "%d",&n);
    for( i=0;i<n;i++) 
    {
        scanf("%d",&c);
        c1^=c;
    }
    printf(]]></description><category></category></item><item><title><![CDATA[车皮编序问题]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/48394471</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/48394471</guid><author>Gig_Face</author><pubDate>Sat, 12 Sep 2015 15:44:00 +0800</pubDate><description><![CDATA[算法与数据结构：算法实验3.1   车皮编序问题]]></description><category></category></item><item><title><![CDATA[1028. D进制的A+B (20)]]></title><link>https://blog.youkuaiyun.com/Gig_Face/article/details/46717435</link><guid>https://blog.youkuaiyun.com/Gig_Face/article/details/46717435</guid><author>Gig_Face</author><pubDate>Wed, 01 Jul 2015 23:59:45 +0800</pubDate><description><![CDATA[输入两个非负10进制整数A和B(30-1)，输出A+B的D (1 
输入格式： 
输入在一行中依次给出3个整数A、B和D。 
输出格式： 
输出A+B的D进制数。 
输入样例：
123 456 8
输出样例：
1103






时间限制 
1000 ms



内存限制 
65536 kB



代码长度限制 
8192 B



判]]></description><category></category></item></channel></rss>