POJ 3641 Pseudoprime numbers

DIY一个SM2262ENG 2TB Nvme固态硬盘,慧荣SM2262EN主控 网友订做的 材料: 慧荣科技的62en,2片1G 三星DDR4缓存,4颗PF29F04T2AOCTH1,8CE B17A。SM2262EN属于慧荣第二代PCIe SSD主控,支持PCIe G3×4通道和全新的NVMe 1.3规范、8个NAND通道设计,并提供完整固件设计,最大循序读取速度高达3.5GB/s,循序写入速度达3.0GB/s,随机读写则高达420K IOPS和420K IOPS。此外,还采用目前最先进的低功耗设计,无论是在待机还是在工作状态下,均展现超低功耗,有效减少30%的功耗,满足用户追求超 阅读详情

Pseudoprime numbers
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11784 Accepted: 5053

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

Output

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

3 2
10 3
341 2
341 3
1105 2
1105 3
0 0

Sample Output

no
no
yes
no
yes
yes

Source


费马定理指出,对于任意的素数 p 和任意的整数 a > 1,满足 ap = a (mod p) 。也就是说,a 的 p 次幂除以 p 的余数等于 a 。p 的某些 (但不是很多) 非素数的值,被称之为以 a 为底的伪素数,对于某个 a 具有该特性。并且,某些 Carmichael 数,对于全部的 a 来说,是以 a 为底的伪素数。

给定 2 < p ≤ 1000000000 且 1 < a < p ,判断 p 是否为以 a 为底的伪素数。

输入

输入包含多个测试用例,以 "0 0" 表示输入结束。每个测试用例,由包含 p 和 a 的一行组成。

输出

对于每个测试用例,如果 p 是以 a 为底的伪素数,则输出 "yes",否则输出 "no" 。




http://poj.org/problem?id=3641


第一次做的时候用的是int 没有用long long 范围比较小,出错了
而且英文题目,还扯了费马小定理,百度了一下,但是也没用到
写两个函数就行了,快速幂,素数判断
/*
	题意:
	判断P是否为素数,是即输出no
	不是就计算a的p次方是否等于a,是就输出yes,否则输出no;
*/


/*
	这个题要求 求出若p不是素数并且p^a是否等于a,若等于则输出yes,否则输出no

*/

#include<stdio.h>
#include<string.h>
#define LL long long
LL int prime(LL int a)
{
	int z=1;
	if(a<2)	return 0;
	for(int i=2;i*i<=a;i++)
	{
		if(a%i==0)
		{
			z=0;
			break;
		}
	}
	return z;
}
LL int pow_mod(LL int a,LL int b,LL int c)
{
	LL int ans=1;
	while(b)
	{
		if(b&1)	ans=ans*a%c;
		a=a*a%c;
		b>>=1;
	}
	return ans;	
}
int main()
{
	LL int p,a;
	while(scanf("%lld %lld",&p,&a)!=EOF)
	{
		if(p==0&&a==0)	break;
		if(prime(p))	printf("no\n");
		else if(pow_mod(a,p,p)==a)	printf("yes\n");
		else printf("no\n");
		
	 } 
	return 0;
}
现在看看原来写的代码发现好啰嗦啊。。。
现在的代码:

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
	return !b?a:gcd(b,a%b);
}
ll pow_mod(ll a,ll b,ll c)//快速幂 
{
	ll ans=1;
	a=a%c; 
	while(b)
	{
		if(b&1)	ans=ans*a%c;
		a=a*a%c;
		b>>=1;
	}
	return ans;
}
ll prime(ll n)//判断素数 
{
	for(int i=2;i<=sqrt(n);i++)
		if(n%i==0)	return 0;
	return n>1;	
}
int main()
{
	ll p,a;
	while(~scanf("%lld %lld",&p,&a)&&(p|a))
	{
		if(!prime(p)&&pow_mod(a,p,p)==a)	printf("yes\n");
		else	printf("no\n");
	 } 
	return 0;
}


平头哥面试——芯片工程师面经 有5个成递增关系的32bit地址值,彼此之间的间隔大小不确定,可以把他们当成parameter来输入模块,现有一个32bit输入数据,要求用尽量少资源去的判断这个数处于哪2个地址之间(地址区间),并且输出区间号。a)后续大概会在9月下旬谈具体的薪资,他们现在也没想好今年应届生的薪资该给多少,到时候如果可以签三方的话可能会跳过意向书环节直接发offer,谈薪资肯定是通过电话来。b. 这边的CPU是自己研制的(玄铁),成都这边主要做存储和网络,都有SOC。a. 成都这边的网络部门具体做的业务有哪些? 阅读详情

相关推荐

Windows11安装Ubuntu 20.04.3 LTS双系统(详细过程)

一、基础环境 听说BIOS模式的传统MBR模式和新式UEFI模式,会对安装双系统产生影响。Windows11应该都是UEFI吧,可以按win+r输入msinfo32回车确认查看BIOS模式,我的电脑是UEFI+双硬盘 Ubuntu安装准备 1、 官网下载Ubuntu 20.04.3 LTS镜像iso文件:https://ubuntu.com/download/desktop 2、 下载官网推荐的U盘启动制作工具rufus: https://rufus.ie/ 3、制作启动u盘 插入U盘, 运行 rufu

学习随笔 9万+

Pseudoprime numbers(快速幂 + 素数)

Pseudoprime numbers 题意 费马小定理指出,对于任何素数p和任何整数a>1,ap ≡a (mod p)。也就是说,如果a的p次幂除以p,余数是a。一些(但不是很多)非素数值p,如果存在整数a>1使得ap = a (mod p),则称p是以a为基的伪素数。(还有一些数被称为Carmichael数,对于所有满足GCD(a, n)=1的正整数a,都是以a为基的伪素数。) 给...

百般回首曾经的相遇,留下的只是一缕残忆 1427

四款办公Agent实测对比:WorkBuddy、千问办公、库库AI、豆包工作怎么选

AI Agent是一种能感知环境、自主决策并执行动作的智能系统,其核心在于工具调用、状态管理与错误恢复构成的执行闭环。在办公场景中,任务边界清晰、容错率高、重复性强,使Agent成为最容易落地的方向。通过调用本地文件、脚本或办公套件,Agent能自动完成文件批量处理、文档摘要、会议纪要等重复劳动,显著提升工作效率。目前WorkBuddy、千问办公、库库AI、豆包工作等产品分别从本地操作、办公套件嵌入、知识处理、轻量助手等路线切入,用户可根据自身场景选择。

weixin_29052717的博客 523

Pseudoprime numbers (快速幂)

Fermat's theorem states that for any prime numberpand for any integera> 1,ap=a(modp). That is, if we raiseato thepthpower and divide byp, the remainder isa. Some (but not very many) ...

Lbo_tao的博客 304

使用kmplot在线进行生存分析

欢迎关注”生信修炼手册”!kmplot是一个生存分析的在线工具,最初是设计用来对肝癌中的miRNA进行生存分析,对应的文章发表在scientific reports上,链接如下https...

庐州月光的博客 1万+

群晖NAS从入门到精通的所有帖子汇总,只要这一篇就够了

本站陆陆续续的转载、原创帖子也有十几二十篇了,从不同的角度对群晖NAS的各种骚操作都有深入的讲解,虽然很多文章都不是讲的很深入,但是针对小白入门已经是足够了,希望对小白来说是一个入门,群晖NAS的大佬们勿喷,哪里写的不对的,请随时留言西安SEO优化站长一定在第一时间纠正修改。

西安SEO优化网站资源 8万+

Pseudoprime numbers(快速幂模+素性测试)

Pseudoprime numbers Fermat's theorem states that for any prime number p and for any integer a &gt; 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some...

codeswarrior的博客 383

Windows上搭建OpenHarmony星闪开发环境:HiSpark Studio全流程指南

嵌入式开发中,环境搭建是项目启动的关键步骤。OpenHarmony作为开源操作系统,其构建体系依赖Python脚本和GN/Ninja工具链,对版本兼容性要求严格。星闪(NearLink)技术作为新一代短距无线通信方案,在时延、并发和抗干扰方面优势显著,广泛应用于智能家居、工业控制等场景。在Windows平台上,通过HiSpark Studio集成开发工具,开发者可高效完成代码编辑、编译、烧录与调试。然而,Python版本冲突、环境变量配置、驱动签名等问题常导致搭建失败。本文详细拆解从系统准备到编译验证的全流

weixin_32658257的博客 344

伪素数判断----Pseudoprime numbers

伪素数定义:伪素数,又叫做伪质数:它满足费马小定理,但其本身却不是素数。最小的伪素数是341。有人已经证明了伪素数的个数是无穷的。事实上,费马小定理给出的是关于素数判定的必要非充分条件。若n能整除2^(n-1)-1,并n是非偶数的合数,那么n就是伪素数。第一个伪素数341 是萨鲁(Sarrus)在1819年发现的。 简单解释一下伪素数:一个数字p本身不是素数但满足费马小定理,即a的p次方模p与a模...

柠檬的博客~~~ 7252

【RT-DETR创新改进】改进前必看!对Ultralytics调参, 1:1还原RT-DETR官方版本实验参数,使用原汁原味的RT-DETR模型,才能高效跑实验、高效涨点、收割小论文、完成大论文!!!

本文针对Ultralytics训练RT-DETR精度不佳问题,提供参数配置优化方案。通过对比官方RT-DETR参数设置,详细指导如何修改Ultralytics版本的关键参数,包括优化器、学习率、权重衰减等6处配置调整。操作步骤涵盖default.yaml文件修改和trainer.py代码调整,确保模型收敛性能达到官方水平,帮助研究者快速提升实验效果。文末附官方源码链接和具体参数配置网址供参考。

Ai缝合怪、助力小伙伴高效发论文 1780

EMB电子机械制动执行器五大核心部件选型解析

本文介绍了EMB电子机械制动执行器五大核心部件的选型方案:采用无刷直流电机作为驱动源,兼具高扭矩和快速响应特性;选用滚珠丝杠传动机构,实现高效旋转-直线转换;行星齿轮减速器满足紧凑空间内的增扭需求;QT450-QT550材质的钳体确保刚度要求;应变片式压力传感器实现夹紧力精确测量。各部件选型综合考虑了性能、可靠性和空间限制,为EMB系统设计提供了完整解决方案。

m0_59012156的博客 1737

POJ 3641: 伪素数

POJ 3641: Pseudoprime numbers

十飏 de FATMOUSE. 379

Open3d (C++) RANSAC拟合圆柱(详细过程版)

实现了一种基于RANSAC算法的圆柱拟合方法。

点云侠的博客 240

【 POJ 3641 】Pseudoprime numbers

Pseudoprime numbersTime Limit: 1000MS Memory Limit: 65536K Total Submissions: 8702 Accepted: 3658 DescriptionFermat’s theorem states that for any prime number p and for any integer a > 1, ap

Nodies 449

IEM模型,地表粗糙度模拟

利用微波反演土壤水分时候,利用IEM模型可以实现对随机地表的模拟

android 蓝牙通讯demo

android 下蓝牙通讯demo,实现了发现蓝牙,使能蓝牙,通过蓝牙发送消息。

poj-3641Pseudoprime numbers

poj-3641Pseudoprime numbers Pseudoprime numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10836 Accepted: 4686 Description Fermat’s theorem states that for any prim

weixin_40433319的博客 261

jre9下载 jdk9

(Java SE Runtime Environment 9.0) 9.0u175 官方最新版 【x86|x64】

poj 3641 Pseudoprime numbers 【快速幂】

Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6645   Accepted: 2697 Description Fermat's theorem states that for any prime number p a

shengweisong 2057

深度学习工具包 DeepLearningToolbox

深度学习工具包Deprecation notice.-----This toolbox is outdated and no longer maintained.There are much better tools available for deep learning than this toolbox, e.g. [Theano](http://deeplearning.net/software/theano/), [torch](http://torch.ch/) or [tensorflow](http://www.tensorflow.org/)I would suggest you use one of the tools mentioned above rather than use this toolbox.Best, Rasmus.DeepLearnToolbox================A Matlab toolbox for Deep Learning.Deep Learning is a new subfield of machine learning that focuses on learning deep hierarchical models of data.It is inspired by the human brain's apparent deep (layered, hierarchical) architecture.A good overview of the theory of Deep Learning theory is[Learning Deep Architectures for AI](http://www.iro.umontreal.ca/~bengioy/papers/ftml_book.pdf)For a more informal introduction, see the following videos by Geoffrey Hinton and Andrew Ng.* [The Next Generation of Neural Networks](http://www.youtube.com/watch?v=AyzOUbkUf3M) (Hinton, 2007)* [Recent Developments in Deep Learning](http://www.youtube.com/watch?v=VdIURAu1-aU) (Hinton, 2010)* [Unsupervised Feature Learning and Deep Learning](http://www.youtube.com/watch?v=ZmNOAtZIgIk) (Ng, 2011)If you use this toolbox in your research please cite [Prediction as a candidate for learning deep hierarchical models of data](http://www2.imm.dtu.dk/pubdb/views/publication_details.php?id=6284)```@MASTERSTHESIS\{IMM2012-06284, author = "R. B. Palm", title = "Prediction as a candidate for learning deep hierarchical models of data", year = "2012",}```Contact: rasmusbergpalm at gmail dot comDirectories included in the toolbox-----------------------------------`NN/` - A library for Feedforward Backpropagation Neural Networks`CNN/` - A library for Convolutional Neural Networks`DBN/` - A library for Deep Belief Networks`SAE/` - A library for Stacked Auto-Encoders`CAE/` - A library for Convolutional Auto-Encoders`util/` - Utility functions used by the libraries`data/` - Data used by the examples`tests/` - unit tests to verify toolbox is workingFor references on each library check REFS.mdSetup-----1. Download.2. addpath(genpath('DeepLearnToolbox'));Example: Deep Belief Network---------------------```matlabfunction test_example_DBNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit RBM and visualize its weightsrand('state',0)dbn.sizes = [100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);figure; visualize(dbn.rbm{1}.W'); % Visualize the RBM weights%% ex2 train a 100-100 hidden unit DBN and use its weights to initialize a NNrand('state',0)%train dbndbn.sizes = [100 100];opts.numepochs = 1;opts.batchsize = 100;opts.momentum = 0;opts.alpha = 1;dbn = dbnsetup(dbn, train_x, opts);dbn = dbntrain(dbn, train_x, opts);%unfold dbn to nnnn = dbnunfoldtonn(dbn, 10);nn.activation_function = 'sigm';%train nnopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.10, 'Too big error');```Example: Stacked Auto-Encoders---------------------```matlabfunction test_example_SAEload mnist_uint8;train_x = double(train_x)/255;test_x = double(test_x)/255;train_y = double(train_y);test_y = double(test_y);%% ex1 train a 100 hidden unit SDAE and use it to initialize a FFNN% Setup and train a stacked denoising autoencoder (SDAE)rand('state',0)sae = saesetup([784 100]);sae.ae{1}.activation_function = 'sigm';sae.ae{1}.learningRate = 1;sae.ae{1}.inputZeroMaskedFraction = 0.5;opts.numepochs = 1;opts.batchsize = 100;sae = saetrain(sae, train_x, opts);visualize(sae.ae{1}.W{1}(:,2:end)')% Use the SDAE to initialize a FFNNnn = nnsetup([784 100 10]);nn.activation_function = 'sigm';nn.learningRate = 1;nn.W{1} = sae.ae{1}.W{1};% Train the FFNNopts.numepochs = 1;opts.batchsize = 100;nn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.16, 'Too big error');```Example: Convolutional Neural Nets---------------------```matlabfunction test_example_CNNload mnist_uint8;train_x = double(reshape(train_x',28,28,60000))/255;test_x = double(reshape(test_x',28,28,10000))/255;train_y = double(train_y');test_y = double(test_y');%% ex1 Train a 6c-2s-12c-2s Convolutional neural network %will run 1 epoch in about 200 second and get around 11% error. %With 100 epochs you'll get around 1.2% errorrand('state',0)cnn.layers = { struct('type', 'i') %input layer struct('type', 'c', 'outputmaps', 6, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %sub sampling layer struct('type', 'c', 'outputmaps', 12, 'kernelsize', 5) %convolution layer struct('type', 's', 'scale', 2) %subsampling layer};cnn = cnnsetup(cnn, train_x, train_y);opts.alpha = 1;opts.batchsize = 50;opts.numepochs = 1;cnn = cnntrain(cnn, train_x, train_y, opts);[er, bad] = cnntest(cnn, test_x, test_y);%plot mean squared errorfigure; plot(cnn.rL);assert(er<0.12, 'Too big error');```Example: Neural Networks---------------------```matlabfunction test_example_NNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);% normalize[train_x, mu, sigma] = zscore(train_x);test_x = normalize(test_x, mu, sigma);%% ex1 vanilla neural netrand('state',0)nn = nnsetup([784 100 10]);opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samples[nn, L] = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.08, 'Too big error');%% ex2 neural net with L2 weight decayrand('state',0)nn = nnsetup([784 100 10]);nn.weightPenaltyL2 = 1e-4; % L2 weight decayopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex3 neural net with dropoutrand('state',0)nn = nnsetup([784 100 10]);nn.dropoutFraction = 0.5; % Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

POJ-3641 Pseudoprime numbers

题意:对于一个非素数p和任意一个数a,若a^p=a(mod p),则输出yes,否则输出no。 思路:判断素数+二分幂。 题目链接:http://poj.org/problem?id=3641 View Code 1 #include <cstdio> 2 #include <cmath> 3 #include <cstdlib...

bancaidun1539的博客 117

poj 3641 Pseudoprime numbers

题目连接 http://poj.org/problem?id=3641 Pseudoprime numbers Description Fermat's theorem states that for any prime numberpand for any integera> 1,ap=a(modp). That is, if we raisea...

dong12276的博客 147

POJ3641 Pseudoprime numbers

题目大意:就是如果P不是素数,能否满足ap =a (modp) 思路:直接素数判断+快速幂取模即可,我开始用的是普通素数判断,后来用了miller_rabin改进版的的素数测试,权当模板吧   AC program:(普通素数判断+快速幂) #include #include #include #include #include using namespace std; int p,a

muxi@Achilles的专栏 885

poj-3641(Pseudoprime numbers)

学习快速幂之后做的第一道题,这里记录一下。 题目大意:给出两个数 a 和 p,如果 p 是一个素数,判断是否满足公式 a^p %p=a,如果满足,则输出yes,否则,输出no。附链接:http://poj.org/problem?id=3641。 大体思路:题目简单,判断素数,然后判断是否满足公式。 以下是ac代码: #include using namespace std; long l

blackturtle的博客 427

POJ 3641 Pseudoprime numbers(快速幂)

嗯... 题目链接:http://poj.org/problem?id=3641 AC代码: 1 #include<cstdio> 2 #include<iostream> 3 4 using namespace std; 5 6 inline bool is_prime(int x){ 7 if(x...

weixin_30408739的博客 334

POJ 3641-Pseudoprime numbers

题目: 题意:

Halo 687

POJ 3641 Pseudoprime numbers

Fermat’s theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) ...

起名好难呀! 188

POJ 3641.Pseudoprime numbers

数论方向的题目,最近开始做做这方面的简单题或者是模板题 题目大意: 根据费马定理,如果p是一个素数,则对任意数字a,有,现在,题目要求找出p[不是素数]时,满足此式的算式 题目链接 其中运用到了快速幂取模的操作(由于数字太过大),这大概也是第一次解决带有mod类型的题,虽然是个超级水题 = = 这种快速幂取模的方法是纯模拟方法,网上其他代码,例如用&判断奇偶的代码质量比我这个好很...

CuriousLiu的博客 333
下一篇: HDU1222 Wolf and Rabbit
陶鸿杰
博客等级 码龄9年 93粉丝 298原创
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付:10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值