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  • Blend One Zero/Zero One

    2020-07-23 11:30:56
    这里是Blend Zero One,而有的Shader是One Zero。这里记录一下他俩的区别 首先是旁边TA巨佬给我的Paper中写道: 还有国外巨佬Bgolus的解释: 这里其实已经解释的很清楚了,我在这里简单翻译一下: Csrc是当前正在...

    最近看Shader的时候会注意到一句话:
    在这里插入图片描述
    这里是Blend Zero One,而有的Shader是One Zero。这里记录一下他俩的区别
    首先是旁边TA巨佬给我的Paper中写道:
    在这里插入图片描述
    还有国外巨佬Bgolus的解释:
    在这里插入图片描述

    这里其实已经解释的很清楚了,我在这里简单翻译一下:
    Csrc是当前正在Rasterizing的某个Pixel经过计算输出的Color,相当于Shader Output
    Cdst是当前位于Fream Buffer中的颜色,也就是该Pixel之前累加的Color。
    如果不进行Blend的话,当Csrc通过Ztest&StencilTest后,Csrc将会覆盖掉Cdst,成为当前Pixel的新颜色。
    但是这里如果开启Blend,就需要进行颜色混合:在Direct3D中,用公式:在这里插入图片描述
    来进行混合。
    Fsrc和Fdst即Blend输入的两个参数。
    如果参数为One Zero:
    那么相当于 Csrc * 1+Cdst * 0,也就是说完全使用当前新绘制的Color。
    反之则不绘制,使用Fream Buffer中的Color。

    如上操作和Alpha通道没有关系,因为Alpha需要另外计算,但公式也是类似的:
    在这里插入图片描述
    这里不展开讲述,以后遇到再细讲。

    展开全文
  • Zero to One(English)

    热门讨论 2015-02-04 03:37:56
    This is a good English book for people in business and management
  • Zero to One_双语

    2015-04-24 16:38:41
    双语。 源于彼得·蒂尔2012 年在斯坦福大学所教授的一门创业课程。期间,他的一个学生布莱克· 马斯特斯记下了详细的笔记,并把它发布到网络上,引来...随后,彼得·蒂尔参与将这份神奇的笔记精编成为《从0到1》一书。
  • Zero to One

    2015-04-07 16:27:43
    从Peter Thiel的课程笔记整理出来的书。
  • 小菜鸡刷CTF(二)

    2019-07-19 14:46:58
    ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ...

    高手进阶区(1)

    接下来就是攻防世界的高手进阶区了,这里的题目讲道理对于我这样的新手还是挺难的,尽力而为吧~(tcl)

    题目1:你猜猜
    题目来源: ISCC-2017
    题目描述:我们刚刚拦截了,敌军的文件传输获取一份机密文件,请君速速破解。
    题目附件:
    504B03040A0001080000626D0A49F4B5091F1E0000001200000008000000666C61672E7478746C9F170D35D0A45826A03E161FB96870EDDFC7C89A11862F9199B4CD78E7504B01023F000A0001080000626D0A49F4B5091F1E00000012000000080024000000000000002000000000000000666C61672E7478740A0020000000000001001800AF150210CAF2D1015CAEAA05CAF2D1015CAEAA05CAF2D101504B050600000000010001005A000000440000000000

    解题:
    题目中的提示性语句就是“一份机密文件”,可以想到也许可以从文件的格式入手。504B0304是zip文件的文件头,因此我们可以使用winhex将其以zip文件的格式保存。
    在这里插入图片描述
    保存后解压时会有一个密码,试一下123456成功了,得到结果:daczcasdqwdcsdzasd
    ————————————————————————————————————
    题目2:enc
    题目来源: 暂无
    题目描述:Fady不是很理解加密与编码的区别 所以他使用编码而不是加密,给他的朋友传了一些秘密的消息。
    题目附件:

    ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO
    ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE
    ONE ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE
    ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ONE
    ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE
    ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE
    ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO
    ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO
    ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
    ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
    ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE
    ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE
    ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO
    ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ONE
    ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO
    ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE
    ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO
    ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO
    ONE ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ZERO ZERO
    ONE ONE ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE
    ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE
    ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE
    ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO
    ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE
    ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ONE ZERO
    ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE
    ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO
    ZERO ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO
    ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE
    ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE
    ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO
    ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE
    ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE
    ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ONE ZERO ONE
    ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO
    ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE
    ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO
    ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE
    ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO
    ONE ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE
    ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
    ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ONE
    ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE
    ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE
    ZERO ZERO ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO
    ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO
    ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE
    ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO
    ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE
    ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE
    ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE
    ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO
    ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE
    ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO
    ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE
    ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO
    ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO
    ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
    ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
    ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE
    ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE
    ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO
    ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ONE
    ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO
    ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE
    ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE
    ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO
    ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO
    ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO
    ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE
    ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE
    ONE ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO
    ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
    ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO
    ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO
    ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO
    ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO
    ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ONE ZERO
    ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE
    ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO
    ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO
    ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE
    ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE
    ZERO ZERO ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO
    ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ONE
    ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO
    ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE
    ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO
    ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO
    ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
    ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
    ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE
    ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE
    ONE ONE ZERO ONE ZERO ZERO ONE ONE ONE ONE ZERO ONE

    解题:
    整个文件里只有zero和one,显然是01编码,不禁感叹他的这位盆友为什么要用这么**的方法将他们替换成英文。将其替换成01编码后,我们可以想到常用的编码方法就是8位二进制编码可以共同编码一个字符,因此我们将它们8位分为一组,找到对应的字符,结果如下:
    Li0gLi0uLiAuIC0uLi0gLS4tLiAtIC4uLS4gLSAuLi4uIC4tLS0tIC4uLi4uIC0tLSAuLS0tLSAuLi4gLS0tIC4uLi4uIC4uLSAuLS0uIC4uLi0tIC4tLiAtLS0gLi4uLi4gLiAtLi0uIC4tLiAuLi4tLSAtIC0tLSAtIC0uLi0gLQ==
    好明显的base64编码~结果为:
    .- .-… . -…- -.-. - …-. - … .---- … — .---- … — … …- .–. …-- .-. — … . -.-. .-. …-- - — - -…- -
    又又又是莫斯密码,解码后就可以得到结果了:ALEXCTFTH15O1SO5UP3RO5ECR3TOTXT

    import base64
    #import morse_talk as mtalk
    with open('zero_one', 'r') as f:
        data = f.read()
    data = data.replace("ZERO","0").replace("ONE","1").replace(' ','').replace('\n','')
    word=''
    for i in range(0, len(data), 8):
        word+=(chr(int(data[i:i+8], 2)))
    word=base64.b64decode(word).decode(encoding='UTF-8')
    s = word.split(" ")
    print(s)
    dict = {'.-': 'A',
            '-...': 'B',
            '-.-.': 'C',
            '-..':'D',
            '.':'E',
            '..-.':'F',
            '--.': 'G',
            '....': 'H',
            '..': 'I',
            '.---':'J',
            '-.-': 'K',
            '.-..': 'L',
            '--': 'M',
            '-.': 'N',
            '---': 'O',
            '.--.': 'P',
            '--.-': 'Q',
            '.-.': 'R',
            '...': 'S',
            '-': 'T',
            '..-': 'U',
            '...-': 'V',
            '.--': 'W',
            '-..-': 'X',
            '-.--': 'Y',
            '--..': 'Z',
            '.----': '1',
            '..---': '2',
            '...--': '3',
            '....-': '4',
            '.....': '5',
            '-....': '6',
            '--...': '7',
            '---..': '8',
            '----.': '9',
            '-----': '0',
            '..--..': '?',
            '-..-.': '/',
            '-.--.-': '()',
            '-....-': '-',
            '.-.-.-': '.'
            }
    for item in s:
        print (dict[item],end='')
    

    ————————————————————————————————————

    题目3:告诉你个秘密
    题目来源: ISCC-2017
    题目描述:暂无
    题目附件:
    636A56355279427363446C4A49454A7154534230526D6843
    56445A31614342354E326C4B4946467A5769426961453067

    解题:
    给出的题目是两行16位数,所以就试着用ASCII码转换了一下,得到了cjV5RyBscDlJIEJqTSB0RmhCVDZ1aCB5N2lKIFFzWiBiaE0g
    然后用base64解码,得到了r5yG lp9I BjM tFhB T6uh y7iJ QsZ bhM
    这神奇的密码想了我好久,最后竟然是!!键盘!!
    每4个键包围了一个字符,最终结果是:TONGYUAN
    ————————————————————————————————————
    题目4:Easy-one
    题目来源: Hack-you-2014
    题目描述:破解密文,解密msg002.enc文件
    题目附件:
    encryptor.c

    #include <stdlib.h>
    #include <stdio.h>
    #include <string.h>
    
    int main(int argc, char **argv) {
    	if (argc != 3) {
    		printf("USAGE: %s INPUT OUTPUT\n", argv[0]);
    		return 0;
    	}
    	FILE* input  = fopen(argv[1], "rb");
    	FILE* output = fopen(argv[2], "wb");
    	if (!input || !output) {
    		printf("Error\n");
    		return 0;
    	}
    	char k[] = "CENSORED";
    	char c, p, t = 0;
    	int i = 0;
    	while ((p = fgetc(input)) != EOF) {
    		c = (p + (k[i % strlen(k)] ^ t) + i*i) & 0xff;
    		t = p;
    		i++;
    		fputc(c, output);
    	}
    	return 0;
    }
    

    msg001

      Hi! This is only test message
    

    msg001.enc

    灄@佇紦矘缑N1i_5徙?辏犆?R
    

    msg002.enc

    獨僂鐙摪陦(?p\}㈩欸1^シ廛T偾?咘芩鬯?h?}荊tKk蝌)??侂?w壙7㎜Z6縌*猙灗襤侊鎟2$萇?旞噓T?场S?&碡涉翾汉€enN疓璃暇侁鴊D諞褲`'^街?[}qh9吽頵槃
    N?z?B??P内?暣?0;}瞌宏-儝錻2奌兜)豿?l飢=?鸦<'硌??锑B璦!a珉腷鈻紭%粵f.厂嚘嶑??鯠z姄{D?RO?'M梮饂柫鹥務Ns潮瞤?務<?r弞?桞y*???~m鴵桘bd嬣桖裪(椋31?|脮Q+
    藟/6,覘I?!?魘蓪湹V?╟uc壂毈汾臔i1?'=}斚R]??
    

    解题:
    显然encryptor.c中是一段加密的密码,而msg001中是一段明文文字 ,msg001.enc中是其加密后的文字,我们要找的flag应该是msg002.enc对应的解密后的明文。
    这道题看上去就是一道写逆代码的题目,但是里面有一个坑,我们尝试后发现msg001加密后并非是msg001.enc中的内容,这是因为它的加密代码中key是错误的,所以我们需要找出正确的key。
    找出key的代码如下:

    #include <stdlib.h>
    #include <stdio.h>
    #include <string.h>
    
    int main(int argc, char **argv) {
    	
    	FILE* input  = fopen("msg001.enc", "rb");
    	if (!input ) {//如果有一个文件无法打开 
    		printf("Error\n");
    		return 0;
    	}
    	char c, p, t = 0;
    	int i=0;
    	char w[] = "Hi! This is only test message\n";  //原来input中的值
        unsigned int j = 0;
        while ((p = fgetc(input)) != EOF) {
            for (j=31;j<125;j++) {//在找k的真正值,只有当能够逆向回w时,真正找到k 
                c = (p - (j ^ t) - i*i) & 0xff;
                if (c == w[i]) {
                    printf("%c",j);
                    t = c;
                    i++;
                    break;
                }
            }
        }
    	return 0;
    }
    

    得到密钥为:VeryLongKeyYouWillNeverGuess
    然后使用密钥去得到解密,代码如下:

    #include <stdlib.h>
     #include <stdio.h>
     #include <string.h>
     
     int main(int argc, char **argv) {
         FILE* input  = fopen("msg002.enc", "rb");
         FILE* output = fopen("msg002.txt", "wb");
         if (!input || !output) {
             printf("Error\n");
             return 0;
         }
        char c, p, t = 0;
        int i = 0;
        char k[] = "VeryLongKeyYouWillNeverGuess";
        i = 0;
        c, p, t = 0;
        int g = 0;
        while ((p = fgetc(input)) != EOF) {
        	//printf("%c",p);
            c = (p - (k[i % strlen(k)] ^ t) - i*i) & 0xff;
            printf("Decrypting %x i=%d t=%d k=%d -> %d\n",p,i,t,(k[i % strlen(k)] ^ t),c);
            t = c;//c是改变之前的c 
            i++;
             printf("%c",c);
             fputc(c, output);
             g++;
             if (g>450) {break;}
        }
     
        return 0;
     }
    

    得到的结果为:
    The known-plaintext attack (KPA) is an attack model for cryptanalysis where the attacker has samples of both the plaintext (called a crib), and its encrypted version (ciphertext). These can be used to reveal further secret information such as secret keys and code books. The term “crib” originated at Bletchley Park, the British World War II decryption operation.
    The flag is CTF{6d5eba48508efb13dc87220879306619}

    展开全文
  • 基于二元矩阵的粗糙集属性约简方法,李华雄,,提出了一种基于二元矩阵的粗糙集属性约简方法。在分辨矩阵中引入二元矩阵描述方法,将属性约简问题转化为二元矩阵覆盖问题。依据
  • Zero or One

    2020-02-03 13:24:19
    Zero or One File: zero.[c|cpp|java] Everyone probably knows the game Zero or One (in some regions in Brazil also known as Two or One), used to determine a winner among three or more players. For those...

    Zero or One
    File: zero.[c|cpp|java]
    Everyone probably knows the game Zero or One (in some regions in Brazil also known as Two or
    One), used to determine a winner among three or more players. For those unfamiliar, the game works
    as follows. Each player chooses a value between zero or one; prompted by a command (usually one of
    the contestants announces “Zero or… One!”), all participants show the value chosen using a hand: if
    the value chosen is one, the contestant shows a hand with an extended index finger; if the value chosen
    is zero, the contestant shows a hand with all fingers closed. The winner is the one who has chosen
    a value different from all others. If there is no player with a value different from all others (e.g. all
    players choose zero, or some players choose zero and some players choose one), there is no winner.
    Alice, Bob and Clara are great friends and play Zerinho all the time: to determine who will buy
    popcorn during the movie session, who will enter the swimming pool first, etc… They play so much
    that they decided make a plugin to play Zerinho on Facebook. But since the don’t know how to
    program computers, they divided the tasks among friends who do know, including you.
    Given the three values chosen by Alice, Bob and Clara, each value zero or one, write a program
    that determines if there is a winner, and in that case determines who is the winner.
    Input
    The input contains a single line, with three integers A, B and C ,indicating respectively the values
    chosen by Alice, Beto and Clara.
    Output
    Your program must output a single line, containing a single character. If Alice is the winner the
    character must be ‘A’, if Beto is the winner the character must be ‘B’, if Clara is the winner the
    character must be ‘C’, and if there is no winner the character must be ‘*’ (asterisc).
    在这里插入图片描述

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    int main(){
        int a,b,c;
        int sum;
        while(~scanf("%d%d%d",&a,&b,&c)){
            sum=0;
            if(a==0)sum++;
            if(b==0)sum++;
            if(c==0)sum++;
            if(sum==3||sum==0)printf("*\n");
            if(sum==1){
                if(a==0)printf("A\n");
                else if(b==0)printf("B\n");
                else if(c==0)printf("C\n");
            }
            if(sum==2){
                if(a==1)printf("A\n");
                else if(b==1)printf("B\n");
                else if(c==1)printf("C\n");
            }
        }
        return 0;
    }
    
    展开全文
  • MATLABfrom Zero to One视频教程 全17讲.rar. rar1
  • zeroOne-crx插件

    2021-04-05 04:29:30
    语言:українська Eduzerone.com教育平台的扩展。 安装“零筒”按钮后,在Eduzerone.com的扩展名为EduzerOne.com... 在安装“零筒”按钮后,UTUBA中会出现“零筒”按钮,该按钮允许单击“向页面添加课程”。
  • 异常信息: Test class should have exactly one public zero-argument constructor 错误原因: @Test方法所在类中,不能存在有参数构造函数,无参构造可以存在 解决方案: 把有参构造器去掉,使用set方法初始化...

    错误背景: 做测试的时候,为了偷懒,使用@Test注解代替main方法来执行代码, 

    public class Demo {
        private  int i;
        public Demo(int i) {
            this.i= i;
        }
    
        private void f(){
            System.out.println("普通方法"+i);
        }
    
        @Test
        public void method(){
            Demo demo = new Demo(25);
            demo2.f();
        }
    }
    

    异常信息:  Test class should have exactly one public zero-argument constructor

    错误原因:  @Test方法所在类中,不能存在有参数构造函数,无参构造可以存在

    解决方案: 把有参构造器去掉,使用set方法初始化变量

                     或者不要使用@Test注解,使用main方法执行.     

                     或者在另外一个类中使用@Test方法 

    展开全文
  • 《从无到有》(即《Zero to One》)是根据彼得·泰尔在斯坦福讲授“初创企业”这门课的视频和讲义的课程笔记的摘编成的一本书《从无到有》
  • ZeroTier One.msi

    2020-08-21 15:41:29
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  • Zero-shot Learning 训练集中没有某个类别的样本,但是如果我们可以学到一个牛逼的映射,这个映射好到我们即使在训练的时候没看到这个类,但是我们在遇到的时候依然能通过这个映射得到这个新类的特征。即: 对于...
  • 硅谷创投教父、PayPal创始人作品,斯坦福大学改变未来的一堂课,为世界创造价值的商业哲学。 在科技剧烈改变世界的今天,想要成功,你必须在一切发生之前研究结局。 你必须找到创新的独特方式,让未来不仅仅...
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    2017-01-14 11:45:48
    本书讲的是一场思维运动:质疑现有观念,从零开始重新审视自己所从事的业务。 1. 最反主流的行动不是抵制潮流,而是在潮流中不丢弃自己的独立思考。 2. 如果你想创造并获得持久的价值,不要跟风建立一个没有特色的...
  • Palatucci, Mark, et al. "Zero-shot learning with semantic output codes."Advances in ...
  • Zero-shot One-shot Few-shot learning 算法

    千次阅读 2019-06-17 21:17:23
    Zero-shot learning 指的是我们之前没有这个类别的训练样本。但是我们可以学习到一个映射X->Y。如果这个映射足够好的话,我们就可以处理没有看到的类了 One-shot learning 指的是我们在训练样本很少,甚至只有一...
  • 为了 “多快好省” 地通往炼丹之路,炼丹师们开始研究 Zero-shot Learning / One-shot Learning / Few-shot Learning。 爱上一匹野马 (泛化能力),可我的家里没有草原 (海量数据) 。 Learning类型 分为: Zero-shot...
  • Zeroone_C#门户网站cms程序
  • 为了 “多快好省” 地通往炼丹之路,炼丹师们开始研究 Zero-shot Learning / One-shot Learning / Few-shot Learning。 爱上一匹野马 (泛化能力),可我的家里没有草原 (海量数据) 。 Learning类型
  • zero__to__one-源码

    2021-03-30 11:19:03
    zero__to__one
  • 第一次编译时遇到了must take either zero or one argument的报错。一下是部分代码 { private: //数据成员 T real; T image; public: Complex(); //默认构造函数模板 Complex(const T &m, const T &...
  • ZeroTier One

    2017-12-07 09:52:30
    ZeroTier One,内网穿透,实现虚拟局域网的组建,配置简单。
  • 为了 “多快好省” 地通往炼丹之路,炼丹师们开始研究 Zero-shot Learning / One-shot Learning / Few-shot Learning。 爱上一匹野马 (泛化能力),可我的家里没有草原 (海量数据) 。 2. Learning类型 (1)...
  • 什么是One-shot Learning 、Zero-shot Learning?

    万次阅读 多人点赞 2018-07-24 16:34:22
    One/zero-shot learning都是用来进行学习分类的算法。 Zero-shot Learing 就是训练样本里没有这个类别的样本,但是如果我们可以学到一个牛逼的映射,这个映射好到我们即使在训练的时候没看到这个类,但是我们在遇到...
  • acme-zero和acme-one 包依存关系与吸收元素形成一个可交换的半定态。 它们满足以下公理: 关联性 对于所有依赖项a , b和c , (a , b) , c = a , (b , c) 实际上,出于这个原因,Cabal语法甚至不允许您使用括号...
  • In this letter, a compact second-order mixed coupling bandpass filter (BPF) with one controllable transmission zero (TZ) near the passband edge is presented using multilayer substrate integrated ...
  • =zero && input_val <=one pytorch 训练时,中间会出现报错 如果是BCEloss,需要加一个 sigmoid 或 sofmax 函数。 其他层: 解决:在每一层cov,使用batch_norm一般可以解决问题
  • MATLABfrom Zero to One视频教程 全17讲.rar

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