codeforces April fool contest 2020

达成成就：AK一场CF（

这次玩的梗我基本都会？Jordan Curve Theorem、元素周期表，不过那首limerick我没看出acoustic。。。

Is it rated?

输出no就行了。

本题用来测试选手何时能够正常登录cf（确信）

Limericks

诗的十行首字母分别是TWO FACTORS。。。不过我比赛的时候并没有看出来。。

不过看到样例之后应该可以联想到分解因数吧。。。

...And after happily lived ever they

...and they lived happily ever after. Separately.

看到输入是个6-bit integer，标题是6个单词的句子的重排，大概能猜到每个单词对应一个bit。。。

样例输入和输出的数二进制表示中1的个数都相等。。。大概能猜到就是把输入的数转成二进制重排后输出。。。

重排的方法就是看标题是怎么重排的，也就是053241。

Elementary!

我可能是看puzzling.SE看多了。。。所以看到elementary就知道是指元素周期表。。。

问题：给出一个单词，能否把它拆成元素的缩写？例如WATSON = W + At + S + O + N就可以，HOLMES就不行，因为L、OL和LM都不是元素周期表里的元素。。。

写了个dp。。。反正数据范围小，随便过。。。

Jordan Smiley

其实不用看标题就知道问题是：输入一个pixel的坐标，问这个pixel在不在曲线内部。。。

smiley是因为曲线画的是一个笑脸。。。Jordan因为Jordan Curve Theorem。。。

由于linux上没有MSPaint.exe，我对其他画图工具都不熟，所以没法做flood fill。。。但是用射线法还是可以判断一个pixel是否在曲线内部。

首先把图像下下来用python读出每个pixel（$960\times 960$的）。。。然后对于输入的坐标$(x,y)$，把它映到图上对应的点$(15x+7,15y+7)$，然后做射线法。具体地说，就是水平向右边打一条射线，看它与多少个pixel相交，如果与奇数个pixel相交则点在内部，偶数个则点在外部。（线宽为$3$是奇数，所以和奇数个pixel相交等价于和曲线交了奇数次。）

输出一个$64\times 64$的矩阵，hardcode进源代码里就是了。

char ans[333][333] ={
"1111111111111111111111111101010000001011111111111111111111111111\n",
"1111111111111111111111010001010101101000001111111111111111111111\n",
"1111111111111111111100010111011100101111011011111111111111111111\n",
"1111111111111111100001010100000111100001010001011111111111111111\n",
"1111111111111111001111110101111100001111010101001111111111111111\n",
"1111111111111101100100000100010001011010011101101011111111111111\n",
"1111111111111000110110111111010111000011001001001001111111111111\n",
"1111111111100000010010010001000100011110011011011100011111111111\n",
"1111111111000000000000111101111101110100110010010110001111111111\n",
"1111111110000000000000000000010000000110000110000000000111111111\n",
"1111111100000000000000000000000000000000000000000000000011111111\n",
"1111111000000000000000000000000000000000000000000000000001111111\n",
"1111111000000000000000000000000000000000000000000000000001111111\n",
"1111110000000000000000000000000000000000000000000000000000111111\n",
"1111100000000000000000000000000000000000000000000000000000011111\n",
"1111100000000000000000000000000000000000000000000000000000011111\n",
"1111000000000000000000000000000000000000000000000000000000001111\n",
"1110000000000000000000000000000000000000000000000000000000000111\n",
"1111100000000000000000000000000000000000000000000000000000000111\n",
"1111111000000000000111011000000000000001110101000000000000011111\n",
"1100111110000000001101110000000000000001011111110000000011111111\n",
"1100001111100000111001011100000000000000010010011010011111110011\n",
"1000000011111011100011000110000000000010110111001111111110000001\n",
"1000000000111110110110010011100000000010010000011011110000000001\n",
"1000000000000100100011111000110000000111011011110001000000000001\n",
"1000000000111110111000100011100000000101000001011101101000000001\n",
"0000000000001000101101101110110000001101101111001000111000000000\n",
"0000000000011101101001000100010000101000100001100011101000000000\n",
"0000000000010000001000011110110101111010110100001110001000000000\n",
"0000000000011111011011000100011101010010010110111011011000000000\n",
"0000000000001001000010010001110100011111000010100000010000000000\n",
"0000000000011100011011011100010110001000011010001110111000000000\n",
"0000000000000111110010001001000100111110110010011011101000000000\n",
"0000000000000010010111101011110101101010010111001001000000000000\n",
"0000000000001111000010101010100001001011010001100011100000000000\n",
"0000000000000101101110001110000111011001111011001110000000000000\n",
"0000000000000000111000111010010010010011010001011011000000000000\n",
"0000000000000000001110100000011111000000011101110000000000000000\n",
"1000000000000000011011110000010100000000000111011000000000000001\n",
"1000000000000000000000000000000111100000000000000000000000000001\n",
"1000000000000000000000000000101101000000000000000000000000000001\n",
"1000000000000000000000000010111000000000000000000000000000000001\n",
"1100000000000000000000000011101110000000000000000000000000000011\n",
"1100000001100000000000000000001000000000000000000000000000000011\n",
"1110000000101000000000000000011100000000000000000000010000000111\n",
"1110000000111011100000000000110100000000000100000101111000000111\n",
"1110000000101010110101110010100110101101001110011111010000000111\n",
"1111000000001110011111011111110011100111111010110001000000001111\n",
"1111100000001011000010000100011000111100010000010100000000011111\n",
"1111100000000001011011011101000010001001011111011100000000011111\n",
"1111110000000011110010010111101110101111000101110000000000111111\n",
"1111111000000000100111000001001010111010011100000000000001111111\n",
"1111111000000000110001101011100011101000110111000000000001111111\n",
"1111111100000000011100001110001001000010000100000000000011111111\n",
"1111111110000000001001111010111111011110100000000000000111111111\n",
"1111111111000000000000101000010010010100110000000000001111111111\n",
"1111111111100000000000001111000011110111100000000000011111111111\n",
"1111111111111000000000000000000001000000000000000001111111111111\n",
"1111111111111100000000000000000000000000000000000011111111111111\n",
"1111111111111111000000000000000000000000000000001111111111111111\n",
"1111111111111111100000000000000000000000000000011111111111111111\n",
"1111111111111111111100000000000000000000000011111111111111111111\n",
"1111111111111111111111000000000000000000001111111111111111111111\n",
"1111111111111111111111111100000000000011111111111111111111111111\n",
};

Again?

祖传奇偶题（前年也有）

当我做完ABCEF题的时候比赛已经过去了一个小时。。。我一看，有6000人过了D，接近过了A的人数？？？猜测本题是傻逼题。。。随手写了个判断最后一位奇偶上去，过了。。。

It's showtime

真正的showtime

又是一道语言题。。。编译器CE的时候会输出what the FUCK DID I DO WRONG。。。谷歌“what the FUCK DID I DO WRONG esolang”可知语言是ArnoldC

接下来就是现学了。。。这语言最难的是输入，需要特意google一遍该怎么输入。。。其他的加减乘除if while在github wiki上都有。。。TIO上也有解释器。。。另外，说的是这语言只支持16位整数，但是读入10w级别的数好像没问题？

这题golf起来可能很有意思（挖坑

Lingua Romana

继续搜Lingua Romana esolang，发现这是个用拉丁文写Perl的方言。。。（我还以为是什么独立的语言。。。nvm

找到了一个该方言的介绍。。。发现和cf给的程序大概能对上。。于是一句一句地翻译就行了。。。

per nextum in unam tum XI conscribementis fac sic
    vestibulo perlegementum da varo.
    morde varo.
    seqis cumula varum.
cis

翻译：

execute it 11 times {
    read a variable from stdin
    chomp it (?)
    push it to a stack
}

然后

per nextum in unam tum XI conscribementis fac sic
    seqis decumulamenta da varo.
    varum privamentum fodementum da aresulto.
    varum tum III elevamentum tum V multiplicamentum da bresulto.
    aresultum tum bresultum addementum da resulto.

翻译：

execute it 11 times: {
    pop from stack a variable
    aresult = sqrt(abs(var))
    bresult = (var ** 3) * 5
    result = aresult + bresult

然后

    si CD tum resultum non praestantiam fac sic
        dictum sic f(%d) = %.2f cis tum varum tum resultum egresso describe.
        novumversum egresso scribe.
    cis

翻译：

    if 400 > result {
        stdout.printf("f(%d) = %.2f", var, res)
        stdout.println()
    }

最后

    si CD tum resultum praestantiam fac sic
        dictum sic f(%d) = MAGNA NIMIS! cis tum varum egresso describe.
        novumversum egresso scribe.        
    cis
cis

翻译：

    if 400 <= result {
        stdout.printf("f(%d) = MAGNA NIMIS!", var)
        stdout.println()
    }
}

于是把这玩意照抄一遍就能AC本题了。。。

坑：codegolf