中文
Jump to navigation
Jump to search
It is Humanized 中文(简体,中国大陆).
Commands
Just speak Chinese.
Programs
Hello, world!
输出"Hello, world!"。
A+B Problem
定义整型变量 A 、 B 和 C 。 读入 A 和 B,然后将 A+B 存储到 C 中。 输出 C 的值。
99 Bottles of Beers
定义整型变量 I ,然后初始化为 99 。 重复执行以下代码块直到 I > 1 这个条件不满足为止: 输出f"{I} bottles of the beers on the wall, \n{I} bottles of the beers. \n" 输出f"Take 1 down, pass it around, \n{I-1} bottles of the beers on the wall. \n\n" 将 I 减去 1。 输出"1 bottle of the beer on the wall, \n1 bottle of the beer. \n" 输出"Take 1 down, pass it around, \n-2147483648 bottles of the beers on the wall. \n\n" 输出"No more bottles of the beers on the wall, \nno more bottles of the beers." 输出"Go to the store and buy some more, \n99 bottles of the beers on the wall. \n\n\n\n"
Guessing Number
导入random库内的所有模块。 定义整型变量 A 、 Answer 和 B 。 以"Please input the number of plays: "作为提示读入 A 。 将 Answer 初始化为 取随机整数(-50, 50)。 重复执行以下代码块直到 A ≥ 0 不满足为止: 以"Please input the number you guessed: "作为提示读入 B 。 如果 B = Answer 则: 输出"That's it! Thank you for playing!"。 打破循环。 否则 如果 B < Answer 则: 输出f"{B} is too small! Try again. \n"。 将 A 减去 1。 否则 如果 B > Answer 则: 输出f"{B} is too big! Try again. \n"。 将 A 减去 1。
Quine
输出这个程序的源代码。
Turing-Completeness
It is turing complete 'cause I tried and successfully maded a simple operating system(not on show) and this Esolang can do Python code:
切换至Python,然后执行两条分界线内的代码。 ———————————分————————————界———————————线—————————————— print("Hello, world", end="!") ———————————分————————————界———————————线——————————————
Or C++:
切换至C++,然后执行两条分界线内的代码。 ———————————分————————————界———————————线—————————————— #include<bits/stdc++.h> using namespace std; unsigned long long a, b, c; char op; int main() { cin >> a >> b >> op; switch (op) { case '+': { cout << a + b; break; } case '-': { cout << a - b; break; } case '*': { cout << a*b; break; } case '/': { if (b != 0) { c = a % b; cout << a / b << "......" << c; } else { cout << "ZeroDivisionError: You can't divide by 0!"; } break; } case '^': { cout << (unsigned long long)(pow(a, b)); break; } case '>': { unsigned long long t = a >> b; cout << t; break; } case '<': { unsigned long long q = a << b; cout << q; break; } default: { cout << "SyntaxError: No such Operation!"; break; } } return 0; } ———————————分————————————界———————————线——————————————