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POJ 3062 Celebrity jeopardy(我的水题之路——原样输出)
阅读量:4069 次
发布时间:2019-05-25

本文共 1776 字,大约阅读时间需要 5 分钟。

Celebrity jeopardy
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 11708   Accepted: 6572

Description

It's hard to construct a problem that's so easy that everyone will get it, yet still difficult enough to be worthy of some respect. Usually, we err on one side or the other. How simple can a problem really be? 
Here, as in Celebrity Jepoardy, questions and answers are a bit confused, and, because the participants are elebrities, there’s a real need to make the challenges simple. Your program needs to prepare a question to be solved --- an equation to be solved --- given the answer. Specifically, you have to write a program which finds the simplest possible equation to be solved given the answer, considering all possible equations using the standard mathematical symbols in the usual manner. In this context, simplest can be defined unambiguously several different ways leading to the same path of resolution. For now, find the equation whose transformation into the desired answer requires the least effort. 
For example, given the answer X = 2, you might create the equation 9 - X = 7. Alternately, you could build the system X > 0; X^2 = 4. These may not be the simplest possible equations. Solving these mind-scratchers might be hard for a celebrity.

Input

Each input line contains a solution in the form <symbol> = <value>

Output

For each input line, print the simplest system of equations which would to lead to the provided solution, respecting the use of space exactly as in the input.

Sample Input

Y = 3X=9

Sample Output

Y = 3X=9

Source

给一个键值对,要求你找出最简单的那个满足这个键值对的方程式,输出格式和输入格式一样。
键值对本身就是最简单的方程式,所以,我们可以原样输出就可以了。(人生好绝望 T_T)
代码(1AC):
#include 
int main(void){ char ch; while((ch = getchar()) != EOF){ putchar(ch); } return 0;}

转载地址:http://jloji.baihongyu.com/

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