NYOJ1253 第七届acm省赛 F Turing equation

最新推荐文章于 2019-05-03 00:32:23 发布

原创最新推荐文章于 2019-05-03 00:32:23 发布 · 293 阅读

0 ·

CC 4.0 BY-SA版权

省赛专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一种特殊的数学挑战——图灵方程，这是一种需要将数字反向读取并验证加法等式的趣味问题。文章提供了详细的算法实现，利用栈来逐个处理等式中的数字，并最终判断等式是否成立。

描述

The fight goes on, whether to store numbers starting with their most significant digit or their least significant digit. Sometimes this is also called the "Endian War". The battleground dates far back into the early days of computer science. Joe Stoy, in his (by the way excellent) book "Denotational Semantics", tells following story:

"The decision which way round the digits run is, of course, mathematically trivial. Indeed, one early British computer had numbers running from right to left (because the spot on an oscilloscope tube runs from left to right, but in serial logic the least significant digits are dealt with first). Turing used to mystify audiences at public lectures when, quite by accident, he would slip into this mode even for decimal arithmetic, and write things like 73+42=16. The next version of the machine was made more conventional simply by crossing the x-deflection wires: this, however, worried the engineers, whose waveforms were all backwards. That problem was in turn solved by providing a little window so that the engineers (who tended to be behind the computer anyway) could view the oscilloscope screen from the back.

You will play the role of the audience and judge on the truth value of Turing's equations.

输入

The input contains several test cases. Each specifies on a single line a Turing equation. A Turing equation has the form "a+b=c", where a, b, c are numbers made up of the digits 0,...,9. Each number will consist of at most 7 digits. This includes possible leading or trailing zeros. The equation "0+0=0" will finish the input and has to be processed, too. The equations will not contain any spaces.

输出

For each test case generate a line containing the word "TRUE" or the word "FALSE", if the equation is true or false, respectively, in Turing's interpretation, i.e. the numbers being read backwards.

样例输入

73+42=16
5+8=13
0001000+000200=00030
0+0=0

样例输出

TRUE
FALSE
TRUE

来源

第七届河南省程序设计大赛

题意是给一个加法式子，将每一项反转，如果结果成立，输出TRUE

#include<iostream>
#include<stack>
#include<string>
using namespace std;

int main(){
	string s;
	int flag;
	int i;
	long long x,y,z;
	stack<char> s1,s2,s3;
	while(cin>>s){
		if(s=="0+0=0")
		break;
		flag=0;
		x=0;y=0;z=0;
		for(i=0;i<s.size();i++){   //拆分因子 
			if(s[i]>='0'&&s[i]<='9'&&flag==0){
				s1.push(s[i]);
			}
			else if(s[i]=='+') flag=1;
			else if(s[i]>='0'&&s[i]<='9'&&flag==1)
			s2.push(s[i]);
			else if(s[i]=='=') flag=2;
			else if(s[i]>='0'&&s[i]<='9'&&flag==2)
			s3.push(s[i]);
		}
		while(s1.size()){
			x=x*10+(s1.top()-'0');
			s1.pop();
		}
		while(s2.size()){
			y=y*10+(s2.top()-'0');
			s2.pop();
		}
		while(s3.size()){
			z=z*10+(s3.top()-'0');
			s3.pop();
		}
//		cout<<x<<y<<z; 
		if(x+y==z)
		cout<<"TRUE"<<endl;
		else
		cout<<"FALSE"<<endl; 
	}
	return 0;
}