FZU 1171 Hard to Believe, but True!

最新推荐文章于 2019-02-15 17:31:12 发布

_Occult_

最新推荐文章于 2019-02-15 17:31:12 发布

阅读量713

点赞数

CC 4.0 BY-SA版权

分类专栏： FZU -----------杂题----------- 文章标签： fzu

本文链接：https://blog.youkuaiyun.com/jtjy568805874/article/details/47376345

-----------杂题----------- 同时被 2 个专栏收录

234 篇文章

订阅专栏

FZU

43 篇文章

订阅专栏

本文介绍了一个有趣的编程挑战，即验证图灵提出的逆向算术谜题的正确性。通过将数字逆向读取并验证加法等式的真伪，读者可以了解到一种特殊的数字处理方式，并通过提供的C++代码实现来加深理解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

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.
[C. Strachey - private communication.]"

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

Input

The input contains several test cases. Each specifies on a single line a Turing equation. A Turing equation has the form "a+b=c", wherea, 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.

Output

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.

Sample Input

73+42=16

5+8=13

10+20=30

0001000+000200=00030

1234+5=1239

1+0=0

7000+8000=51

0+0=0

Sample Output

True

False

True

True

False

False

True

True

把数字翻转看等式是否成立

#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=1005;
char s[maxn];

int main()
{
	while (~scanf("%s",s))
	{
		int a=0,b=0,c=0,i,j;
		for (i=0,j=1;s[i]!='+';i++,j*=10) a+=j*(s[i]-'0');
		for (i++,j=1;s[i]!='=';i++,j*=10) b+=j*(s[i]-'0');
		for (i++,j=1;s[i];i++,j*=10) c+=j*(s[i]-'0');
		if (a+b==c) printf("True\n"); else printf("False\n");
		if (!strcmp(s,"0+0=0")) break;
	}
	return 0;
}