问题 : Turing equation

本文介绍了一道关于图灵方程的问题,通过逆序数字来判断等式的真假,并提供了C语言实现的源代码。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

问题 : Turing equation

时间限制: 1 Sec  内存限制: 128 MB

http://218.198.32.182/problem.php?cid=1069&pid=5

题目描述

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


解题思路:

就是讲每个数取逆序数字,然后相加,看是否成立。 输出0+0=0的时候结束。


源代码:

# include <stdio.h>
# include <string.h>


int main(void)
{
	char a[100];
	while (~ scanf("%s", a))
	{
		int i, d = strlen(a), c, j, k;
		int sum = 0, sum1 = 0, sum2 = 0;
		c = 1;
		for (i = 0; a[i] != '+'; i ++)
		{
			sum += (a[i] - '0')*c;
			c *= 10;
		}	
		c = 1;
		for (j = i+1; a[j] != '='; j ++)
		{
			sum1 += (a[j] - '0')*c;
			c *= 10;
		}
		c = 1;
		for (k = j+1; a[k] != '\0'; k ++)
		{
			sum2 += (a[k] - '0')*c;
			c *= 10;
		}
		if (sum == 0 && sum1 == 0 && sum2 == 0)
		break;
		if (sum + sum1 == sum2)
		printf("TRUE\n");
		else
		printf("FALSE\n");
		
	}
	return 0;
}

内容概要:本文探讨了在MATLAB/SimuLink环境中进行三相STATCOM(静态同步补偿器)无功补偿的技术方法及其仿真过程。首先介绍了STATCOM作为无功功率补偿装置的工作原理,即通过调节交流电压的幅值和相位来实现对无功功率的有效管理。接着详细描述了在MATLAB/SimuLink平台下构建三相STATCOM仿真模型的具体步骤,包括创建新模型、添加电源和负载、搭建主电路、加入控制模块以及完成整个电路的连接。然后阐述了如何通过对STATCOM输出电压和电流的精确调控达到无功补偿的目的,并展示了具体的仿真结果分析方法,如读取仿真数据、提取关键参数、绘制无功功率变化曲线等。最后指出,这种技术可以显著提升电力系统的稳定性与电能质量,展望了STATCOM在未来的发展潜力。 适合人群:电气工程专业学生、从事电力系统相关工作的技术人员、希望深入了解无功补偿技术的研究人员。 使用场景及目标:适用于想要掌握MATLAB/SimuLink软件操作技能的人群,特别是那些专注于电力电子领域的从业者;旨在帮助他们学会建立复杂的电力系统仿真模型,以便更好地理解STATCOM的工作机制,进而优化实际项目中的无功补偿方案。 其他说明:文中提供的实例代码可以帮助读者直观地了解如何从零开始构建一个完整的三相STATCOM仿真环境,并通过图形化的方式展示无功补偿的效果,便于进一步的学习与研究。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值