F题 Turing equation
来源:互联网 发布:2200万淘宝买家资料 编辑:程序博客网 时间:2024/06/05 03:51
Turing equation
时间限制:1000 ms | 内存限制:65535 KB
难度:1
- 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=165+8=130001000+000200=000300+0=0
- 样例输出
TRUEFALSETRUE
解题思路:
题意是说给你一个等式,判断每个数从右到左装换之后等式是否成立,成立为输出TRUE,否则输出FALSE
我的代码:
#include<bits/stdc++.h>using namespace std;int main(){ char a[30]; while(scanf("%s",a)!=EOF) { int r[3]={0},ok=0,i,k=0,j,b[3]={0}; for(i=0;i<=strlen(a);i++) { if(a[i]=='+'||a[i]=='='||i==strlen(a)) { for(j=i-1;j>=k;j--) r[ok]=r[ok]*10+(a[j]-'0'); b[ok]=i-1-k+1; ok++; k=i+1; } } if(b[0]==1&&b[1]==1&&b[2]==1&&r[0]==0&&r[1]==0&&r[2]==0)//判断每个数是否为一位数且为0 break; else { if(r[0]+r[1]==r[2]) printf("TRUE\n"); else printf("FALSE\n"); } } return 0;}
0 0
- F题 Turing equation
- F.Turing equation
- zzuoj 10399: F.Turing equation
- 【第七届河南省赛】F.Turing equation
- Turing equation
- Turing equation
- Turing equation
- Turing equation
- NYOJ1253 第七届acm省赛 F Turing equation
- 河南省第七届省赛 问题 F: Turing equation 水题
- NYOJ 1253 Turing equation
- 第七届河南省赛 zzuoj 10399: F.Turing equation (模拟)
- 练习场 1253 Turing equation
- 河南省第七届ACM【Turing equation】
- zzuoj--10399--Turing equation(模拟)
- 山东省第八届ACM省赛 F 题(quadratic equation)
- quadratic equation 17年省赛F题,SDUT 3898
- F——quadratic equation
- Android自定义控件系列之应用篇——圆形进度条
- JQuery日期插件datepicker的使用
- 机器视觉光源知识总结(二)
- C++struct结构
- 调整数组顺序使奇数位于偶数前面
- F题 Turing equation
- soc时钟系统
- 网站备案期间何为闭站
- NYOJ 55 懒省事的小明
- CSDN-markdown编辑器使用
- Maven
- vb.net 教程 5-15 图像处理之内存处理 4
- 2017.4.23 一元三次方程求解 思考记录
- 虚函数与纯虚函数