Hard to Believe, but True!
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3537   Accepted: 2024

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", 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.

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

Source

分析:

思路比较简单

自己的做法:

 #include<string>
#include<cstring>
#include<iostream>
using namespace std;
int main(){//
string s;
int a[],b[],c[];
while(cin>>s){
if(s=="0+0=0"){ //注意
cout<<"True"<<endl;
break;
}
int i=;
int j=;
memset(a,,sizeof(a));
memset(b,,sizeof(b));
memset(c,,sizeof(c));
while(s[i]!='+'){
a[j++]=s[i++]-'';
}
j=;
i++;
while(s[i]!='='){
b[j++]=s[i++]-'';
}
j=;
i++;
while(s[i]){
c[j++]=s[i++]-'';
//cout<<c[j-1]<<endl;
}
for(i=;i<=;i++){
a[i+]+=(a[i]+b[i])/;
a[i]=(a[i]+b[i])%;
}
for(i=;i<;i++){
if(a[i]!=c[i])
break;
}
if(i==)
cout<<"True"<<endl;
else
cout<<"False"<<endl;
}
return ;
}

网上的代码:

学习点:

1.string.find(char a):返回字符a在字符串中的位置(从0开始)

2.string.substr(a,b):返回字符串从a开始的b个字符的字符子串

 #include <iostream>
#include <string>
using namespace std;
int trans(string s) {
int a=;
for (int i=s.length()-;i>=;i--)
a=a*+s[i]-'';
return a;
}
int main() {
string s,s1,s2,s3;
while (cin >> s) {
if (s=="0+0=0") {
cout << "True" << endl;
break;
}
bool flag=true;
int p1=s.find("+");
int p2=s.find("=");
s1=s.substr(,p1);
s2=s.substr(p1+,p2-p1-);
s3=s.substr(p2+,s.length()--p2);
if (trans(s1)+trans(s2)!=trans(s3)) flag=false;
if (flag) cout << "True" << endl;
else cout << "False" << endl;
} return ;
}

poj 2572 Hard to Believe, but True!的更多相关文章

  1. POJ 2572

    #include<stdio.h> #include<iostream> #include<string> using namespace std; int mai ...

  2. 【POJ 2572 Advertisement】

    Time Limit: 1000MSMemory Limit: 10000K Total Submissions: 947Accepted: 345Special Judge Description ...

  3. poj 1417 True Liars(并查集+背包dp)

    题目链接:http://poj.org/problem?id=1417 题意:就是给出n个问题有p1个好人,p2个坏人,问x,y是否是同类人,坏人只会说谎话,好人只会说实话. 最后问能否得出全部的好人 ...

  4. POJ 1417 - True Liars - [带权并查集+DP]

    题目链接:http://poj.org/problem?id=1417 Time Limit: 1000MS Memory Limit: 10000K Description After having ...

  5. True Liars POJ - 1417

    True Liars After having drifted about in a small boat for a couple of days, Akira Crusoe Maeda was f ...

  6. POJ 1417 True Liars(种类并查集+dp背包问题)

    题目大意: 一共有p1+p2个人,分成两组,一组p1,一组p2.给出N个条件,格式如下: x y yes表示x和y分到同一组,即同是好人或者同是坏人. x y no表示x和y分到不同组,一个为好人,一 ...

  7. POJ 1417 True Liars

    题意:有两种人,一种人只会说真话,另一种人只会说假话.只会说真话的人有p1个,另一种人有p2个.给出m个指令,每个指令为a b yes/no,意思是,如果为yes,a说b是只说真话的人,如果为no,a ...

  8. POJ 2965. The Pilots Brothers' refrigerator 枚举or爆搜or分治

    The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22286 ...

  9. POJ 1753. Flip Game 枚举or爆搜+位压缩,或者高斯消元法

    Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37427   Accepted: 16288 Descr ...

随机推荐

  1. Android-Activity临时数据的保存

    Activity临时数据的保存是非常重要的,例如:一款小说APP应用,读者使用这款APP看到了223页,用户也没有去记看了多少页: 突然去接了个电话,或者开启的应用程序太多了,可能会导致这款APP应用 ...

  2. 【模板模式】 Template Pattern

    模板模式 又叫模板方法模式,在一个方法中定义一个算法的骨架,而将一些步骤延迟到子类中.模板方法使得子类可以在不改变算法结构的情冴下,重新定义算法中的某些步骤(这个我觉得很抽象,很抽象) e:学会说“不 ...

  3. 【转】C#中使用aria2c进行下载并显示进度条

    [转自] C#中使用aria2c进行下载并显示进度条 - 云平台知识库 - 博客园https://www.cnblogs.com/littlehb/p/5782714.html 正则表达式的生成网站: ...

  4. mongodb driver2.5环境注意事项

    mongodb driver2.5环境注意事项 一.问题: 如果使用vs2012开发就会报这个错误: 未能加载文件或程序集“System.Runtime.InteropServices.Runtime ...

  5. hihocoder1580 Matrix

    题目链接:(vjudge)戳我 从今天开始不咕咕地填坑啦 考虑一般的求最大子矩阵和...我们一般都是DP,或者直接上悬线法递推. 下面附一个DP的代码: #include<iostream> ...

  6. [ZJOI2008] 树的统计Count

    题目链接:戳我 树链剖分. 注意一点就是维护最大值的时候最好写成下面代码里那个样子,要不然会因为可能左右区间没有的问题有奇奇怪怪的锅. 代码如下: #include<iostream> # ...

  7. jquery源码解析:jQuery延迟对象Deferred(工具方法)详解2

    请接着上一课继续看. $.Deferred()方法中,有两个对象,一个是deferred对象,一个是promise对象. promise对象有以下几个方法:state,always,then,prom ...

  8. jquery源码解析:expando,holdReady,ready详解

    jQuery的工具方法,其实就是静态方法,源码里面就是通过extend方法,把这些工具方法添加给jQuery构造函数的. jQuery.extend({       //当只有一个对象时,就把这个对象 ...

  9. <转>jmeter JDBC Request之Query Type

    本博客转载自:http://www.cnblogs.com/imyalost/category/846346.html 个人感觉不错,对jmeter讲解非常详细,担心以后找不到了,所以转发出来,留着慢 ...

  10. Source Insight 查看函数调用关系

    研究海思SAMPLE时,使用Source Insight查看源码,函数调用关系查看: 选中一个函数,右键选择"Show in Relation Window" 显示界面的相关设置: ...