PAT 1060. Are They Equal

本文介绍了一种算法,用于判断两个浮点数在给定有效数字位数的情况下是否被视为相等。通过去除前导零和小数点,将数字转换为标准形式,并比较其前N位有效数字来实现。

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If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*105 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line "YES" if the two numbers are treated equal, and then the number in the standard form "0.d1...dN*10^k" (d1>0 unless the number is 0); or "NO" if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:
3 12300 12358.9
Sample Output 1:
YES 0.123*10^5
Sample Input 2:
3 120 128
Sample Output 2:
NO 0.120*10^3 0.128*10^3

代码分析

这道题首先是考有效数字的表达,在自己进行测试检验是要注意输入数据是小数的情况,最后一个测试点输入数据是0.000和00.0000的情况,给出下面几组测试数据:
3 12300 12358.9
YES 0.123*10^5


1 12300 12358.9
YES 0.1*10^5


1 1.2300 1.23589
YES 0.1*10^1


5 1.2300 1.23589
NO 0.12300*10^1 0.12358*10^1


4 0.01234 0.012345
YES 0.1234*10^-1


5 0.01234 0.012345
NO 0.12340*10^-1 0.12345*10^-1


5 0.1234 0.12345
NO 0.12340*10^0 0.12345*10^0


0 0.11 0
YES 0.*10^0或者YES 0.0*10^0,都可以AC,测试点应该没有这个例子


1 0.1 0
NO 0.1*10^0 0.0*10^0


1 0.0 0.1
NO 0.0*10^0 0.1*10^0


1 0.0 0.000
YES 0.0*10^0


1 00.0 0.000
YES 0.0*10^0


4 00.0 0.000
YES 0.0000*10^0


5 00.0 0.000
YES 0.00000*10^0


1 05.0 5.000
YES 0.5*10^1


1 00.01 0.010
YES 0.1*10^-1
#include<iostream>
#include<algorithm> 
using namespace std;
int main(){
    string a,b;
    int n,t1,t2;
    cin>>n>>a>>b;
    auto it=find(a.begin(),a.end(),'.');
    if(it!=a.end()){
      t1=-(a.end()-1-it);
      a.erase(it); 
    }
    else t1=0;
    int i;
    for(i=0;i<a.size();i++){
        if(a[i]!='0') break;
    }
    a=a.substr(i,a.size()-i);
    it=find(b.begin(),b.end(),'.');
    if(it!=b.end()){
       t2=-(b.end()-1-it);
       b.erase(it);
    }
    else t2=0;
    for(i=0;i<b.size();i++)
        if(b[i]!='0') break;
    b=b.substr(i,b.size()-i);
    t1+=a.size();   t2+=b.size();
    if(a=="")   t1=0;
    if(b=="")   t2=0;
    if(a.size()<n)  a.insert(a.end(),n-a.size(),'0');
    if(b.size()<n)  b.insert(b.end(),n-b.size(),'0');
    if(t1==t2&&a.substr(0,n)==b.substr(0,n))
           cout<<"YES 0."<<a.substr(0,n)<<"*10^"<<t1<<endl;
       else
           cout<<"NO 0."<<a.substr(0,n)<<"*10^"<<t1<<" 0."<<b.substr(0,n)<<"*10^"<<t2;
}

转载于:https://www.cnblogs.com/A-Little-Nut/p/8365385.html

(c++题解,代码运行时间小于200ms,内存小于64MB,代码不能有注释)It’s time for the company’s annual gala! To reward employees for their hard work over the past year, PAT Company has decided to hold a lucky draw. Each employee receives one or more tickets, each of which has a unique integer printed on it. During the lucky draw, the host will perform one of the following actions: Announce a lucky number x, and the winner is then the smallest number that is greater than or equal to x. Ask a specific employee for all his/her tickets that have already won. Declare that the ticket with a specific number x wins. A ticket can win multiple times. Your job is to help the host determine the outcome of each action. Input Specification: The first line contains a positive integer N (1≤N≤10 5 ), representing the number of tickets. The next N lines each contains two parts separated by a space: an employee ID in the format PAT followed by a six-digit number (e.g., PAT202412) and an integer x (−10 9 ≤x≤10 9 ), representing the number on the ticket. Then the following line contains a positive integer Q (1≤Q≤10 5 ), representing the number of actions. The next Q lines each contain one of the following three actions: 1 x: Declare the ticket with the smallest number that is greater than or equal to x as the winner. 2 y: Ask the employee with ID y all his/her tickets that have already won. 3 x: Declare the ticket with number x as the winner. It is guaranteed that there are no more than 100 actions of the 2nd type (2 y). Output Specification: For actions of type 1 and 3, output the employee ID holding the winning ticket. If no valid ticket exists, output ERROR. For actions of type 2, if the employee ID y does not exist, output ERROR. Otherwise, output all winning ticket numbers held by this employee in the same order of input. If no ticket wins, output an empty line instead. Sample Input: 10 PAT000001 1 PAT000003 5 PAT000002 4 PAT000010 20 PAT000001 2 PAT000008 7 PAT000010 18 PAT000003 -5 PAT102030 -2000 PAT000008 15 11 1 10 2 PAT000008 2 PAT000001 3 -10 1 9999 1 -10 3 2 1 0 3 1 2 PAT000001 3 -2000 Sample Output: PAT000008 15 ERROR ERROR PAT000003 PAT000001 PAT000001 PAT000001 1 2 PAT102030
08-12
### 使用 `Assert.AreEqual()` 比较索引值 `Assert.AreEqual()` 是单元测试框架中的一个重要方法,用于验证两个对象是否相等。当比较数组或列表的第一个元素 (`index[0]`) 时,可以按照如下方式使用该函数: ```csharp // 假设有一个整数数组 numbers 和预期的第一个元素 expectedFirstElement int[] numbers = { 1, 2, 3 }; int expectedFirstElement = 1; // 验证第一个元素是否等于预期值 Assert.AreEqual(expectedFirstElement, numbers[0]); ``` 此代码片段会检查 `numbers` 数组的第一个元素是否确实为 `expectedFirstElement` 所指定的值[^1]。 对于更复杂的数据结构,比如多维数组或其他集合类,则需注意访问特定位置的方式可能有所不同。例如,在处理二维数组的情况下,应该这样写断言语句来对比某个具体坐标的数值: ```csharp double[,] matrix = new double[,] { { 1.0, 2.0 }, { 3.0, 4.0 } }; double expectedResultAtPosition = 2.0; Assert.AreEqual(expectedResultAtPosition, matrix[0, 1]); // 行列下标从0开始计数 ``` 另外需要注意的是,如果被测条件不满足,即实际值与期望值不符,那么 `Assert.AreEqual()` 将抛出异常并停止当前测试案例执行流程。这有助于快速定位问题所在之处[^2]。 在某些情况下,可能会遇到浮点数精度误差导致的失败情况。为了避免这种情况发生,可以在调用 `AreEqual` 方法时提供第三个参数作为允许的最大绝对差额(tolerance),从而实现近似匹配而非严格意义上的完全相同: ```csharp float actualValue = 0.1f + 0.2f; // 可能不会精确得到 0.3 float expectedValue = 0.3f; float tolerance = 0.0001f; Assert.AreEqual(expectedValue, actualValue, tolerance); ```
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