PAT-A 1060 Are They Equal (25 分)

根据机器保存浮点数的精度判断两个数是否相等。输入包含三个数:精度N,浮点数A和B。输出相等则打印'YES'及标准形式,不等则打印'NO'及两数的标准形式。

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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 0.123×10​5​​ 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 10​100​​, 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.d[1]...d[N]*10^k (d[1]>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

思路:

关键是找到小数点移动的方向和距离。规律如下


浮点数      小数点下标       第一个非0数的下标       指数
12345         5                          0                               5 (5-0)
123.45        3                          0                               3 (3-0)
0.00123·     1                          4                               -2 (1 - 4 + 1)
0.000345    1                          5                               -3  (1 - 5 + 1)

 

坑点:0要写成0.000*10^0的形式。
  

/*
浮点数      小数点下标      第一个非0数的下标       指数
12345       5               0                       5 (5-0)
123.45      3               0                       3 (3-0)
0.00123·    1               4                       -2 (1 - 4 + 1)
0.000345    1               5                       -3  (1 - 5 + 1)
*/

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>

using namespace std;

// 返回第一个小数点的位置
int indexOfPoint(char num[])
{
    for(int i = 0; i < strlen(num); i++)
    {
        if(num[i] == '.')
            return i;
    }
    return strlen(num);
}

//返回第一个非零整数的位置
int indexOfFirst(char num[])
{
    for(int i = 0; i < strlen(num); i++)
    {
        if(num[i] > '0' && num[i] <= '9')
            return i;
    }
    return -1; //返回-1表示是0
}

// 得到标准格式
string get(char n1[], int n)
{
    int point; // 小数点的下标
    int first; // 第一个非零数字的下标

    //先计算第一个浮点数
    point = indexOfPoint(n1); //小数点的位置
    first = indexOfFirst(n1); //第一个非零数字的位置

    string s1 = "0.";
    if(first != -1)
    {
        // 标准格式的指数
        int index = point - first;
        index = index > 0 ? index : index + 1;

        // 计算有效位上的数字
        int i = 0;
        while(i < n)
        {
            if(first < strlen(n1) && n1[first] != '.')
            {
                s1 += n1[first];
                i++;
            }
            else if(first >= strlen(n1))
            {
                s1 += "0";
                i++;
            }
            first++;
        }
        s1 += "*10^";
        s1 += to_string(index);
    }
    else
    {
        for(int i = 0; i < n; i++)
            s1 += "0";
        s1 += "*10^";
        s1 += "0"; // 0.000要求为0.000*10^0
    }
    return s1;
}
int main()
{
    char n1[110], n2[110];
    int n;

    scanf("%d", &n);
    scanf("%s", n1);
    scanf("%s", n2);

    string s1 = get(n1, n);
    string s2 = get(n2, n);
    if(s1 == s2)
        printf("YES %s\n", s1.c_str());
    else
        printf("NO %s %s\n", s1.c_str(), s2.c_str());
    return 0;
}

 

(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
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