【PAT甲级】1060 Are They Equal

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🔑中文翻译：它们是否相等
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1060 Are They Equal

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.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
1

Sample Output 1:

YES 0.123*10^5
1

Sample Input 2:

3 120 128
1

Sample Output 2:

NO 0.120*10^3 0.128*10^3
1

思路

这道题主要难点就在于将给定的浮点数转换成 $0.x\ast 10^y$ 的形式，我们这里来讲如何处理字符串：

1. 先找到 . ，如果是整数需要人为在后面加上一个 . ，因为我们要得到 . 的位置 $k$ ，用于分隔两边的数字。
2. 将 . 两边的数字即字符串合并，然后去除多余的前缀 $0$ 。
3. 判断去除完 $0$ 后字符串是否为空，要将 $k$ 也置 $0$ ，防止其越界。
4. 判断有效位数 $n$ 与字符串的长度的大小，如果字符串位数大于有效位数则要进行截断，反之补 $0$ 。
5. 返回时，注意字符串的格式。

代码

#include
using namespace std;

int n;
string a, b;

string change(string str, int n)
{
    //找到'.'，如果没有则加在后面并且找到其所在位置
    int k = str.find(".");
    if (k == -1)   str += ".", k = str.find(".");

    //合并'.'两边的字符串，并去除前缀0
    string s = str.substr(0, k) + str.substr(k + 1);
    while (s.size() && s[0] == '0')  s = s.substr(1), k--;

    if (s.empty())   k = 0;    //字符串为空k也要置0
    if (s.size() > n)  s = s.substr(0, n);    //字符串长度大于有效位要截断
    else s += string(n - s.size(), '0');       //反之要补0

    return "0." + s + "*10^" + to_string(k);
}

int main()
{
    cin >> n >> a >> b;

    a = change(a, n);
    b = change(b, n);

    if (a == b)    cout << "YES " << a << endl;
    else cout << "NO " << a << " " << b << endl;

    return 0;
}
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