PAT甲级1060 Are They Equal (25 分)_pat are they equal-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_42306885/article/details/98650380

本文探讨了在计算机中如何判断两个浮点数在特定有效数位下是否相等的问题，通过科学计数法进行精确比较，介绍了获取阶数和尾数的有效算法。

1060 Are They Equal (25 分)

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

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

题目大意：

给定有效数位n与浮点数a、b。请问两浮点数是否相等，如果是则输出YES，并输出有效形式。

思路：

其实问题就是在问科学计数法 + 排除掉有效数位外的数。

算法包括了两部分，一部分是“如何获得阶数”、第二部分是“如果获得尾数”，即有效数位问题。

先解决阶数问题：

假设有123.4444，小数点前的数是正阶数，即此时阶数ans = 3。写成科学计数法0.1234444*10^4。

假设有 0000.1234，小数点前虽然有数，但要排除先导0，即应当为0.1234,

假设有0.001234，小数点后有0，则此时阶数应当为 ans = -2，写成科学计数法为0.1234 * 10 ^-2。

可以知道阶数获得方法：如果有先导0，则排除掉先导0，如果此时是小数点，则删除掉小数点，然后再排除掉小数点后的0，并且每次都让阶数减一。

如果排除掉先导0后不是小数点，那么就去找小数点，小数点前每位数都让阶数加一。

遇到小数点后记得删除小数点。

再解决有效数位的问题：

假设现在是有效数位为3，123.4444在第一步处理后0.1234444*10^4。实际上是（s = 1234444，阶数ans = 4），超过有效数位的部分直接删除掉。

0.001234，写成科学计数法为0.1234 * 10 ^-2，有效数位为5， s = 1234 , s.size() < 5,在s 后面补0就可以了。

参考代码：

#include<iostream>
#include<string>
using namespace std;
int n;
string a, b;
int change(string &s){
	int ans = 0, idx = 0;
	while(s.front() == '0') s.erase(s.begin());
	if(s.front() == '.'){
		s.erase(s.begin());
		while(s.front() == '0') s.erase(s.begin()), ans--;
	}else{
		while(s[idx] != '.' && idx < s.size())	idx++, ans++;
		if(idx < s.size())	s.erase(idx, 1);
	}
	if(s.size() == 0){
		while(s.size() < n)	s += '0';
		return 0;
	}
	if(s.size() > n)	s.erase(s.begin() + n, s.end());
	while(s.size() < n)	s += '0';
	return ans;
}
int main(){
	cin >> n >> a >> b;
	int num1 = change(a), num2 = change(b);
	if(a == b && num1 == num2){
		printf("YES 0.%s*10^%d", a.c_str(), num1);
	}else{
		printf("NO 0.%s*10^%d", a.c_str(), num1);
		printf(" 0.%s*10^%d", b.c_str(), num2);
	}
	return 0;
}