1060 Are They Equal (25 分) 复杂字符串处理

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 0.123×10​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

不如分类讨论   多复制点代码而已  思路简单点 不怎么耗费时间  否则没有好的测试用例+打断点  很难调好

坑点：题目没有说清楚 开头可以有好多个0   值为0时指数为0  

string.erase(m,1);删除下标为m的元素千万不能忘记",1"限定个数  这不是set  否则m后面的全部被删除了 
s.substr() 是错的 必须赋值回给s即 s=s.substr() 

比较难过的测试用例:

3 0.000 0
4 123.5678 123.5
5 0010.013 10.012
4 00100.000000012 100.000000013
4 0000.000000123 0.0000001230
3 0.0520 0.0521

#include<iostream>
#include<string>
using namespace std;

int n;
string a, b;

string ZERO() {//值为0时的返回值
	string ans = "0.";
	for (int i = 0;i<n;i++) ans = ans + "0";
	ans = ans + "*10^0";//肯定是0次  不然肯定会特殊说明的
	return ans;
}

//分类讨论还简单些
string stdard(string s) {	//00123.567
	//预处理 去掉前导0 小数点要留下  一步一步简化处理
	int T = s.find_first_not_of("0");
	if (T == string::npos) return ZERO();
	else s=s.substr(T);//去掉开头的0   剩下的只有  .7895 .056 78955  789.55 

	int m = s.find('.');//小数点位置 
	int k = s.find_first_not_of(".0");//第一个有效数字下标位置 	
	if (k == -1) return ZERO();
	int p;//数量级
	if (m == -1) p = s.length();//整数  12345   12345
	else if (m>k) {//整数+小数 12345.0456   12345.0456
		p = m - k;
		s.erase(m,1);//q去掉小数点   ",1"千万别忘了 这不是set 
	}
	else {//纯小数   .0046
		p = -(k - m - 1);
		s = s.substr(k);//开头的.0肯定要去掉   别忘记赋值给s
	}
	string d = s.substr(0, n);
	while (d.length()<n) d = d + "0";//不够n位 末尾添0 
	char sign[3];sprintf(sign, "%d", p);
	return "0." + d + "*10^" + sign;
}

int main() {
	//freopen("in.txt", "r", stdin);
	cin >> n >> a >> b;
	a = stdard(a);b = stdard(b);
	if (a == b) cout << "YES " << a;
	else cout << "NO " << a << " " << b;
	return 0;
}

 

来源：CSDN

作者：树叶子_

链接：https://blog.csdn.net/hza419763578/article/details/100173816