Are They Equal (25)

Are They Equal (25)

难度： ⭐⭐⭐
题目连接
题目描述
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.

输入描述:
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.

输出描述:
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.

输入例子:
3 12300 12358.9

输出例子:
YES 0.123*10^5

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大意
给出两个浮点数f1, f2 和一个整数n,判断这两个浮点数在类似于科学记数法的形式下是否相等，并输出这两个浮点数用这种计数法表示的形式。这种计数法只保存前面n位有效数字和指数，形式为0.xxxxx * 10 ^ k。

分析
首先分析规律，然后用字符串处理的方法解决。要表示这种计数法只要得到一个浮点数的前面n为有效数字和指数即可。然而这样的方法对程序的健壮性有较高的要求，因为问题的输入案例往往有很多你意料不到的输入，有时不看失败样例的话很难猜到是什么情况导致答案不对。另外比较坑的是不足n位要在后面补0题目没有说明。

MYCODE

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
using namespace std;

//得到指数值
int getExp(string s){
	int common = -1;
	for (int i = 0; i < s.length(); i++){	//寻找逗号位置
		if (s[i] == '.') {
			common = i;
			break;
		}
	}
	int isNum = -1;
	for (int i = 0; i < s.length(); i++){	//寻找第一个有效数字位置
		if (s[i]>'0' && s[i] <= '9'){
			isNum = i;
			break;
		}
	}
	if (isNum == -1) return 0;					//有可能为 00000.0
	if (common == -1 && isNum != 0) return 1;	//如00001这样的数
	if (common == -1) return s.length();
	int ans = common - isNum;
	if (ans >= 0) return ans;
	return ans + 1;
}

//得到前n为有效数字
string getUse(string s, int n){
	int sss = n;
	string temp = "";
	int i = 0;
	for (; i < s.length() && (s[i] == '0' || s[i] == '.'); i++);
	for (; i < s.length() && n; i++){
		if (s[i] != '.'){ temp += s[i]; n--; }
	}
	if (temp == "") temp = "0";
	temp = "0." + temp;
	while (temp.length() <= sss+1) temp += "0";
	return temp;
}

struct mynum {
	string use;
	int exp;
	void init(string s, int n){
		this->use = getUse(s, n);
		this->exp = getExp(s);
	}
	void printf(){
		cout << use << "*10^" << exp;
	}
};

int main(){
	string s1,s2;
	mynum t1, t2;
	int n;
	cin >> n >> s1 >> s2;
	t1.init(s1,n);
	t2.init(s2,n);
	if (t1.exp == t2.exp && t1.use == t2.use){
		cout << "YES ";
		t1.printf();
		cout << endl;
	}
	else{
		cout << "NO ";
		t1.printf();
		cout << " ";
		t2.printf();
		cout << endl;
	}
}

//  99 0 0.000
//  5 00000001 1