PAT 1060 Are They Equal （25 分）

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.

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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.

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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

解析

把普通数字 ->科学记数法
要注意前导0 ，比如给出0012345，000.0124。
分两种情况：
① $0.a_1a_2a_3a_4a_5a_6.....$
② $b_mb_{m-1}.....b_3b_2b_1.a_1a_2a_3a_4a_5a_6.....$

#include<string>
#include<iostream>
#include<utility>
using namespace std;
pair<string,int> change(const string& a) {
	string result="0.";
	int K=0;
	int len = a.size(),i=0;
	for (; i < len; i++) {     //跳过前面的0
		if (a[i] != '0') {
			break;
		}
	}   
	//now a[i] may '.'   or number
	//a maybe is 0000,
	if (i != len) {
		if (a[i] == '.') {     // .
			bool allzero = true;
			for (i++; i < len; i++) {
				if (a[i] == '0')
					K--;
				else {
					allzero = false;
					break;
				}
			}
			// a maybe 000.0000
			if (i != len)
				result += a.substr(i);
			if (allzero)
				K = 0;
		}
		else {                // number
			bool flag = false;
			for (; i < len; i++) {
				if (a[i] != '.') {
					result += a[i];
					if (!flag)
						K++;
				}
				else
					flag = true;
			}
		}
	}
	return make_pair(result,K);
}
int main()
{
	int N;
	string A, B;
	pair<string, int> PA, PB;
	cin >> N >> A >> B;
	PA = change(A);
	PB = change(B);
	PA.first.resize(N+2,'0');
	PB.first.resize(N+2, '0');
	if (PA.first == PB.first && PA.second==PB.second) {
		cout << "YES ";
		cout << PA.first << "*10^" << PA.second<<endl;
	}
	else {
		cout << "NO ";
		cout << PA.first << "*10^" << PA.second <<" ";
		cout << PB.first << "*10^" << PB.second << endl;
	}
}