【C++】PAT (advanced level)1060. Are They Equal (25)

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本文介绍了一种算法，用于判断两个浮点数在一个仅保存指定有效数字位数的计算机系统中是否被视为相等。该算法通过简单的截断而非四舍五入来比较数值，并详细解释了输入输出规范及样例。

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1060. Are They Equal (25)

时间限制

50 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.d₁...d_N*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.

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

做烦躁了。放一放

//#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<vector>
#include<map>
//#include<iomanip>
using namespace std;

int main(){
	freopen("in.txt","r",stdin);
	int N,n;
	char A[101],B[101];
	string a,b;
	cin>>N>>A>>B;
	n=0;
	int locf=-1,locr=-1;
	int an,bn;
	bool ff=false,fr=false;
	int ccdig=0;
	while(A[n]!='\0'){
		if(A[n]=='.'){
			locf=n;
		}else if(!ff&&A[n]=='0'){
			;
		}else if(ccdig<N){
			a.push_back(A[n]);
			ccdig++;
			if(fr==false){
				fr=true;
				locr=n;
			}
		}
		n++;
	}
	if(locf==-1){
		locf=n;
	}
	an=locf-locr;
	if(locr==-1){
		an=0;
	}
	
	n=0;
	locf=-1,locr=-1;
	ccdig=0;
	ff=false,fr=false;
	while(B[n]!='\0'){
		if(B[n]=='.'){
			locf=n;
		}else if(!ff&&B[n]=='0'){
			;
		}else if(ccdig<N){
			b.push_back(B[n]);
			ccdig++;
			if(fr==false){
				fr=true;
				locr=n;
			}
		}
		n++;
	}
	if(locf==-1){
		locf=n;
	}
	bn=locf-locr;
	if(locr==-1){
		bn=0;
	}
	while(a.size()<N){
		a.push_back('0');
	}
	while(b.size()<N){
		b.push_back('0');
	}
	if(an==bn&&a==b){
		cout<<"YES ";
		cout<<"0."<<a<<"*10^"<<an;
	}else {
		cout<<"NO ";
		cout<<"0."<<a<<"*10^"<<an<<" 0."<<b<<"*10^"<<bn;
	}
	


	system("pause");
	return 0;
}