PAT 1060 Are They Equal (25分)

2023-10-05 15:42

文章标签 25 pat 1060 equal

本文主要是介绍PAT 1060 Are They Equal (25分)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805413520719872

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 0.123×10^5 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.

输入
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.

输出
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.

样例输入
3 12300 12358.9

样例输出
YES 0.123*10^5

代码

#include<iostream>
#include<string>
using namespace std;int n;
string deal(string s, int &e) {int k = 0;while(s.length() > 0 && s[0] == '0')s.erase(s.begin());if(s[0] == '.') {s.erase(s.begin());while(s.length() > 0 && s[0] == '0'){s.erase(s.begin());e--;}}else {while(k < s.length() && s[k] != '.') {k++;e++;}if(k < s.length())s.erase(s.begin() + k);}if(s.length() == 0)e = 0;int num = 0;k = 0;string res;while(num < n) {if(k < s.length())res += s[k++];elseres += '0';num++;}return res;
}int main() {string s1, s2, s3, s4;cin >> n >> s1 >> s2;int e1 = 0, e2 = 0;s3 = deal(s1, e1);s4 = deal(s2, e2);if(s3 == s4 && e1 == e2)cout<<"YES 0."<<s3<<"*10^"<<e1<<endl;elsecout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl;return 0;
}