题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805413520719872
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.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
思路
去掉前导0。
先计算指数:分情况讨论:①、纯小数;②、有整数部分。
再求本体部分。
#include<iostream>
using namespace std;
int n;
string deal(string s,int &e)
{
while(s[0]=='0'&&s.length()>0)
{
s.erase(s.begin());
}
if(s[0]=='.')
{
s.erase(s.begin());
while(s[0]=='0'&&s.length()>0)
{
s.erase(s.begin());
e--;
}
}
else
{
int k=0;
//while(s[k]!='0')
while(s[k]!='.'&&k<s.length())
{
k++;
e++;
}
if(s[k]=='.') s.erase(s.begin()+k);
}
if(s.length()==0)
{
e=0;
}
int num=0;
string ans;
while(num<n)
{
if(num<s.length())
{
ans+=s[num];
num++;
}
else
{
ans+='0';
num++; //******别忘了num++
}
}
return ans;
}
int main()
{
string s1,s2,s3,s4;
cin>>n>>s1>>s2;
int e1=0;
int e2=0;
s3=deal(s1,e1);
s4=deal(s2,e2);
if(s3==s4&&e1==e2)
{
cout<<"YES 0."<<s3<<"*10^"<<e1;
}
else
{
cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2;
}
return 0;
}