若题目中出现错误,建议使用如下的测试用例:
4 0000 0000.00
4 00123.5678 0001235
3 0.0520 0.0521
4 00000.000000123 0.0000001230
4 00100.00000012 100.00000013
5 0010.013 10.012
4 123.5678 123.5
3 123.5678 123
4 123.0678 123
3 0.000 0
对应的正确输出应该为:
YES 0.0000*10^0
NO 0.1235*10^3 0.1235*10^4
NO 0.520*10^-1 0.521*10^-1
YES 0.1230*10^-6
YES 0.1000*10^3
NO 0.10013*10^2 0.10012*10^2
YES 0.1235*10^3
YES 0.123*10^3
YES 0.1230*10^3
YES 0.000*10^0
代码(算法笔记优化):
#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++];
else res += '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;
}else{
cout << "NO 0." << s3 << "*10^" << e1 <<" 0." << s4 << "*10^" << e2 <<endl;
}
return 0;
}