【未完成】【笨方法學PAT】1060 Are They Equal (25 分)
一、題目
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
二、題目大意
科學計數法。
三、考點
string
四、注意
1、最後一個測試點沒有通過;
2、參考:https://www.liuchuo.net/archives/2289。
五、程式碼
#include<iostream>
#include<string>
using namespace std;
int n, m;
string getString(string s) {
string t = "0.", s1, s2;
int i = 0, k;
bool flag_dot = false;
for (i = 0; i < s.length(); ++i) {
if (s[i] == '.') {
flag_dot = true;
break;
}
}
s1 = s.substr(0, i);
k = i;
if (flag_dot == true) {
s2 = s.substr(i + 1);
s1 += s2;
}
i = 0;
while (i < s1.length() && s1[i] == '0') {
i++;
k--;
}
s1 = s1.substr(i);
if (n >= s1.length()) {
t += s1;
for (int i = 0; i < n - s1.length(); ++i)
t += '0';
}
else {
for (int i = 0; i < n; ++i)
t += s1[i];
}
t = t + "*10^" + to_string(k);
return t;
}
int main() {
//read
cin >> n;
m = n;
string s1, s2;
cin >> s1 >> s2;
//solve
s1 = getString(s1);
s2 = getString(s2);
//output
if (s1 == s2)
cout << "YES " << s1 << endl;
else
cout << "NO " << s1 << " " << s2 << endl;
system("pause");
return 0;
}