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
题意：
给出两个数，问将他们写成保留N位小数的科学计数法后是否相等。如果想等，则输出“YES”，并给出该转换结果；如果不相等，则输出“NO”，并分别给出两个数的转换结果
参考代码：
#include<iostream>
#include<string>
using namespace std;
int n; //有效位数
string deal(string s, int& e) {
int k = 0; //s的下标
while(s.length() > 0 && s[0] == '0') {
s.erase(s.begin()); //去掉s的前导零
}
if(s[0] == '.') { //若去掉前导零后是小数点，则说明s是小于1的小数
s.erase(s.begin()); //去掉小数点
while(s.length() > 0 && s[0] == '0') {
s.erase(s.begin()); //去掉小数点后非零位前的所有零
e--; //每去掉一个0，指数e减一
}
}
else{ //若去掉前导零后不是小数点，则找到后面的小数点删除
while(k<s.length() && s[k] != '.') { //寻找小数点
k++;
e++; //只要不遇到小数点，就让指数e++
}
if(k < s.length()) { //while结束后k < s.length(),说明遇到了小数点
s.erase(s.begin() + k); //把小数点删除
}
}
if(s.length() == 0) {
e = 0; //如果去除前导零后s的长度变为0,则说明这个数是0
}
int num = 0;
k = 0;
string res;
while(num < n) { //只要精度还没有到n
if(k < s.length()){
res += s[k++]; //只要还有数字，就加到res末尾
}
else{
res += '0'; //否则res末尾添加0
}
num++;
}
return res;
}
int main() {
string s1, s2, s3, s4;
cin >> n >> s1 >> s2;
int e1 = 0,e2 = 0; //e1,e2为s1与s2的指数
s3 = deal(s1,e1);
s4 = deal(s2,e2);
if(s3 == s4 && e1 == e2) { //若主体相同且指数相同，则输出“YES”
cout<<"YES 0."<<s3<<"*10^"<<e1<<endl;
}
else{
cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl;
}
return 0;
}