1060 Are They Equal (25分)

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

 坑点太多

刚开始写的只有17分。。。。

有效数字： 用string的 find_first_not_of(“0”) 找到第一个有效数字的位置pos。对于0.001这样中间有小数点的情况，只要一开始用 find() 函数求出并记住小数点位置，然后用 erase() 删掉小数点即可。
从pos开始的n个数都是有效数字，可以用string的构造函数把这个区间的字符拷贝过来，不足n位的话用循环在后面补0。
指数： 因为这个转换是把小数点移到第一个有效数字前，所以
指数 = 小数点实际位置 - 第一个有效数字位置
注意前导零这个坑点

​​想要看结题思路的看这篇文章​​

代码：

//#include <bits/stdc++.h>
//using namespace std;
//int n, lena, lenb;
//string a, b;
//int getPos(string s){
//  int ind, len = s.length();
//  for(ind = 0; ind < len; ind++){
//    if(s[ind] == '.'){
//      return ind;
//    }
//  }
//  return ind;
//}
//int main(){
//  cin >> n >> a >> b;
//  lena = a.length();
//  lenb = b.length();
//  int slen1 = getPos(a), slen2 = getPos(b);
  cout << slen1 << " " << slen2 << endl;
//  int flag = 0;
//  for(int i = 0; i < n; i++){
//    if(a[i] != b[i]){
//      flag = 1;
//      break;      
//    }
//  }
//  if(flag || slen1 != slen2){
//    cout << "NO ";
//    cout << "0." << a.substr(0, n) << "*10^" << slen1 << " ";
//    cout << "0." << b.substr(0, n) << "*10^" << slen2; 
//  }else{
//    cout << "YES ";
//    cout << "0." << a.substr(0, n) << "*10^" << slen1 << endl; 
//  }
//  return 0;
//}
//17分  
#include<bits/stdc++.h>
#define ll long long
using namespace std;

int n;
string a, b;

int main(){
  cin >> n >> a >> b;
  int pa, pb, pos1, pos2; //  pa，pb表示两个数的小数点实际位置，pos1、2表示第一个有效数字的位置
  if(a.find(".") == a.npos)
    a += ".0";
  if(b.find(".") == b.npos) //  补上小数点以便同一计算
    b += ".0";
  pa = a.find("."), pb = b.find(".");

  string aa = a, bb = b;  //  aa和bb最终将存放每个数的所有有效数字（仅数字）
  aa.erase(aa.find("."), 1);
  bb.erase(bb.find("."), 1);  //  去掉小数点
  pos1 = aa.find_first_not_of("0"), pos2 = bb.find_first_not_of("0");

  aa = pos1 == (int)aa.npos? "0" : string(aa, pos1, n);
  bb = pos2 == (int)bb.npos? "0" : string(bb, pos2, n);// 注意判断数是0的情况！
  //  位数不足有效数字个数则用0补
  while(aa.length() < n) 
    aa += '0';
  while(bb.length() < n)   
    bb += '0';
  int k1 = pa - pos1, k2 = pb - pos2; //  指数
  //数是0的情况，10的指数规定为0
  if(aa == string(n, '0'))
    k1 = 0;
  if(bb == string(n, '0'))  
    k2 = 0;
  string resa = "0." + string(aa, 0, n) + "*10^" + to_string(k1);
  string resb = "0." + string(bb, 0, n) + "*10^" + to_string(k2);
  if(resa == resb)
    cout << "YES " << resa << endl;
  else
    cout << "NO " << resa << ' ' << resb << endl;

  return 0;
}