【算法笔记】A1060 Are They Equal

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 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 (<) 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 1, 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

　　比较两个数的科学计数法是否相同，相同输出YES，否则输出NO。本题不需要考虑四舍五入。

思路

　　把接收到的两个数删除掉前导的0。然后分为两种情况：一种是.00XXXX形式；另一种是XXXX.XXXX形式。

　　如果是.00XXXX形式，删除小数点和小数点后面的0，直到碰到不为0的数，每删一个0，指数减1。

　　如果是XXXX.XXXX形式，遍历字符串直到碰到小数点，把小数点删除。每遍历一个数，指数加1。

　　最后比较改变后的字符串和指数，输出结果。注意输入  000.0  时输出 YES 0.000*^

code：

 #include<bits/stdc++.h>
 using namespace std;
 string num1, num2;
 int n;
 static string turn(string a, int &e){//e前面加&，值才会被修改
     string result="0.";
     while(a.size()>&&a[]==''){//删除前导0
         a.erase(a.begin());
     }
     if(a[]=='.'){//0.XXXXXX形式
         a.erase(a.begin());
         while(a.size()> && a[]==''){
             a.erase(a.begin());
             e--;
         }
     }else{//XXXX.XXXX形式
         int k = ;
         while(k < a.size() && a[k]!='.'){
             e++;
             k++;
         }
         if(k<a.size()) a.erase(a.begin()+k);
     }
     if(a.size()==) e = ;//删除0和小数点后长度为0
     int num = , k = ;
     while(num < n){
         if(k < a.size()) result += a[k++];
         else result += '';
         num++;
     }
     return result;
 }
 int main(){
     int e1 = , e2 = ;
     cin>>n>>num1>>num2;
     string s1 = turn(num1,e1);
     string s2 = turn(num2,e2);
     if(s1 == s2 && e1 == e2)
         cout<<"YES "<<s1<<"*10^"<<e1;
     else
         cout<<"NO "<<s1<<"*10^"<<e1<<" "<<s2<<"*10^"<<e2;
     return ;
 }