【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 3 000.0 0 时输出 YES 0.000*10^0 

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()>0&&a[0]=='0'){//删除前导0
        a.erase(a.begin());
    }
    if(a[0]=='.'){//0.XXXXXX形式
        a.erase(a.begin());
        while(a.size()>0 && a[0]=='0'){
            a.erase(a.begin());
            e--;
        }
    }else{//XXXX.XXXX形式
        int k = 0;
        while(k < a.size() && a[k]!='.'){
            e++;
            k++;
        }
        if(k<a.size()) a.erase(a.begin()+k);
    }
    if(a.size()==0) e = 0;//删除0和小数点后长度为0
    int num = 0, k = 0;
    while(num < n){
        if(k < a.size()) result += a[k++];
        else result += '0';
        num++;
    }
    return result;    
}
int main(){
    int e1 = 0, e2 = 0;
    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 0;
}