点击链接PAT甲级-AC全解汇总

题目：
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 0.123×10^​5​​ 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 10^​100​​ , 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
1

Sample Output 1:

YES 0.123*10^5
1

Sample Input 2:

3 120 128
1

Sample Output 2:

NO 0.120*10^3 0.128*10^3
1

题意：
判断在k位有效位的时候科学记数法的表达是否一样

算法笔记的代码：

#include
#include
#include
using namespace std;
int n;
string deal(string s,int& e){
    int k=0;
    while(s.length()>0&&s[0]=='0'){
        s.erase(s.begin());
    }
    if(s[0]=='.'){                       //  0.***
        s.erase(s.begin());
        while(s.length()>0&&s[0]=='0'){ //    0.000***
            s.erase(s.begin());
            e--;
        }                               // ->0.000[***]
    }else{                              //    ***.***
        while(k<s.length()&&s[k]!='.'){
            k++;
            e++;
        }                           // ***.***
        if(k<s.length()){           // s[k]=='.'
            s.erase(s.begin()+k);
        }                           //[***].[***]
    }
    if(s.length()==0){
        e=0;
    }                               // 0.0
    int num=0;
    k=0;
    string res;
    while(num<n){
        if(k<s.length())    //还有非零数
            res+=s[k++];
        else                //用零补齐
            res+='0';
        num++;              //有效位
    }
    return res;
}
int main(){
    string s1,s2,s11,s22;
    cin>>n>>s1>>s2;
    int e1=0,e2=0;
    s11=deal(s1,e1);
    s22=deal(s2,e2);
    if(s11==s22&&e1==e2){
        cout<<"YES 0."<<s11<<"*10^"<<e1<<endl;
    }else{
        cout<<"NO 0."<<s11<<"*10^"<<e1<<" 0."<<s22<<"*10^"<<e2<<endl;
    }
    return 0;
}
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