1073 Scientific Notation (20 分)

2024-01-09 12:20:58

Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9].[0-9]+E[+-][0-9]+ which means that the integer portion has exactly one digit, there is at least one digit in the fractional portion, and the number and its exponent‘s signs are always provided even when they are positive.

Now given a real number

Input Specification:

Each input contains one test case. For each case, there is one line containing the real number

Output Specification:

For each test case, print in one line the input number

Sample Input 1:

+1.23400E-03

Sample Output 1:

0.00123400

Sample Input 2:

-1.2E+10

Sample Output 2:

-12000000000

题解：找到E的位置，分别处理尾数和指数即可

#include<bits/stdc++.h>
using namespace std;

int main(){
    string s,s1,s2;
    cin>>s;
    int pos;
    if(s[0]==‘-‘){
        printf("-");
        s.erase(s.begin());
    }
    else{
        s.erase(s.begin());
    }
    for(int i=0;i<s.length();i++){
        if(s[i]!=‘E‘){
            s1+=s[i];
        }
        else{
            pos=i;
            s2=s.substr(i+1);
            break;
        }
    }
    
    int sum=0;
    for(int i=1;i<s2.length();i++){
        sum=sum*10+(s2[i]-‘0‘);
    }
    if(sum==0){
        cout<<s1<<endl;
        return 0;
    }
    if(s2[0]==‘+‘){
        cout<<s1[0];
        if(s1.length()-2>sum){
            for(int i=2,j=1;i<s1.length();i++,j++){
                if(j==sum){
                    printf("%c",s1[i]);
                    printf(".");
                }
                else{
                    printf("%c",s1[i]);
                }
            }
            printf("\n");
        }
        else{
            cout<<s1.substr(2);
            for(int i=0;i<sum-s1.length()+2;i++){
                printf("0");
            }
            printf("\n");
        }
    }
    else{
        printf("0.");
        for(int i=0;i<sum-1;i++){
            printf("0");
        }
        for(int i=0;i<s1.length();i++){
            if(s1[i]==‘.‘){
                continue;
            }
            else{
                printf("%c",s1[i]);
            }
        }
        printf("\n");
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;

int main(){
    string s,s1,s2;
    cin>>s;
    int pos;
    if(s[0]==‘-‘){
        printf("-");
        s.erase(s.begin());
    }
    else{
        s.erase(s.begin());
    }
    for(int i=0;i<s.length();i++){
        if(s[i]!=‘E‘){
            s1+=s[i];
        }
        else{
            pos=i;
            s2=s.substr(i+1);
            break;
        }
    }
    
    int sum=0;
    for(int i=1;i<s2.length();i++){
        sum=sum*10+(s2[i]-‘0‘);
    }
    if(sum==0){
        cout<<s1<<endl;
        return 0;
    }
    if(s2[0]==‘+‘){
        cout<<s1[0];
        if(s1.length()-2>sum){
            for(int i=2,j=1;i<s1.length();i++,j++){
                if(j==sum){
                    printf("%c",s1[i]);
                    printf(".");
                }
                else{
                    printf("%c",s1[i]);
                }
            }
            printf("\n");
        }
        else{
            cout<<s1.substr(2);
            for(int i=0;i<sum-s1.length()+2;i++){
                printf("0");
            }
            printf("\n");
        }
    }
    else{
        printf("0.");
        for(int i=0;i<sum-1;i++){
            printf("0");
        }
        for(int i=0;i<s1.length();i++){
            if(s1[i]==‘.‘){
                continue;
            }
            else{
                printf("%c",s1[i]);
            }
        }
        printf("\n");
    }
    return 0;
}

注：测试点2，3考察点：

当指数与小数部分相等时，注意“.”，例如：+1.234E+03 ===> 1234(T) 1234.(F)

易错数据：

+3.1415E+004 //3.1415

-3.1415926E+4 //-3.1415.926

+3.1415926E-01　　　//0.31415926

-3.1415926E-0005　　//-0.000031415926