1073. Scientific Notation (20)

最新推荐文章于 2019-01-22 22:54:45 发布

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本文介绍如何将科学计数法表示的实数转换为常规数字表示形式，并保留所有有效数字，包括尾随的零。通过分析输入的科学计数法格式，使用字符串处理而非数值运算来确保精度。

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Scientific Notation (20)
时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
HOU, Qiming

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 A in scientific notation, you are supposed to print A in the conventional notation while keeping all the significant figures.

Input Specification:

Each input file contains one test case. For each case, there is one line containing the real number A in scientific notation. The number is no more than 9999 bytes in length and the exponent’s absolute value is no more than 9999.

Output Specification:

For each test case, print in one line the input number A in the conventional notation, with all the significant figures kept, including trailing zeros,
Sample Input 1:

+1.23400E-03

Sample Output 1:

0.00123400

Sample Input 2:

-1.2E+10

Sample Output 2:

-12000000000

这题必须要用字符串来做，而不是代入数中，否则最多只能拿18分。因为过大的数和过小的数可能会超过double的表示范围。
代码来自

Alex_512
感谢他的分享。

#include<cstdio>
#include<cstring>
int main()
{
    int flag=0,len,exp=0;
    char num[10010];
    scanf("%s",num);
    len=strlen(num);
    if(num[0]=='-')
        printf("-");
    while(num[flag]!='E')
        flag++;
    for(int i=flag+2;i<len;i++)
    {
        exp=exp*10+num[i]-'0';
    }
    if(num[flag+1]=='-')
    {
        printf("0.");
        for(int i=0;i<exp-1;i++)
            printf("0");
        for(int i=1;i<flag;i++)
        {
            if(num[i]=='.')
                continue;
            printf("%c",num[i]);
        }
    }
    else
    {
        for(int i=1;i<flag;i++)
        {
            if(num[i]=='.')
                continue;
            printf("%c",num[i]);
            if(i==exp+2&&flag-3!=exp)
                printf(".");
        }
        for(int i=0;i<exp-(flag-3);i++)
            printf("0");       
    }

}

另附错误代码：

#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
int main()
{
    freopen("in.txt","r",stdin);
    char c,a;
    long int k1,k2,l=0;
    long double k;
    long double n;
    while(scanf("%c",&c)){
    cin>>k1>>a>>k2>>a>>n;
    k=k2;
    while(k>1){
        k/=10;
        l++;
    }
    if(c=='-') cout<<c;
    k+=k1;
    k=pow(10,n)*k;
    if(k<1){
          l=l-n;
          cout.precision(l);
          cout.setf(ios::fixed);
          cout<< k << endl; 
    }
    else{
        if(l<=n) printf("%.lf",k);
        else
        {
          l=l-n;
          cout.precision(l);
          cout.setf(ios::fixed);
              cout<< k << endl; 
        }
    }
    }
    return 0;
}