1073 Scientific Notation （20 分）

最新推荐文章于 2021-08-05 11:12:27 发布

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最新推荐文章于 2021-08-05 11:12:27 发布

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原文链接：http://www.cnblogs.com/Mered1th/p/10426822.html

本文介绍了一个字符串处理问题，旨在将科学计数法表示的实数转换为常规计数法，并保留所有有效数字，包括末尾的零。通过分析题目要求，文章提供了一段详细的代码实现，用于解决这一挑战。

1073 Scientific Notation （20 分）

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


分析：字符串处理题，按题意改就行了。代码改了好多次。。。可能只有我自己能看懂了吧。。20分的题debug了半小时。。。

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-24-16.27.04
 6 * Description : A1073
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 
20 int main(){
21 #ifdef ONLINE_JUDGE
22 #else
23     freopen("1.txt", "r", stdin);
24 #endif
25     string s,ans="";
26     cin>>s;
27     if(s[0]=='-'){
28         printf("-");
29     }
30     s.erase(s.begin());
31     int i,len=s.length(),k,p;
32     for(i=0;i<len;i++){
33         if(s[i]>='0' &&s[i]<='9'){
34             ans+=s[i];
35         }
36         else if(s[i]=='.'){
37             k=i; //记录小数点位置
38             continue;
39         }
40         else if(s[i]=='E') {
41             p=i;  //记录E的位置
42             break;
43         }
44     }
45     bool flag=true;
46     if(s[i+1]=='-') flag=false;
47     s.erase(i,2); //删除E和指数符号
48     string e="";
49     for(i;i<len-1;i++){
50         e=e+s[i];
51     }
52     int E=stoi(e);
53     if(flag==false){
54         printf("0.");
55         for(int j=0;j<E-1;j++){
56             printf("0");
57         }
58         cout<<ans;
59     }
60     if(flag==true){
61         if(p-k-1==E){
62             cout<<ans;
63         }
64         else if(p-k-1<E){
65             cout<<ans;
66             for(int j=0;j<E-1;j++){
67                 printf("0");
68             }
69         }
70         else{
71             int a=(p-k-1)-E+2,m;
72             for(m=0;m<a;m++){
73                 printf("%c",ans[m]);
74             }
75             printf(".");
76             for(;m<ans.size();m++){
77                 printf("%c",ans[m]);
78             }
79         }
80     }
81     return 0;
82 }