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PAT (Basic Level) Practice (中文) 1024 科学计数法

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn = 100000;
 6 char a[maxn];
 7 int main(){
 8     int evalue=0,pos=0,flag=0,ep,len,pointp;
 9     int count=0;
10     scanf("%s",a);  
11 
12     len = strlen(a);
13     for(int i=1;i<len;i++){
14         if(a[i]=='.'){pointp=i;}
15         if(a[i]=='E') {ep = i;break;}
16     }
17     for(int i=ep+2;i<len;i++){
18         evalue = evalue * 10 + (a[i]-'0');
19     }
20     if(a[ep+1]=='+') pos = 1;
21     if(a[0]=='-') printf("%c",a[0]);
22     if(pos==1){
23         count=0;
24         for(int i=1;i<ep;i++){
25             if(i>pointp){
26                 if(count<evalue) {printf("%c",a[i]);count++;}
27                 else if(count==evalue) printf(".%c",a[i]);
28                 else printf("%c",a[i]);
29             }
30             else if(i<pointp) printf("%c",a[i]);
31         }
32         if(count<evalue){
33             for(int i=count;i<evalue;i++) printf("0");
34         }
35     }
36     else{
37         printf("0.");
38         for(int i=1;i<evalue;i++) printf("0");
39         for(int i=1;i<ep;i++)
40             if(i!=pointp)
41                 printf("%c",a[i]);      
42     }
43     
44 
45 
46     return 0;
47 }
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posted on 2020-04-16 15:51  chenxi16  阅读(168)  评论(0编辑  收藏  举报