題目描述

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.

翻譯：科學計數法是一種讓科學家可以很輕鬆的處理非常大的數字或非常小的數字的方法。其符號用正則表示式表示為[+-][1-9]”.”[0-9]+E[+-][0-9]+，即整數部分有且只有一位數字，小數部分至少有一位數字，並且它的數字和指數符號總是會提供即使它們為正數。
現在給你一個用科學計數法表示的實數A，你需要按照傳統格式輸出A並保留所有的有效數字。

IPNUT FORMAT

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.

翻譯：每個輸入檔案包含一組測試資料。對於每組輸入資料，第一行包括一個用科學計數法表示的實數A。數字位數的長度不超過9999，並且指數的絕對值不超過9999。

OUTNUT FORMAT

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,

翻譯：對於每組測試資料，輸出一行傳統格式下的數字A，並保留所有的有效數字，包括尾部0。

Sample Input 1:

+1.23400E-03

Sample Output 1:

0.00123400

Sample Input 2:

1.2E+10

Sample Output 2:

12000000000

解題思路

這道題主要是字串的處理，如果指數為負則新增前導0，為正數時判斷輸出位數是否達到指數，如果未達到在末尾輸出0。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#define INF 99999999
using namespace std;
char s[10010];
int main(){
    scanf("%s",s);
    int length=strlen(s);
    int flag=0,wei=0,symbol=0;
    for(int i=0;i<length;i++){
        if(i==0&&s[i]!='+')printf("-");
        if(flag&&s[i]=='-')symbol=1;
        if(flag&&s[i]>='0'&&s[i]<='9')wei=wei*10+s[i]-'0';
        if(s[i]=='E')flag=i;
    }
    if(symbol==1){
        printf("0.");
        for(int i=1;i<wei;i++)printf("0");
        for(int i=1;i<flag;i++){
            if(s[i]!='.')printf("%c",s[i]);
        }
    }
    else {
        int f=0;
        for(int i=1;i<flag;i++){
            if(i>wei+2&&!f)printf("."),f=1;
            if(s[i]!='.')printf("%c",s[i]);
        }
        if(wei)
        for(int i=0;i<wei-flag+3;i++){
            printf("0");
        }
    }
    printf("\n");
    return 0;
}