PAT (Advanced Level) Practice — 1073 Scientific Notation （20 分）_pat (advanced level) practice答案-优快云博客

本文介绍了一个程序，用于将科学计数法表示的实数转换为常规计数法，同时保留所有有效数字，包括尾随的零。程序通过定位'E'字符和解析指数来处理正负指数的情况。

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805395707510784

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 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

：题意 找到E所在的位置以及指数的值，分别对指数为正值还是负值进行模拟。

#include<iostream>
#include<cstring>
using namespace std;
char s[10011]; 
int main(){
    while(cin>>s){
    	if(s[0]=='-'){
    		printf("-");
		}
		int len=strlen(s);
		int pos=0;
		while(s[pos]!='E'){
			pos++;
		}
		int zs=0;
		for(int i=pos+2;i<len;i++){
			zs=zs*10+s[i]-'0';
		}
		if(s[pos+1]=='-'){
			printf("0.");
			for(int i=0;i<zs-1;i++){
				printf("0");
			}
			for(int i=1;i<pos;i++){
				if(s[i]=='.'){
					continue;
				}else{
					printf("%c",s[i]);
				}
			}
		}else{
			for(int i=1;i<pos;i++){
				if(s[i]=='.'){
					continue;
				}
				if(i==zs+3&&pos-3>zs){
					printf(".");
				}
				printf("%c",s[i]);
			}
			for(int i=1;i<=zs-pos+3;i++){
				printf("0");
			}
		}
		printf("\n");
	} 
	return 0;
}