Ural 1001. Reverse Root(没有输入结束的坑爹题)-白红宇的个人博客

Ural 1001. Reverse Root(没有输入结束的坑爹题)

发布日期：2021-11-06 16:56:40 浏览次数：5 分类：技术文章

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1001. Reverse Root

Time limit: 2.0 second

Memory limit: 64 MB

The problem is so easy, that the authors were lazy to write a statement for it!

Input

The input stream contains a set of integer numbers

A_i (0 ≤

A_i ≤ 10

¹⁸). The numbers are separated by any number of spaces and line breaks. A size of the input stream does not exceed 256 KB.

Output

For each number

A_i from the last one till the first one you should output its square root. Each square root should be printed in a separate line with at least four digits after decimal point.

Sample

input	output
1427 0 876652098643267843 5276538	2297.0716936297014.11640.000037.7757

input

output

0   876652098643267843 5276538

2297.0716936297014.11640.000037.7757

AC代码：

#include 
    
     #include 
     
      double a[300000],p;int main(){	int i=0,n;	while(scanf("%lf",&p)!=EOF)//没有输入结束的标志		a[i++]=sqrt(p);	for(n=i-1;n>=0;n--)		printf("%.4lf\n",a[n]);	return 0;}

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