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Ural 1001 - Reverse Root

时间:2014-01-16 00:10:40      阅读:615      评论:0      收藏:0      [点我收藏+]
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 Ai (0 ≤ Ai ≤ 1018). 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 Ai 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
  
   
2297.0716
936297014.1164
0.0000
37.7757
Problem Author: Prepared by Dmitry Kovalioff
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// Ural Problem 1001. Reverse Root
// Verdict: Accepted  
// Submission Date: 22:34:56 12 Jan 2014
// Run Time: 0.328s
//  
// 版权所有(C)acutus。(mail: acutus@126.com) 
//  
// [解题方法]  
// 简单题,直接按题意输入输出即可
// 注意:数组开大点,要使用全局数组

#include<stdio.h>

double a[300000];

void solve(void)
{
    int i;
    double t;
    i = 0;
    while(scanf("%lf",&t) != EOF) a[i++] = t;
    i--;
    while(i >= 0) {
        printf("%.4lf\n", sqrt(a[i]));
        i--;
    }
}

int main(void)
{
    solve();
    return 0;
}
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Ural 1001 - Reverse Root

原文:http://www.cnblogs.com/acutus/p/3516782.html

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