Ural 1001 - Reverse Root
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
// 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; }
作者:acutus
本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利.
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步