湖南大学ACM程序设计新生杯大赛(同步赛)L - Liao Han
题目描述
Small koala special love LiaoHan (of course is very handsome boys), one day she saw N (N<1e18) guys standing in a row, because the boys had some strange character,The first time to Liao them will not be successful, but the second time will be successful; third times to Liao them will not be successful, but the fourth time will be successful;....... By analogy (toxic psycho)
So the little koala burst odd thought of a fancy up method. The N is the sum of handsome boys, and labeled from 1 to N, starting from the first guy, whose number is 1, she will Liao all the boys whose number is Multiple of 1, then number 2 and all the boys whose number is Multiple of 2(toxic method)
Later, little Kola found that some handsome guys did not get her Liao, she asked the smart you to help her look for how many handsome guys did not successfully be Liaoed?
输入描述:
Multiple groups of Case. (no more than 10000 groups)
Each group of data is 1 lines, with an integer N per line.
The N is 0 at the end of the input.
输出描述:
Each group of Output1 rows, a number of 1 lines representing the number of boys with no Liao to the group of data.
输入
1
2
3
4
0
输出
1
1
1
2
题解
规律。
暴力打表后会发现有规律,答案就是$\sqrt{n}$,比赛的时候写了个二分...
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; long long n; bool check(long long x) {
if((2LL + x) * x >= n) return 1;
return 0;
} int main() {
while(~scanf("%lld", &n)) {
if(n == 0) break;
long long L = 1, R = 1e9, ans;
while(L <= R) {
long long mid = (L + R) / 2;
if(check(mid)) ans = mid, R = mid - 1;
else L = mid + 1;
}
printf("%lld\n", ans);
}
return 0;
}
湖南大学ACM程序设计新生杯大赛(同步赛)L - Liao Han的更多相关文章
- 湖南大学ACM程序设计新生杯大赛(同步赛)J - Piglet treasure hunt Series 2
题目描述 Once there was a pig, which was very fond of treasure hunting. One day, when it woke up, it fou ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)A - Array
题目描述 Given an array A with length n a[1],a[2],...,a[n] where a[i] (1<=i<=n) is positive integ ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)B - Build
题目描述 In country A, some roads are to be built to connect the cities.However, due to limited funds, ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)I - Piglet treasure hunt Series 1
题目描述 Once there was a pig, which was very fond of treasure hunting. The treasure hunt is risky, and ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)E - Permutation
题目描述 A mod-dot product between two arrays with length n produce a new array with length n. If array ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)D - Number
题目描述 We define Shuaishuai-Number as a number which is the sum of a prime square(平方), prime cube(立方), ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)H - Yuanyuan Long and His Ballons
题目描述 Yuanyuan Long is a dragon like this picture? I don’t know, ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)G - The heap of socks
题目描述 BSD is a lazy boy. He doesn't want to wash his socks, but he will have a data structure called ...
- 湖南大学ACM程序设计新生杯大赛(同步赛)C - Do you like Banana ?
题目描述 Two endpoints of two line segments on a plane are given to determine whether the two segments a ...
随机推荐
- Installing kubectl
Installing kubectl Kubernetes uses a command-line utility called kubectl for communicating with the ...
- webpack中Module build failed: Unknown word (2:1)
在新建的webpack.config.js文件中配置好style-loader和css-loader,注意顺序为:style-loader,css-loader,less-loader,postcss ...
- swift开发常用代码片段
// 绑定事件 cell.privacySwitch.addTarget(self, action: #selector(RSMeSettingPrivacyViewController.switch ...
- LintCode 156: Merge Interval
LintCode 156: Merge Interval 题目描述 给出若干闭合区间,合并所有重叠的部分. 样例 给出的区间列表 => 合并后的区间列表: [ [ [1, 3], [1, 6], ...
- JavaScript事件和方法
单击一个超链接触发事件 1.用a标签的onclick <a href="#" onclick="js代码"> 这种写法呢,存在一种弊端,就是点击后会 ...
- Perl6多线程1 Thread : new / run
先看一个小例子: ) { #默认参数 say $name; } sub B(:name($name)) { #默认参数为 any say $name; } A(); A(); B(); B(name ...
- Mysql储存过程7: case
#用在储存过程中: create procedure k() begin declare number int; )); case number then select '>0'; else s ...
- ahttp
# -*- coding: utf-8 -*- # @Time : 2018/8/20 14:35 # @Author : cxa # @File : chttp.py # @Software: Py ...
- C/C++——C语言库函数大全
本文转载自:https://blog.csdn.net/yanfan0916/article/details/6450442###; 1. 分类函数: ctype.h int isalpha(int ...
- /proc/mounts介绍
现在的 Linux 系统里一般都有这么三个文件:/etc/fstab,/etc/mtab,和 /proc/mounts,比较容易让人迷惑.简单解释一下. /etc/fstab 是只读不写的,它提供的是 ...