CodeForces 743C Vladik and fractions (数论)
题意:给定n,求三个不同的数满足,2/n = 1/x + 1/y + 1/z。
析:首先1是没有解的,然后其他解都可以这样来表示 1/n, 1/(n+1), 1/(n*(n+1)),这三个解。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = (1<<20) + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
} int main(){
while(cin >> n){
if(n == 1) printf("-1\n");
else printf("%d %d %d\n", n, n+1, n*(n+1));
}
return 0;
}
CodeForces 743C Vladik and fractions (数论)的更多相关文章
- Codeforces 743C - Vladik and fractions (构造)
Codeforces Round #384 (Div. 2) 题目链接:Vladik and fractions Vladik and Chloe decided to determine who o ...
- Codeforces Round #384 (Div. 2) C. Vladik and fractions 构造题
C. Vladik and fractions 题目链接 http://codeforces.com/contest/743/problem/C 题面 Vladik and Chloe decided ...
- codeforces 811E Vladik and Entertaining Flags(线段树+并查集)
codeforces 811E Vladik and Entertaining Flags 题面 \(n*m(1<=n<=10, 1<=m<=1e5)\)的棋盘,每个格子有一个 ...
- [CodeForces - 1225D]Power Products 【数论】 【分解质因数】
[CodeForces - 1225D]Power Products [数论] [分解质因数] 标签:题解 codeforces题解 数论 题目描述 Time limit 2000 ms Memory ...
- Bash and a Tough Math Puzzle CodeForces 914D 线段树+gcd数论
Bash and a Tough Math Puzzle CodeForces 914D 线段树+gcd数论 题意 给你一段数,然后小明去猜某一区间内的gcd,这里不一定是准确值,如果在这个区间内改变 ...
- 【44.64%】【codeforces 743C】Vladik and fractions
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- Codeforces Round #384 (Div. 2) C. Vladik and fractions(构造题)
传送门 Description Vladik and Chloe decided to determine who of them is better at math. Vladik claimed ...
- Codeforces 396B On Sum of Fractions 数论
题目链接:Codeforces 396B On Sum of Fractions 题解来自:http://blog.csdn.net/keshuai19940722/article/details/2 ...
- vjudge I - Vladik and fractions 一道小学生的提。
原题链接:https://vjudge.net/contest/331993#problem/I Vladik and Chloe decided to determine who of them i ...
随机推荐
- windbg学习---.browse打开一个新的command 窗口
.browse r eax .browse <command>将会显示新的命令浏览窗口和运行给出的命令
- Ajax完整结构和删除
1.ajax完整结构 注意:(1)最后一个没有"," (2)ajax对网速要求高,最好有各种提示和使用按钮(可使其失效,防止重复加载) $.ajax({ url: "aj ...
- android 对View的延时更换内容
一.当ImageView按下时可以跟换一张按下效果的图片进行显示,使用postDelayed即可以让view在规定时间后执行run()中的内容 img.setImageResource(R.drawa ...
- log4j相对路径找不到,处理方法
http://blog.csdn.net/u012345283/article/details/40821833?utm_source=tuicool&utm_medium=referral
- Canvas 实现图片剪切
用户上传头像然后截图的需求很常见,很多做法是把图像发送到后端,把裁剪后的结果发送给浏览器,这种方式会增加处理时延.最近正好学习了HTML5里的canvas,发现它的图片处理功能比较强大,就打算用can ...
- zabbix3.0安装部署文档
zabbix v3.0安装部署 摘要: 本文的安装过程摘自http://www.ttlsa.com/以及http://b.lifec-inc.com ,和站长凉白开的<ZABBIX从入门到精通v ...
- SpringMVC保存数据到mysql乱码问题
SpringMVC保存数据到mysql乱码问题 乱码问题常见配置 一.web.xml配置过滤器 <filter> <filter-name>encoding-filter< ...
- Spark RDD aggregateByKey
aggregateByKey 这个RDD有点繁琐,整理一下使用示例,供参考 直接上代码 import org.apache.spark.rdd.RDD import org.apache.spark. ...
- Coursera 机器学习课程 机器学习基础:案例研究 证书
完成了课程1 机器学习基础:案例研究 贴个证书,继续努力完成后续的课程:
- 浅谈Oracle表之间各种连接
Oracle表之间的连接分为三种: 1.内连接(自然连接) 2.外连接 2.1.左外连接(左边的表不加限制,查询出全部满足条件的结果) 2.2.右外连接(右边的表不加限制,查询出全部满足条件的结果) ...