题目传送门 题目描述 请找出一组合法解使得$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{n}$成立. 其中$x,y,z$为正整数且互不相同. 输入格式 一个整数$n$. 输出格式 一组合法的解$x,y,z$,用空格隔开. 若不存在合法的解,输出“-1”. 样例 样例输入 样例输出 2 3 6 数据范围与提示 对于$100%$的数据满足$n\leqslant {10}^4$ 要求答案中$x,y,z\leqslant 2\times {10}^9$ 提供…

C. Vladik and fractions 题目链接 http://codeforces.com/contest/743/problem/C 题面 Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct posit…

Codeforces Round #384 (Div. 2) 题目链接:Vladik and fractions Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer \(n\) he can represent fraction \(\frac{2}{n}\) as a sum of three distinct posi…

Description 请找出一组合法的解使得\(\frac {1}{x} + \frac{1}{y} + \frac {1}{z} = \frac {2}{n}\)成立 其中\(x,y,z\)为正整数并且互不相同 Input 一个整数\(n\) Output 一组合法的解\(x, y ,z\),用空格隔开 若不存在合法的解,输出\(-1\) 首先,最先容易想到的是令\(x,y,z\)其中一个数为\(n\),那么我们现在的问题就变成了求解这个式子. \[ \frac{1}{x}+\frac{1}…

传送门 Description Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction   as a sum of three distinct positive fractions in form . Help Vladik with that, i.e for a g…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as…

原题链接:https://vjudge.net/contest/331993#problem/I Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form . Hel…

ACM思维题训练集合 To confuse the opponents, the Galactic Empire represents fractions in an unusual format. The fractions are represented as two sets of integers. The product of numbers from the first set gives the fraction numerator, the product of numbers…

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 <…

题目 构造一组 $x, y, z$,使得对于给定的 $n$,满足 $\frac{1}{x}  + \frac{1}{y} + \frac{1}{z} =  \frac{2}{n}$. 分析: 样例二已经暴露了此题的本质. 显然 $n, (n+1), n(n+1)$ 为一组合法解.特殊地,当 $n=1$ 时,无解,因为此时 $n+1$ 与 $n(n+1)$ 相等(也可以证明没有其他形式的解). #include<bits/stdc++.h> using namespace std; int n;…