题目链接:hdu 1299 Diophantus of Alexandria 题意: 给你一个n,让你找1/x+1/y=1/n的方案数. 题解: 对于这种数学题,一般都变变形,找找规律,通过打表我们可以发现这个答案只与这个数的因子有关. n=a1^p1*a2^p2*...*an^pn ans=((1+2*p1)*(1+2*p2)*...*(1+2*pn)+1)/2 #include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;++i)…

Diophantus of Alexandria Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2269    Accepted Submission(s): 851 Problem Description Diophantus of Alexandria was an egypt mathematician living in Ale…

思路: 因为x,y必须要大与n,那么将y设为(n+k);那么根据等式可求的x=(n2)/k+n;因为y为整数所以k要整除n*n; 那么符合上面等式的x,y的个数就变为求能被n*n整除的数k的个数,且k>=n; 那么k就在n到n*n之间,不过这样不则么好求. 因为x,y具有对称性,从y=(n+k)和x=(n2)/k+n;也可以看出(n*n/k,和k等价因为k*(n*n/k)=n*n): 由于任何数都可以拆成素数的乘积,形式为a1^x1*a2^x2*a3*^x3......; 那么因数的个数为(1+…

1/x + 1/y = 1/n 1<=n<=10^9给你 n 求符合要求的x,y有多少对 x<=y// 首先 x>n 那么设 x=n+m 那么 1/y= 1/n - 1/(n+m)// 1/y = m/(n*n+n*m) 所以满足 n*n|m 的m都是可以的 // 问题转化成求n*n 的约数个数 因数分解 然后用求约数个数公式// 最后答案记得 除 2 因为重复算了#include <iostream> #include <algorithm> #inclu…

hdoj 1299 Diophantus of Alexandria 链接:http://acm.hdu.edu.cn/showproblem.php?pid=1299 题意:求 1/x + 1/y = 1/n (x <= y) 的组数. 思路:转化为一个数的因子个数. 因为x,y,z 都是整数,令 y = n+k (倒数和相等,x,y 明显大于 n),带入式子可得 x = n*n / k + n :所以 x 的组数就与k相关了,只要 k 满足是 n*n 的约数,组数就 +1.假设 n = (p…

Description Diophantus of Alexandria was an egypt mathematician living in Alexandria. He was one of the first mathematicians to study equations where variables were restricted to integral values. In honor of him, these equations are commonly called d…

Diophantus of Alexandria was an egypt mathematician living in Alexandria. He was one of the first mathematicians to study equations where variables were restricted to integral values. In honor of him, these equations are commonly called diophantine e…

给一个数n问有多少种x,y的组合使$\frac{1}{x}+\frac{1}{y}=\frac{1}{n},x<=y$满足,设y = k + n,代入得到$x = \frac{n^2}{k} + n$,也就是求n^2的因子数量 /** @Date : 2017-09-08 10:45:12 * @FileName: HDU 1299 数论 分解.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link…

Diophantus of Alexandria Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2326    Accepted Submission(s): 887 Problem Description Diophantus of Alexandria was an egypt mathematician living in Ale…

Diophantus of Alexandria Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3242 Accepted Submission(s): 1287 Problem DescriptionDiophantus of Alexandria was an egypt mathematician living in Alexandri…