Diophantus of Alexandria Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3210    Accepted Submission(s): 1269 Problem Description Diophantus of Alexandria was an egypt mathematician living in Al…

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…

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…

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…

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…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…

http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…

http://acm.hdu.edu.cn/showproblem.php?pid=4859 题目大意: 在一个矩形周围都是海,这个矩形中有陆地,深海和浅海.浅海是可以填成陆地的. 求最多有多少条方格线满足两侧分别是海洋和陆地 这道题很神 首先考虑一下,什么情况下能够对答案做出贡献 就是相邻的两块不一样的时候 这样我们可以建立最小割模型,可是都说是最小割了 无法求出最大的不相同的东西 所以我们考虑转化,用总的配对数目 - 最小的相同的对数 至于最小的相同的对数怎么算呢? 我们考虑这样的构造方法:…