本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! 题目链接:HDU4910 正解:线性筛+$Miller-Rabin$ 解题报告: 我也是醉了,开始用$Pollard-rho$一直TLE,只好给成线性筛然后除,结果又一直$WA$,突然发现我的预处理只做到了$10^5$然而应该做到$10^6$… 我打了表,…

Problem about GCD Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 470    Accepted Submission(s): 77 Problem Description Given integer m. Find multiplication of all 1<=a<=m such gcd(a, m)=1 (cop…

Problem Description Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Here you can assume that there are enough coins for both kinds.Please calculate the maximal value that you cannot pay and the total number that you cannot pay. Inp…

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs. If z liters of water is measurable, you must…

分析:(别人写的) 对于所有(l, r)区间,固定右区间,所有(li, r)一共最多只会有log个不同的gcd值, 可以nlogn预处理出所有不同的gcd区间,这样区间是nlogn个,然后对于询问离线处理, 用类似询问区间不同数字的方法,记录每个不同gcd最后出现的位置,然后用树状数组进行维护 注:我是看了这段文字会的,但是他的nlogn预处理我不会,我会nlog^2n的 dp[i][j]代表以i为右端点,向左延伸2^j个点(包括i)的gcd,然后因为这样的gcd满足递减,所以可以二分找区间 代…

传送门 今天的签到题. 有一个很显然的结论,gcd(n∗m,k)≤2gcd(n*m,k)\le 2gcd(n∗m,k)≤2. 本蒟蒻是用的行列式求三角形面积证明的. 如果满足这个条件,就可以直接构造出一个满足限制的直角三角形了. 代码: #include<bits/stdc++.h> #define ll long long using namespace std; inline ll read(){ ll ans=0; char ch=getchar(); while(!isdigit(ch…

首先 m = 1 时 ans = 0对于 m > 1 的 情况 由于 1 到 m-1 中所有和m互质的数字,在 对m的乘法取模 运算上形成了群 ai = ( 1<=a<m && gcd(a,m) == 1 ) 所以 对于 a 必然存在b = a^(-1) = inv(a) 使得 a * b = 1 (mod m) 这里存在两种情况 a != b 那么最后的连乘式中a b均出现一次,相乘得1 a == b 那么最后的连乘式中只出现一个a 实际上所有 a = inv(a) 的…

传送门 •题意 已知 $a,b$,求满足 $x+y=a\ ,\ LCM(x,y)=b$ 条件的 $x,y$: 其中,$a,b$ 为正整数,$x,y$ 为整数: •题解 关键式子:设 $a,b$ 为正整数,如果有 $GCD(a,b)=1$,则有 $GCD(a+b,ab)=1$: 证明可以看这里[…

传送门 •题意 一直整数$a,b$,有 $\left\{\begin{matrix}x+y=a\\ LCM(x*y)=b \end{matrix}\right.$ 求$x,y$ •思路 解题重点:若$gcd(p,q)=1$,则$gcd(p+q,pq)=1$ 设$gcd(x,y)=g$,令$p=\frac{x}{g},q=\frac{y}{g}$,$p,q$互素 则$\left\{\begin{matrix}x+y=p*g+q*g=(p+q)g=a\\ LCM(x,y)=\frac{xy}{g}=…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5869 Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting.…

Different GCD Subarray Query Problem Description   This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Problem JGCD Extreme (II)Input: Standard Input Output: Standard Output Given the value of N, you will have to find the value of G. The definition of G is given below: Here GCD(i,j) means the…

Problem A GCD Input: Standard Input Output: Standard Output Given the value of N, you will have to find the value of G. The definition of G is given below: Here GCD(i,j) means the greatest common divisor of integer i and integer j. For those who have…

II U C   ONLINE   C ON TEST  Problem D: GCD LCM Input: standard input Output: standard output The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smalle…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studyin…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 221    Accepted Submission(s): 58 Problem Description This is a simple problem. The teacher gives Bob a list of probl…

Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5869 Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 This is a simple problem. The teacher gives Bob a list of problems about GCD (Great…

Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as…

树状数组... Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1541    Accepted Submission(s): 599 Problem Description This is a simple problem. The teacher gives Bob a lis…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1328    Accepted Submission(s): 504 Problem Description This is a simple problem. The teacher gives Bob a list of pro…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 828    Accepted Submission(s): 300 Problem Description  and aN. In other words, ai,ai+1,⋯,aj−1,aj is a subarray of a,…

Problem Different GCD Subarray Query 题目大意 给定n个数的序列,有q个询问,每次询问一个区间中所有子区间所形成不同的gcd的数量. 解题分析 由于固定一个数为右端点,所能形成的gcd共有logn,所以可以预处理出每个数为右端点所能形成的gcd,相同gcd取左端点靠右的. 然后将询问离线,按照r从小到大排序.处理gcd重复的方法是将相同的gcd值保留左端点较大的. 参考程序 #include <map> #include <set> #inclu…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 681    Accepted Submission(s): 240 Problem Description This is a simple problem. The teacher gives Bob a list of prob…

I I U C   O N L I N E   C  Problem D: GCD LCM Input: standard input Output: standard output The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smallest…

ProjectEuler_做题记录 简单记录一下. problem 441 The inverse summation of coprime couples 神仙题.考虑答案为: \[\begin{array}{c} S(n) & = & \sum_{i = 1} ^ n \sum_{p = 1} ^ i \sum_{q = p + 1} ^ i \frac {1}{pq}[p + q \geq i][gcd(p, q) = 1] \\ & = & \sum_{i = 1}…

*注意:这套题目应版权方要求,不得公示题面. 从这里开始 Problem A XOR Problem B GCD Problem C SEG 表示十分怀疑出题人水平,C题数据和标程都是错的.有原题,差评. Problem A XOR 题目大意 最小异或生成树 出门左拐Codeforces 888G. Code #include <iostream> #include <cstdlib> #include <cstdio> #include <vector>…

Problem D: GCD Time Limit: 1 Sec  Memory Limit: 1280 MBSubmit: 179  Solved: 25[Submit][Status][Web Board] Description XiaoMingfoundthecomputetimeof Input The first line is an positive integer T. ( Output In each test case, output the compute result o…

 1.burnside定理,polya计数法 这个专题我单独写了个小结,大家可以简单参考一下:polya 计数法,burnside定理小结 2.置换,置换的运算 置换的概念还是比较好理解的,<组合数学>里面有讲.对于置换的幂运算大家可以参考一下潘震皓的那篇<置换群快速幂运算研究与探讨>,写的很好. *简单题:(应该理解概念就可以了) pku3270 Cow Sorting http://acm.pku.edu.cn/JudgeOnline/problem?id=3270 pku…

Problem J GCD Extreme (II) Input: Standard Input Output: Standard Output Given the value of N, you will have to find the value of G. The definition of G is given below:   Here GCD(i,j) means the greatest common divisor of integer i and integer j. For…