LCM Walk Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5584 Description A frog has just learned some number theory, and can't wait to show his ability to his girlfriend. Now the frog is sitting on a grid map o…

A frog has just learned some number theory, and can't wait to show his ability to his girlfriend. Now the frog is sitting on a grid map of infinite rows and columns. Rows are numbered 1,2,⋯from the bottom, so are the columns. At first the frog is sit…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5584 题意:(x, y)经过一次操作可以变成(x+z, y)或(x, y+z)现在给你个点(ex, ey)输出有多少种可能的起点,这些起点经过若干次操作能变成(ex, ey). 思路:我们考虑其中的一次变换,现在为(x, y)(y > x)那么它显然是由(x, y - z)变换来的.其中z = lcm(x, y - z),lcm(x, y - z) = x*(y-z)/gcd(x, y - z).…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5584 题意: 分析: 这题比赛的时候卡了很久,一直在用数论的方法解决. 其实从终点往前推就可以发现,整个过程中的点的gcd都是一样的,利用这个性质倒着搜索一遍就好了. 相同的gcd均为gcd(6,10) 以(6,10)为例,假设倒数第二个点到达(6−x∗gcd,10),那么x=(6−x∗gcd)∗10/gcd,设6=k1∗gcd, 10=k2∗gcd,那么x满足x=k1∗k2/(k2+1),每次只…

Problem Description A frog has just learned some number theory, and can't wait to show his ability to his girlfriend. Now the frog ,,⋯ from the bottom, so are the columns. At first the frog is sitting at grid (sx,sy), and begins his journey. To show…

没用运用好式子...想想其实很简单,首先应该分析,由于每次加一个LCM是大于等于其中任何一个数的,那么我LCM加在哪个数上面,那个数就是会变成大的,这样想,我们就知道,每个(x,y)对应就一种情况. 第二个突破口是,那个式子,我们可以想一想,是不是可以把数进行拆分,我们发现 a=x*k,b=y*k;其中k=gcd(a,b) 并且 x和y互质,这样带入式子,这样我们就把(x*k,y*k)推到了(x*k,x*y+x*y*k),化简即k *(x,(x+1)*y),gcd仍然是k,反过来,我们只需要保证…

LCM Walk Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 47    Accepted Submission(s): 31 Problem Description A frog has just learned some number theory, and can't wait to show his ability to hi…

题目链接:LCM Walk Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 491    Accepted Submission(s): 254 Problem Description A frog has just learned some number theory, and can't wait to show his abili…

http://acm.hdu.edu.cn/showproblem.php?pid=5584 题意: 现在有坐标(x,y),设它们的最小公倍数为k,接下来可以移动到(x+k,y)或者(x,y+k).现在给出终点坐标,求有多少个起点可以通过这种变化方式得到终点. 思路: 现在假设我们处于(x,y)这个坐标上,x和y的最大公约数为k,x和y用k来表示的话可以表示为x=$m_{1}$,y=$m_{2}$. 那么接下来可以得到($m_{1}$k,$m_{2}$k+$m_{1}$$m_{2}$k)或者 (…

题目链接: L - LCM Walk HDU - 5584 题目大意:首先是T组测试样例,然后给你x和y,这个指的是终点.然后问你有多少个起点能走到这个x和y.每一次走的规则是(m1,m2)到(m1+lcm(m1,m2),m2)或者(m1,m2+lcm(m1,m2)). 具体思路: lcm(m1,m2)=m1*m2/(gcd(m1,m2)).然后m1就能表示成t1*gcd(m1,m2),m2能表示成t2*gcd(m1,m2).然后(m1,m2)就能走到(t1*gcd(m1,m2),t2*gcd(…

GCD?LCM! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 316    Accepted Submission(s): 200 Output T lines, find S(n) mod 258280327. Sample Input 8 1 2 3 4 10 100 233 11037 Sample Output 1 5 1…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2982    Accepted Submission(s): 1305 Problem Description Given two positive integers G and L, could you tell me how many solutions of…

Least Common Multiple (HDU - 1019) [简单数论][LCM][欧几里得辗转相除法] 标签: 入门讲座题解 数论 题目描述 The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7…

gcd即最大公约数,lcm即最小公倍数. 首先给出a×b=gcd×lcm 证明:令gcd(a,b)=k,a=xk,b=yk,则a×b=x*y*k*k,而lcm=x*y*k,所以a*b=gcd*lcm. 所以求lcm可以先求gcd,而求gcd的方法就是辗转相除法,也叫做欧几里德算法,核心为gcd(m,n)=gcd(n,m%n) 证明:令 k=gcd(m,n),则 k|m 并且 k|n; 令 j=gcd(n, m mod n), 则j|n 并且 j|(m mod n); 对于m, 可以用n 表示为…

HDU 1005 Number Sequence(数论) Problem Description: A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multipl…

UVA.12716 GCD XOR (暴力枚举 数论GCD) 题意分析 题意比较简单,求[1,n]范围内的整数队a,b(a<=b)的个数,使得 gcd(a,b) = a XOR b. 前置技能 XOR的性质 GCD 由于题目只给出一个n,我们要求对数,能做的也始终暴力枚举a,b,这样就有n^2的复杂度,由于n很大,根本过不了. 于是我们就想用到其中一些性质,如XOR 与GCD,不妨假设 a xor b = c,并且根据题意还知道, gcd(a,b) = c,也就说明c一定是a的因子,所以在枚举的…

题目链接:hdu 5381 The sum of gcd 将查询离线处理,依照r排序,然后从左向右处理每一个A[i],碰到查询时处理.用线段树维护.每一个节点表示从[l,i]中以l为起始的区间gcd总和.所以每次改动时须要处理[1,i-1]与i的gcd值.可是由于gcd值是递减的,成log级,对于每一个gcd值记录其区间就可以.然后用线段树段改动,可是是改动一个等差数列. #include <cstdio> #include <cstring> #include <vecto…

七夕节 (HDU - 1215) [简单数论][找因数] 标签: 入门讲座题解 数论 题目描述 七夕节那天,月老来到数字王国,他在城门上贴了一张告示,并且和数字王国的人们说:"你们想知道你们的另一半是谁吗?那就按照告示上的方法去找吧!" 人们纷纷来到告示前,都想知道谁才是自己的另一半.告示如下: 数字N的因子就是所有比N小又能被N整除的所有正整数,如12的因子有1,2,3,4,6. 你想知道你的另一半吗? Input 输入数据的第一行是一个数字T(1<=T<=500000)…

input T 1<=T<=1000 x y output 有多少个起点可以走n(n>=0)步走到(x,y),只能从(x,y)走到(x,y+lcm(x,y))/(x+lcm(x,y),y) 标准解:从(x,y0)走到(x,y),则设x=ag,y0=bg,g=gcd(x,y0),有y=bg+abg=(a+1)bg,因为a,b互质,a,(a+1)互质,所以a和(a+1)b互质,所以若可以从(x,y0)走到(x,y),有gcd(x,y0)=gcd(x,y),然后将x和y中gcd(x,y)除去之…

A frog has just learned some number theory, and can't wait to show his ability to his girlfriend. Now the frog is sitting on a grid map of infinite rows and columns. Rows are numbered 1,2,⋯ from the bottom, so are the columns. At first the frog is si…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5584 给一个坐标(ex, ey),问是由哪几个点走过来的.走的规则是x或者y加上他们的最小公倍数lcm(x, y). 考虑(ex, ey)是由其他点走过来的,不妨设当走到(x,y)时候,gcd(x, y)=k,x=k*m1, y=k*m2. 下一步有可能是(x, y+x*y/gcd(x, y))或者是(x+x*y/gcd(x,y), y). 用k和m1,m2来表示为(k*m1, k*m2+m1*m2…

Big Number 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1722 ——每天在线,欢迎留言谈论. 题目大意: 给你两个数 n1,n2 . 然后你有一块蛋糕,提前切好,使得不管来 n1 还是 n2 个人都能够当场平均分配. 求 “提前切好” 的最小蛋糕块数. 知识点: (请无视)公式:N = a + b + gcd(a, b) : 思路: (勿无视)先份成p块,然后再拼到一起,再从原来开始的地方,将蛋糕再分成q份,中间肯定会出现完全重合的块…

题目描述: Given two positive integers a and b,find suitable X and Y to meet the conditions: X+Y=a Least Common Multiple (X, Y) =b Input Input includes multiple sets of test data.Each test data occupies one line,including two positive integers a(1≤a≤2*10^…

wls 有一个整数 n,他想将 1 − n 这 n 个数字分成两组,每一组至少有一个数,并且使得两组数字的和的最大公约数最大,请输出最大的最大公约数. Input 输入一行一个整数 n. 2 ≤ n ≤ 1, 000, 000, 000 Output 输出一行一个整数表示答案. Sample Input 6 Sample Output 7 思路: 我们求1到n的sum和为sum, 分成的两组的sum和分别是sum1和sum2, 那么根据题意我们知道 sum1+sum2=sum 所求的答案就是 g…

HDU4497 GCD and LCM 如果 \(G \% L != 0\) ,那么输出 \(0\) . 否则我们有 \(L/G=(p_1^{r_1})\cdot(p_2^{r_2})\cdot(p_3^{r_3})\cdots(p_m^{r_m})\) . 我们又有: \[ x=(p_1^{i_1})\cdot(p_2^{i_2})\cdot(p_3^{i_3})\cdots(p_m^{i_m}) \\ y=(p_1^{j_1})\cdot(p_2^{j_2})\cdot(p_3^{j_3})…

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}=…

A/B Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7238    Accepted Submission(s): 5754 Problem Description 要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973) = 1).   Input 数据的第一行是一…

GT and numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1818    Accepted Submission(s): 490 Problem Description You are given two numbers N and M. Every step you can get a new N in the wa…

#include<bits/stdc++.h> #define ll long long using namespace std; /* ll gcd(ll a, ll b) {//非递归版 ll t; while(b) { t = a % b; a = b; b = t; } return a; } */ ll gcd(ll a, ll b) {//递归版 if(b == 0)return a; else return gcd(b, a % b); } int main() { ll a,…

CRB and Candies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 947    Accepted Submission(s): 442 Problem Description CRB has N different candies. He is going to eat K candies.He wonders how ma…