题意: 给你一个欧拉函数值 phi(n),问最小的n是多少. phi(n) <= 100000000 , n <= 200000000 解题思路: 对于欧拉函数值可以写成 这里的k有可能是等于0的,所以不能直接将phi(n)分解质因子.但是可以知道(Pr - 1)是一定存在的,那就直接枚举素数,满足phi(n) % (Pr-1)的都加进去,然后对这些素数进行爆搜...说到底还是暴力啊...想不到什么巧妙的办法了,最后需要注意的是,一遍枚举完各个素数后phi(n)除后还剩now,现在要判断(no…

题目链接:uva 10837 - A Research Problem 题目大意:给定一个phin.要求一个最小的n.欧拉函数n等于phin 解题思路:欧拉函数性质有,p为素数的话有phip=p−1;假设p和q互质的话有phip∗q=phip∗phiq 然后依据这种性质,n=pk11(p1−1)∗pk22(p2−1)∗⋯∗pkii(pi−1),将全部的pi处理出来.暴力搜索维护最小值,尽管看上去复杂度很高,可是由于对于垒乘来说,增长很快,所以搜索范围大大被缩小了. #include <cstdi…

题目:给出n,求gcd(1,2)+gcd(1,3)+gcd(2,3)+gcd(1,4)+gcd(2,4)+gcd(3,4)+...+gcd(1,n)+gcd(2,n)+...+gcd(n-1,n) 此题和UVA 11426 一样,不过n的范围只有20000,但是最多有20000组数据. 当初我直接照搬UVA11426,结果超时,因为没有预处理所有的结果(那题n最多4000005,但最多只有100组数据),该题数据太多了额... 思路:令sum(n)=gcd(1,n)+gcd(2,n)+...+g…

题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70017#problem/O 题意是给你n,求所有gcd(i , j)的和,其中1<=i <j <n. 要是求gcd(n , x) = y的个数的话,那么就是求gcd(n/y , x/y) = 1的个数,也就是求n/y的欧拉函数.这里先预处理出欧拉函数,然后通过类似筛法的技巧筛选出答案累加起来. #include <iostream> #include &l…

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 trouble understanding summation notation, the meaning of G i…

传送门 Longge's problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7327   Accepted: 2416 Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms.…

Given the value of N, you will have to find the value of G. The definition of G is given below:G =i<N∑i=1j∑≤Nj=i+1GCD(i, j)Here GCD(i, j) means the greatest common divisor of integer i and integer j.For those who have trouble understanding summation no…

题意: 题目背景略去,将这道题很容易转化为,给出n求,n以内的有序数对(x, y)互素的对数. 分析: 问题还可以继续转化. 根据对称性,我们可以假设x<y,当x=y时,满足条件的只有(1, 1). 设f(n)为 集合S{(x, y) | x<y且x.y互素} 的个数,则所求答案为2f(n)+1 f(n)表达式为: ,其中φ(n)为欧拉函数 这里有欧拉函数的一些介绍 #include <cstdio> ; ], sum[maxn + ]; void phi_table(int n)…

题意: 这道题和POJ 3090很相似,求|x|≤a,|y|≤b 中站在原点可见的整点的个数K,所有的整点个数为N(除去原点),求K/N 分析: 坐标轴上有四个可见的点,因为每个象限可见的点数都是一样的,所以我们只要求出第一象限可见的点数然后×4+4,即是K. 可见的点满足gcd(x, y) = 1,于是将问题转化为x∈[1, a], y∈[1, b],求gcd(x, y) = 1的个数. 类比HDU 1695可以用莫比乌斯反演来做,我还写了普通的和分块加速的两份代码,交上去发现运行时间相差并不…

分析:枚举每个数的贡献,欧拉函数筛法 #include <cstdio> #include <iostream> #include <ctime> #include <vector> #include <cmath> #include <map> #include <queue> #include <algorithm> #include <cstring> using namespace std;…

题意:给一个N,和公式 求G(N). 分析:设F(N)= gcd(1,N)+gcd(2,N)+...gcd(N-1,N).则 G(N ) = G(N-1) + F(N). 设满足gcd(x,N) 值为 i 的且1<=x<=N-1的x的个数为 g(i,N). 则F(N)  = sigma{ i * g(i,N) }. 因为gcd(x,N) == i 等价于 gcd(x/i, N/i)  == 1,且满足gcd(x/i , N/i)==1的x的个数就是 N/i 的欧拉函数值.所以g(i,N) 的值…

对每个n,答案就是(phi[2]+phi[3]+...+phi[n])*2+1,简单的欧拉函数应用. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<list> #i…

题意:给一个 n,m,统计 2 和 n!之间有多少个整数x,使得x的所有素因子都大于M. 析:首先我们能知道的是 所有素数因子都大于 m 造价于 和m!互质,然后能得到 gcd(k mod m!, m!) = 1,也就是只要能求出不超过 m!且和 m! 互质的个数就好,也就是欧拉函数呗,但是,,,m!也非常大,根本无法用筛选法进行,但是可以通过递推进行,根据欧拉公式,能知道n! 和 (n-1)! 如果n为中素数,那么它们的素因子肯定是一样的,如果n是素数,那么就会多一项,所以我们能够得到递推式.…

Longge's problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6383   Accepted: 2043 Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now…

链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2435 题意: 给定正整数N和M,统计2和N!之间有多少个整数x满足:x的所有素因子都大于M(2≤N≤1e7,1≤M≤N,N-M≤1e5).输出答案除以100000007的余数.例如,N=100,M=10时答案为43274465. 分析: 因为M≤N,所以N!是M!的整数倍.“所有素…

send a table When participating in programming contests, you sometimes face the following problem: You knowhow to calcutale the output for the given input values, but your algorithm is way too slow to everpass the time limit. However hard you try, yo…

Longge's problem   Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i,…

https://vjudge.net/problem/UVA-10837 求最小的n,使phi(n)=m #include<cstdio> #include<algorithm> #define N 10011 int prime[N],cnt; bool v[N],vis[N]; int p[N],tot,ans; void pre_prime() { ;i<N;i++) { if(!v[i]) prime[++cnt]=i; ;j<=cnt;j++) { if(i*…

发现自己搜索真的很弱,也许做题太少了吧.代码大部分是参考别人的,=_=|| 题意: 给出一个phi(n),求最小的n 分析: 回顾一下欧拉函数的公式:,注意这里的Pi是互不相同的素数,所以后面搜索的时候要进行标记. 先找出所有的素数p,满足(p - 1)整除题目中所给的phi(n) 然后暴搜.. 素数打表打到1e4就够了,如果最后剩下一个大素数单独进行判断. #include <cstdio> #include <cmath> #include <cstring> #i…

链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1155 题意: 在满足|x|≤a,|y|≤b(a≤2000,b≤2000000)的网格中,除了原点之外的整点(即x,y坐标均为整数的点)各种着一棵树.树的半径可以忽略不计,但是可以相互遮挡.求从原点能看到多少棵树.设这个值为K,要求输出K/N,其中N为网格中树的总数. 分析: 显然4…

链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1761 题意: 有一道比赛题目,输入两个整数x.y(1≤x,y≤n),输出某个函数f(x,y).有位选手想交表(即事先计算出所有的f(x,y),写在源代码里),但表太大了,源代码超过了比赛限制,需要精简.那道题目有一个性质,即很容易根据f(x,y)算出f(x*k, y*k)(k是任意…

/** 题目:Help Tomisu UVA - 11440 链接:https://vjudge.net/problem/UVA-11440 题意:给定正整数N和M, 统计2和N!之间有多少个整数x满足,x的所有素因子都大于M (2<=N<=1e7, 1<=M<=N, N-M<=1E5) 输出答案除以1e8+7的余数. 思路: lrjP338 由于x的所有素因子都>M:那么x与M!互质. 根据最大公约数的性质,对于x>y,x与y互质,那么x%y与y也互质. 由于N…

pid=26358">https://uva.onlinejudge.org/index.phpoption=com_onlinejudge&Itemid=8&category=279&page=show_problem&problem=3937 题目:http://acm.bnu.edu.cn/v3/external/124/12493.pdf 大致题意:圆上有偶数n个点.每m个点连起来.最后能够把全部点串联起来就合法.问有多少个m能够完毕串联,串联后形状…

Send a Table Input: Standard Input Output: Standard Output When participating in programming contests, you sometimes face the following problem: You know how to calcutale the output for the given input values, but your algorithm is way too slow to ev…

Longge's problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6918   Accepted: 2234 Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now…

Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N.…

Longge's problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9190   Accepted: 3073 Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now…

题目: Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. "Oh…

6322.Problem D. Euler Function 题意就是找欧拉函数为合数的第n个数是什么. 欧拉函数从1到50打个表,发现规律,然后勇敢的水一下就过了. 官方题解: 代码: //1004-欧拉函数水题 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; typede…

Description Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. …