题意: 求sigma phi(n) 思路: 线性递推欧拉函数 (维护前缀和) //By SiriusRen #include <cstdio> using namespace std; #define maxn 1000005 #define int long long int n,p[maxn+100],s[maxn+100],phi[maxn+100],tot; void Phi(){ for(int i=2;i<=maxn;i++){ if(!s[i])p[++tot]=i,phi…

A - Farey Sequence Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2478 Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 &l…

http://poj.org/problem?id=3090 法雷级数 法雷级数的递推公式非常easy:f[1] = 2; f[i] = f[i-1]+phi[i]. 该题是法雷级数的变形吧,答案是2*f[i]-1. #include <stdio.h> #include <iostream> #include <map> #include <set> #include <stack> #include <vector> #inclu…

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.…

上三角行恰好是[1,n-1]的欧拉函数 http://www.luogu.org/problem/show?pid=2158#sub //#pragma comment(linker, "/STACK:167772160") #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <queue> #include…

Relatives AC代码 Relatives Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16186   Accepted: 8208 Description Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relativel…

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…

找出N*N范围内可见格点的个数. 只考虑下半三角形区域,可以从可见格点的生成过程发现如下规律: 若横纵坐标c,r均从0开始标号,则 (c,r)为可见格点 <=>r与c互质 证明: 若r与c有公因子1<b<min(r,c),则(c/b, r/b)在线段(0, 0)(c, r)上,则(c, r)不是可见格点.(充分性) 若r与c互质,显然线段上不存在整点,则(c, r)不是可见格点.(必要性) φ(n)表示不超过n且与n互素的正整数的个数,称为n的欧拉函数值 也就是横坐标增1后纵坐标合…

链接:http://poj.org/problem?id=3090 题意:在坐标系中,从横纵坐标 0 ≤ x, y ≤ N中的点中选择点,而且这些点与(0,0)的连点不经过其它的点. 思路:显而易见,x与y仅仅有互质的情况下才会发生(0,0)与(x,y)交点不经过其它的点的情况,对于x,y等于N时,能够选择的点均为小于等于N而且与N互质的数,共Euler(N)个,而且不重叠.所以能够得到递推公式aa[i]=aa[i]+2*Euler(N). 代码: #include <iostream> #i…

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,…