题目链接:http://poj.org/problem?id=2480 题目大意: 题解: 我一直很欣赏数学题完美的复杂度 #include<cstring> #include<algorithm> #include<cstdio> #include<iostream> #include<cmath> using namespace std; typedef long long ll; <<)+; ll n; int main() {…

思路:首先给出几个结论: 1.gcd(a,b)是积性函数: 2.积性函数的和仍然是积性函数: 3.phi(a^b)=a^b-a^(b-1); 记 f(n)=∑gcd(i,n),n=p1^e1*p2^e2……; 则 f(n)=∑d*phi(n/d) (d是n的约数)           =∑(pi*ei+pi-ei)*pi^(ei-1). 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include&…

题意:求∑gcd(i,n),1<=i<=n思路:f(n)=∑gcd(i,n),1<=i<=n可以知道,其实f(n)=sum(p*φ(n/p)),其中p是n的因子.为什么呢?原因如下:1到n中有m个数字和n拥有公共的最大因子p,那么就需要把m*p加入答案中.问题是如何计算m的个数.因为假设某个数i与n的最大公约数为p,那么gcd(i,n) = p,可以得到gcd(i/p,n/p)=1.也就是说,有多少个i,就有多少个i/p与n/p互质.那么显然m即为n/p的欧拉函数φ(n/p). 知…

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…

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. …

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.…

/** 大意: 计算f(n) = ∑ gcd(i, N) 1<=i <=N. 思路: gcd(i,x*y) = gcd(i,x) * gcd(i, y ) 所以gcd 为积性函数 又因为积性函数的和函数 也是积性函数(具体数学,了解即可) f(n) = f(p1^a1 * p2^a2 * p3^a3*......* pn^an ) = f(p1^a1) * f(p2^a2) * f(p3* a3) ...... 现在我们先单独考虑一个 f(p1^a1) f(p^k)=1*φ(p^k)+ p*φ…

Happy 2006 Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 8359   Accepted: 2737 Description Two positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9...are al…

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…

题意:问从(0,0)到(x,y)(0≤x, y≤N)的线段没有与其他整数点相交的点数. 解法:只有 gcd(x,y)=1 时才满足条件,问 N 以前所有的合法点的和,就发现和上一题-- [poj 2478]Farey Sequence(数论--欧拉函数 找规律求前缀和) 求 x/y,gcd(x,y)=1 且 x<y 很像.   而由于这里 x可等于或大于y,于是就求 欧拉函数的前缀和*2+边缘2个点+对角线1个点. 1 #include<cstdio> 2 #include<cst…