题目: GCD Again Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 125 Accepted Submission(s): 84   Problem Description Do you have spent some time to think and try to solve those unsolved problem afte…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1370 题意:给你n个整数,第i个整数为Xi.定义phi(k)为k的欧拉函数值,设pi为满足phi(pi)>=Xi的最小整数,题目就是要求sum(p1,p2,p3,...,pn) 思路:对任意x,有prime[i]<=x<prime[i+1]必定有EulerPhi[x]<=prime[i],要满足phi(p)>=x那么p必定为x后面的第一个素数,进行素数打表即可.…

Code: #include<cstdio> using namespace std; const int maxn=4000005; const int R=4000002; const int N=4000002; long long sumv[maxn],f[maxn]; int phi[maxn],prime[maxn],vis[maxn]; void solve(){ phi[1]=1; int cnt=0; for(int i=2;i<=R;++i){ if(!vis[i])…

This article is made by Jason-Cow.Welcome to reprint.But please post the writer's address. http://www.cnblogs.com/JasonCow/ /* ans= sigma(for each prime<=n) { { simga(for i=1 to n/now prime) phi[i] }*2 -1 } */ #include <cstdio> int GI(){ ,c=getch…

D - (例题)欧拉函数性质 Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Description Bamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popul…

1.HDU 2824   The Euler function 2.链接:http://acm.hdu.edu.cn/showproblem.php?pid=2824 3.总结:欧拉函数 题意:求(a,b)间的欧拉函数值的和. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio>…

题意:求1--n中满足gcd(x,y)的值为质数的数对(x,y)的数目 ( (x,y)和(y,x)算两个 ) sol: 设p[i]是一个质数,那么以下两个命题是等价的: 1.gcd(x,y)=1 2.gcd(x*p[i],y*p[i])=p[i] eg:gcd(36,25)=1,gcd(36*7,25*7)=7 所以对于1--n的所有质数p[i],求一下1<=x,y<=n/p[i]中所有gcd(x,y)=1的数对的数目即可. 如何求1--r范围内所有互质数对的数目? 考虑欧拉函数φ(x)=1.…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=2818 题意:给定整数N,求1<=x,y<=N且Gcd(x,y)为素数的数对(x,y)有多少对. 其实就是一个转化问题,求gcd(x, y) = k, 1 <= x, y <= n的对数等于求gcd(x, y) = 1, 1 <= x, y <= n/k的对数.那么接下来我们就只要枚举每个素数k=prime[i]了,然后用到欧拉函数就可以求出来了,Σ( 2*Σ(…

给定整数N,求1<=x,y<=N且Gcd(x,y)为素数的数对(x,y)有多少对. 如果两个数的x,y最大公约数是z,那么x/z,y/z一定是互质的 然后找到所有的素数,然后用欧拉函数求一下前缀和就行 #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; ; const int INF=0x3f3f3…

题目描述 给出 $n$ 和 $p$ ,求 $(\sum\limits_{i=1}^n\sum\limits_{j=1}^nij\gcd(i,j))\mod p$ . $n\le 10^{10}$ . 题解 欧拉函数(欧拉反演)+杜教筛 推式子: $$\begin{align}&\sum\limits_{i=1}^n\sum\limits_{j=1}^nij\gcd(i,j)\\=&\sum\limits_{i=1}^n\sum\limits_{j=1}^nij\sum\limits_{d|…