题目求[A,B]区间内与N互质数的个数. 可以通过求出区间内与N互质数的个数的前缀和,即[1,X],来得出[A,B]. 那么现在问题是求出[1,X]区间内与N互质数的个数,考虑这个问题的逆问题:[1,X]区间内与N不互质数的个数. 于是就可以先处理出N的所有质因数{p0,p1,p2,...,pn}. 而[1,X]能被pi整除的数有$\lfloor \frac X{p_i} \rfloor$个,再利用容斥原理除掉质因数公倍数重复计数的部分就能求出不互质个数. 最后X减去不互质个数就是互质个数了.…

Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 626    Accepted Submission(s): 234  Problem Description Given a number N, you are asked to count the number of integers between A and B…

也许更好的阅读体验 \(\mathcal{Description}\) \(t\)组询问,每次询问\(l,r,k\),问\([l,r]\)内有多少数与\(k\)互质 \(0<l<=r<=10^{15},k<=10^{9},t<=100\) \(\mathcal{Solution}\) 考虑 容斥 先求\([l,r]\)内出有多少数与\(k\)不互质,再用总数减去即可 将\(k\)质因子分解为\(p_1^{k_1}·p_2^{k_2}·...p_n^{k_n}\) 在\([l,…

Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently,…

容斥原理实现的关键在于:组合遍历,即如何遍历2^n种组合. 容斥原理的三种写法: DFS 队列数组 位数组 #include<stdio.h> #include<iostream> #include<stdlib.h> #include<string.h> using namespace std; const int maxn = 32000; bool isPrime[maxn]; int prime[maxn / 4], psize;//线性筛法必须用数…

Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4090    Accepted Submission(s): 1619 Problem Description Given a number N, you are asked to count the number of integers between A and B…

题:http://acm.hdu.edu.cn/showproblem.php?pid=4135 题意:求[A,B]与N互质的数的个数 #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> using namespace std; typedef long long ll; ; ll a[M]; ll A,B,N; int tot; void init(){ tot…

题目分析: 这种题目标题写莫比乌斯反演会不会显得太恐怖了,那就容斥算了. gcd不为1的肯定可以开根.所以把根式结果算出来就行了. 辣鸡题目卡我精度. 代码: #include<bits/stdc++.h> using namespace std; ; long long n; ]; void init(){ ;i<=;i++){ int p = i; mu[i] = -; ;j*j<=p;j++){ ; ) p/=j,cnt++; && cnt != ) mu[i…

#include <cstdio> #include <string.h> #include <cmath> using namespace std; #define MAXSIZE 40000 #define LL long long int prim[MAXSIZE]; LL A,B,n,odd,even; int k;//k记录素数的个数 void findPrim(LL n) { k=; ; i*i<=n; i++) //对n进行素因子分解 {//筛法求素…

题意: 求小于n (1 ≤ n ≤ 10^8)的数中,与n互质的数的四次方和. 知识点: 差分: 一阶差分: 设  则    为一阶差分. 二阶差分: n阶差分:     且可推出    性质: 1. 2. 差分序列: 给你一列数 a[i][1],a[i][2],a[i][3],a[i][4],a[i][5]…… 那么a[i][j]=a[i-1][j+1]-a[i-1][j], 即后一行是上一行相邻两项的差(第一行除外). 如果给你一个多项式, 比如 f(x)=(x+1)*(x+2)*……*(x…