Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,…

题目给一个数字集合,问有多少个小于n的正整数能被集合里至少一个元素整除. 当然是容斥原理来计数了,计算1个元素组合的有几个减去2个元素组合的LCM有几个加上3个元素组合的LCM有几个.注意是LCM. 而[1,n]中能被x整除的数字有$ \lfloor \frac nx \rfloor$个,因为设有$t$个,$x \times t \leqslant n$. 计算多个数LCM利用:$lcm(a,b)=a/gcd(a,b)\times b $,$lcm(a,b,c)=lcm(a,lcm(b,c))$…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5664    Accepted Submission(s): 1630 Problem Description   Now you get a number N, and a M-integers set, you shoul…

How many integers can you find Time Limit : 12000/5000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 95   Accepted Submission(s) : 30 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description   Now…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6434    Accepted Submission(s): 1849 Problem Description   Now you get a number N, and a M-integers set, you shou…

//设置m,Q小于n可以设置如何几号m随机多项整除 //利用已知的容斥原理 //ans = 数是由数的数目整除 - 数为整除的两个数的数的最小公倍数 + 由三个数字... #include<cstdio> #include<cstring> #include<iostream> using namespace std ; const int maxn = 110 ; typedef __int64 ll ; int a[maxn] ; int len ; int n ,…

题意 就是给出一个整数n,一个具有m个元素的数组,求出1-n中有多少个数至少能整除m数组中的一个数 (1<=n<=10^18.m<=20) 题解 这题是容斥原理基本模型. 枚举n中有多少m中元素的个数,在结合LCM考虑容斥. #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<algorithm> using namespa…

题目链接: http://acm.hdu.edu.cn/showproblem.php? pid=1796 题目大意: 给你一个整数N.和M个整数的集合{A1.A2.-.Am}.集合内元素为非负数(包括零),求小于N的 正整数(1~N-1)中,能被M个整数的集合中随意一个元素整除的正整数个数. 比如N = 12.M = {2,3},在1~N-1中,能被2整除的数为{2,4,6.8.10},能被3整除的数为 {3.6,9}.则所求集合为{2,3,4.6,8,9,10},共7个,则答案为7. 思路:…

题目链接Hdu4135 Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1412    Accepted Submission(s): 531 Problem Description Given a number N, you are asked to count the number of integers betwe…

这个分类怎么觉得这么水呢.. 这个分类做到尾的模板集: //gcd int gcd(int a,int b){return b? gcd(b, a % b) : a;} //埃氏筛法 O(nlogn) int prime[MAX_N]; bool is_prime[MAX_N];//i是不是素数 int sieve() { ; ; i <= n; i++) is_prime[i] = true; is_prime[] = is_prime[] = false; ; i <= n; i++) {…