lightoj1197区间素数筛】的更多相关文章

模板题,不过好像有点问题,当a==1的时候,答案把一也算进去了,要减去 #include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cassert> #include<iomanip> #include<cstdlib…
题意: 给定一个区间a,b,a-b>=100000,1<=a<=b<=231,求出给定a,b区间内的素数的个数 区间素数筛 (a+i-1)/ ii向上取整,当a为 i 的整数倍时,直接从a开始标记,当a不是 i 的整数倍时,得出 i 的整数倍且大于a,如果a小于等于 i 时,从2i开始标记 for(ll i=2;i*i<b;++i) { if(is_prime_small[i]) { for(ll j=2*i;j*j<b;j+=i) { is_prime_small[j…
The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no prop…
Description The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is…
The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no prop…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1197 给你a和b求a到b之间的素数个数. 先在小区间素数筛,大区间就用类似素数筛的想法,把a到b之间不是素数的标记出来.因为b-a最多1e5的大小,所以每组数据的时间复杂度最多就o(1e5 log1e5). #include <iostream> #include <cstdio> #include <cstring> using namespace std…
Help Hanzo 题意:求a~b间素数个数(1 ≤ a ≤ b < 231, b - a ≤ 100000).     (全题在文末) 题解: a~b枚举必定TLE,普通打表MLE,真是头疼.. b - a ≤ 100000 是关键. 类似素数筛的方法: 1.初始化vis[]=0 ; 2.素数的倍数vis[]=1; 3.  b较小时,素数筛解决   b很大时,素数筛的vis[]会MLE,此时用vis2[i-a]保存vis[i]就不会MLE 了.. #include<iostream>…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1197 题意:给你两个数 a b,求区间 [a, b]内素数的个数, a and b (1 ≤ a ≤ b < 231, b - a ≤ 100000). 由于a和b较大,我们可以筛选所有[2, √b)内的素数,然后同时去筛选掉在区间[a, b)的数,用IsPrime[i-a] = 1表示i是素数: ///LightOj1197求区间素数的个数; #include<stdio.h&g…
http://lightoj.com/volume_showproblem.php?problem=1197 题目大意: 就是给你一个区间[a,b]让你求这个区间素数的个数 但a.b的值太大没法直接进行素数筛选(没法开那么大的数组),我们可以将a当做0,将b当做b-a 这样求[a,b]之间就变成了求[0, b - a]之间,这样就可以开数组来筛选 下图是代码式子j = j + prime[i] - a % prime[i]的由来 #include<stdio.h> #include<ma…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=6069 题意: 给出 l, r, k.求:(lambda d(i^k))mod998244353,其中 l <= i <= r, d(i) 为 i 的因子个数. 思路:若 x 分解成质因子乘积的形式为 x = p1^a1 * p2^a2 * ... * pn^an,那么 d(x) = (a1 + 1) * (a2 + 1) * ... * (an + 1) .显然 d(x^k) = (a1 * k…