Codeforces 385C Bear and Prime Numbers 其实不是多值得记录的一道题,通过快速打素数表,再做前缀和的预处理,使查询的复杂度变为O(1). 但是,我在统计数组中元素出现个数时使用了map,以至于后面做前缀和的累加时,每次都要对map进行查询,以至于TLE.而自己一直没有发现,以为是欧拉筛对于这道题还不够优,于是上网搜题解,发现别人的做法几乎一样,但是却能跑过,挣扎了许久才想起是map的原因.map的内部实现是一颗红黑树,每次查询的复杂度为O(logN),在本来时…
385C - Bear and Prime Numbers 思路:记录数组中1-1e7中每个数出现的次数,然后用素数筛看哪些能被素数整除,并加到记录该素数的数组中,然后1-1e7求一遍前缀和. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset((a),(b),sizeof(a)) const int INF=0x3f…
Recently, the bear started studying data structures and faced the following problem. You are given a sequence of integers x1, x2, ..., xn of length n and m queries, each of them is characterized by two integers li, ri. Let's introduce f(p) to represe…
第一眼看这道题目的时候觉得可能会很难也看不太懂,但是看了给出的Hint之后思路就十分清晰了 Consider the first sample. Overall, the first sample has 3 queries. The first query l = 2, r = 11 comes. You need to count f(2) + f(3) + f(5) + f(7) + f(11) = 2 + 1 + 4 + 2 + 0 = 9. The second query comes…
题意翻译 给你一串数列a.对于一个质数p,定义函数f(p)=a数列中能被p整除的数的个数.给出m组询问l,r,询问[l,r]区间内所有素数p的f(p)之和. 题目描述 Recently, the bear started studying data structures and faced the following problem. You are given a sequence of integers x1,x2,...,xn x_{1},x_{2},...,x_{n} x1,x2,.…
A - Bear and Prime 100 思路:任何一个合数都可以写成2个以上质数的乘积.在2-100中,除了4,9,25,49外都可以写成两个以上不同质数的乘积. 所以打一个质数加这四个数的表:{2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41,43,47,49},询问19次,如果能被整出两次以上,说明是合数,否则是质数. #include<bits/stdc++.h> using namespace std; #define ll long long…
