CF1316C [Primitive Primes] 给出两个多项式\(a_0+a_1x+a_2x^2+\dots +a_{n-1}x^{n-1}\)和\(b_0+b_1x+b_2x^2+ \dots +a_{m-1}x^{m-1}\),每个多项式的系数的\(\gcd\)都为1 给出质数\(p\),求在这两个多项式乘积的系数中,任意一个不能被\(p\)整除的系数 比赛的时候没想到,居然了想用线段树维护,什么都想用线段树..然后悲惨掉分 设乘积的系数序列为\(c\) 我们可以设在这两个序列中,第一…

数学题 在f(x)和g(x)的系数里找到第一个不是p的倍数的数,然后相加就是答案 为什么? 设x1为f(x)中第一个不是p的倍数的系数,x2为g(x)...... x1+x2前的系数为(a[x1+x2]*b[0]+a[x1+x2-1]*b[1]..)+a[x1]*b[x2]+(.....a[0]*b[x1+x2]) 除了a[x1]b[x2]外,其余的都含有p的倍数,所以和肯定不是p的倍数 #include<bits/stdc++.h> using namespace std; ; int ar…

It is Professor R's last class of his teaching career. Every time Professor R taught a class, he gave a special problem for the students to solve. You being his favourite student, put your heart into solving it one last time. You are given two polyno…

题意:给你两个一元多项式\(f(x)\)和\(g(x)\),保证它们每一项的系数互质,让\(f(x)\)和\(g(x)\)相乘得到\(h(x)\),问\(h(x)\)是否有某一项系数不被\(p\)整除. 题解:这题刚开始看了好久不知道怎么写,但仔细看题目给的条件可能会看出一点门道来,我们知道,\(c_{i}\)是由\(f(x)\)和\(g(x)\)中多个某两项积的和来得到的(\(c_{i}=a_{0}b{i}+a_{1}b_{i-1}+...+a_{i}b_{0}\)),那么我们只要找到第一个不…

题面 It is Professor R's last class of his teaching career. Every time Professor R taught a class, he gave a special problem for the students to solve. You being his favourite student, put your heart into solving it one last time. You are given two pol…

CodeCraft-20 (Div. 2) A. Grade Allocation 思路 : 无脑水题 代码 #include<iostream> #include<algorithm> #include<vector> using namespace std; const int Len = 1e3 + 5; int ar[Len]; int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);…

[TOC] 背景概述 在OpenGL绘制图形时,可能需要绘制多个并不相连的图形.这样的情况下这几个图形没法被当做一个图形来处理.也就需要多次调用 DrawArrays 或 DrawElements. 如果图形很多,可能会需要用一个循环来调用: for (int i = 0; i < num_objects; i++) { glDrawArrays(GL_TRIANGLES, object[n]->first_vertex, object[n]->vertex_count); } 每一次调…

Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. References: How Many Primes Are There? Sieve of Eratosthenes Credits:Special thanks to @mithmatt for adding this problem and creating all test…

Given an entity with an attribute firstName, Core Data automatically generates firstName, setFirstName:, primitiveFirstName, and setPrimitiveFirstName. The primary use of primitive accessors is to prevent key-value observing notifications from being…

坏味道--基本类型偏执(Primitive Obsession) 特征 使用基本类型而不是小对象来实现简单任务(例如货币.范围.电话号码字符串等). 使用常量编码信息(例如一个用于引用管理员权限的常量USER_ADMIN_ROLE = 1 ). 使用字符串常量作为字段名在数组中使用. 问题原因 类似其他大部分坏味道,基本类型偏执诞生于类初建的时候.一开始,可能只是不多的字段,随着表示的特性越来越多,基本数据类型字段也越来越多. 基本类型常常被用于表示模型的类型.你有一组数字或字符串用来表示某个实…