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…

D. Mike and Feet time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are…

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…

B. Our Tanya is Crying Out Loud time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Right now she actually isn't. But she will be, if you don't solve this problem. You are given integers n, k, A an…

A. Olympiad 给出n个数,让你找出有几个非零并且不重复的数 所以用stl的set //#define debug #include<stdio.h> #include<math.h> #include<cmath> #include<queue> #include<stack> #include<string> #include<cstring> #include<string.h> #include…

C. Phone Numbers time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output And where the are the phone numbers? You are given a string s consisting of lowercase English letters and an integer k. Find th…

构造题 话说挺水的题..当时怎么就WA到自闭呢.. 先把每个位置按照最低要求填满,也就是相差1..然后从最后一位开始把剩下的数加上,直到不能加为止. #include <bits/stdc++.h> #define INF 0x3f3f3f3f #define full(a, b) memset(a, b, sizeof a) using namespace std; typedef long long ll; inline int lowbit(int x){ return x &…

数论 gcd 看到这个题其实知道应该是和(a+k)(b+k)/gcd(a+k,b+k)有关,但是之后推了半天,思路全无. 然而..有一个引理: gcd(a, b) = gcd(a, b - a) = gcd(b, b - a) (b > a) 证明一下: 令 gcd(a, b) = c, (b > a) 则有 a % c = 0, b % c = 0 那么 (a - b) % c = 0 令 gcd(a, b - a) = c', 假设c' != c 则有 a % c' = 0, (b - a…

题解: A 喵哈哈村的跳棋比赛 题解:其实我们要理解题意就好了,画画图看看这个题意.x<y,那么就交换:x>y,那么x=x%y. 如果我们经过很多次,或者y<=0了,那么就会无限循环. #include<iostream> #include<cstdio> #include<cstring> using namespace std; long long x,y; void work(){ while(1){ if(x==0){ cout<<…