题目链接: B. Kyoya and Permutation time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Let's define the permutation of length n as an array p = [p1, p2, ..., pn] consisting of n distinct integers…
D. Optimal Number Permutation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have array a that contains all integers from 1 to n twice. You can arbitrary permute any numbers in a. Let…
A. Hanoi tower Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description you the conditions of this task. There are 3 pivots: A, B, C. Initially, n disks of different diameter are placed on the pivot A: the smallest dis…
BOPCTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86686#problem/D Description You invented a new chess figure and called it BOPC. Suppose it stands at the square grid at the point with coordinates …
一看就是找规律的题.只要熟悉异或的性质,可以秒杀. 为了防止忘记异或的规则,可以把异或理解为半加运算:其运算法则相当于不带进位的二进制加法. 一些性质如下: 交换律: 结合律: 恒等律: 归零律: 典型应用:交换a和b的值:a=a^b^(b=a); #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<…
Another Rock-Paper-Scissors Problem 题目连接: http://codeforces.com/gym/100015/attachments Description Sonny uses a very peculiar pattern when it comes to playing rock-paper-scissors. He likes to vary his moves so that his opponent can't beat him with hi…
题目描述: Malek Dance Club time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fu…
大致题意: 有一个W*H的长方形,有n个人,分别站在X轴或Y轴,并沿直线向对面走,第i个人在ti的时刻出发,如果第i个人与第j个人相撞了 那么则交换两个人的运动方向,直到走到长方形边界停止,问最后每个人的坐标. 题解: 两个人要相撞,当且仅当Xi-Ti=Xj-Tj,所以可以将Xi-Ti分组,对于同一组里的人会互相碰撞,不同组的不会,这样就只要考虑同一组里 的人,画个图可以发现,规律 然后根据规律化坐标即可 #include<cstdio> #include<cstring> #in…
Tree and Permutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description There are N vertices connected by N−1 edges, each edge has its own length.…
一边长为a的直角三角形,a^2=c^2-b^2.可以发现1.4.9.16.25依次差3.5.7.9...,所以任何一条长度为奇数的边a,a^2还是奇数,那么c=a^2/2,b=c+1.我们还可以发现,1.4.9.16.25.36各项差为8.12.16.20,偶数的平方是4的倍数,那么c=a^2/4-1,b=a^2/4+1. #include <iostream> using namespace std; int main() { long long n; cin>>n; n*=n;…