D. Robot Vacuum Cleaner time limit per test 1 second memory limit per test 256 megabytes Problem Description Pushok the dog has been chasing Imp for a few hours already. Fortunately, Imp knows that Pushok is afraid of a robot vacuum cleaner. While mo…
CF922 CodeForces Round #461(Div.2) 这场比赛很晚呀 果断滚去睡了 现在来做一下 A CF922 A 翻译: 一开始有一个初始版本的玩具 每次有两种操作: 放一个初始版本进去,额外得到一个初始版本和一个复制版本 放一个复制版本进去,额外得到两个复制版本 一开始有\(1\)个初始版本,是否能恰好得到\(x\)个复制版本和\(y\)个初始版本 Solution 傻逼题 要特判一些特殊情况(没有\(1A\)...) #include<iostream> #includ…
题目链接:http://codeforces.com/contest/922 B. Magic Forest time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Imp is in a magic forest, where xorangles grow (wut?) A xorangle of order n is such a…
A - Cloning Toys /* 题目大意:给出两种机器,一种能将一种原件copy出额外一种原件和一个附件, 另一种可以把一种附件copy出额外两种附件,给你一个原件, 问能否恰好变出题目要求数量的原件和附件 题解:注意当附件需求不为0的时候,原件需求必须大于1 */ #include <cstdio> #include <algorithm> int main(){ int a,b; scanf("%d%d",&a,&b); if((a-…
C. Robot Breakout time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output n robots have escaped from your laboratory! You have to find them as soon as possible, because these robots are experimental,…
C. Cave Painting time limit per test 1 second memory limit per test 256 megabytes Problem Description Imp is watching a documentary about cave painting. Some numbers, carved in chaotic order, immediately attracted his attention. Imp rapidly proposed…
B. Magic Forest time limit per test 1 second memory limit per test 256 megabytes Problem Description Imp is in a magic forest, where xorangles grow (wut?) A xorangle of order n is such a non-degenerate triangle, that lengths of its sides are integers…
A. Cloning Toys time limit per test 1 second memory limit per test 256 megabytes Problem Description Imp likes his plush toy a lot. Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, h…
Magic Forest 题意:就是在1 ~ n中找三个值,满足三角形的要求,同时三个数的异或运算还要为0: , where  denotes the bitwise xor of integers x and y. 我已开始没想到a^b^c=0相当于c = a^b; 这样的转化就可以是三次的变成二次的: #include <cstdio> #include <algorithm> #include <iostream> using namespace std; ];…
Codeforces Round #552 (Div. 3) 题目链接 A. Restoring Three Numbers 给出 \(a+b\),\(b+c\),\(a+c\) 以及 \(a+b+c\) 这四个数,输出一种合法的 \(a,b,c\).   可以发现,前面的两个数加起来减去最后的 \(a+b+c\),答案就出来一个.最后这样求出\(a,b,c\)即可. 代码如下: Code #include <bits/stdc++.h> using namespace std; typede…