[AtCoder3856]Ice Rink Game - 模拟】的更多相关文章

Problem Statement An adult game master and N children are playing a game on an ice rink. The game consists of K rounds. In the i-th round, the game master announces: Form groups consisting of Ai children each! Then the children who are still in the g…
[题意]n个人进行游戏,每轮只保留最大的a[i]倍数的人,最后一轮过后剩余2人,求最小和最大的n,或-1.n<=10^5. [算法]递推||二分 [题解]令L(i),R(i)表示第i轮过后的最小人数和最大人数. 令X(i)和Y(i)表示区间[L(i),R(i)]中最小的a[i]倍数和最大的a[i]倍数. L(i-1)=X(i),R(i-1)=Y(i)+a[i]-1. X(i)=L(i)/a[i](上取整),Y(i)=R(i)/a[i](下取整). 答案为L(0),R(0). #include<…
题面在这里! 倒着维护可以取的范围区间,判一下可不可能无解即可. #include<bits/stdc++.h> #define ll long long using namespace std; const int N=1e5+5; int n,a[N]; ll L,R; int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",a+i); L=R=2; for(int i=…
An adult game master and N children are playing a game on an ice rink. The game consists of K rounds. In the i-th round, the game master announces: Form groups consisting of Ai children each! Then the children who are still in the game form as many g…
A - Move and Win 题解 看两个人相遇的时候谁先手即可,相遇之后第一个移动的人必输 代码 #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define pdi pair<db,int> #define mp make_pair #define pb push_back #define enter putchar('\n') #def…
A - Move and Win Time limit : 1sec / Memory limit : 512MB Score : 300 points Problem Statement A game is played on a strip consisting of N cells consecutively numbered from 1 to N. Alice has her token on cell A. Borys has his token on a different cel…
Description You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice. The level of the cave where y…
题意:给定初始数量的冰激凌,然后n个操作,如果是“+”,那么数量就会增加,如果是“-”,如果现有的数量大于等于要减的数量,那么就减掉,如果小于, 那么孩子就会离家.问你最后剩下多少冰激凌,和出走的孩子数量. 析:多水的一个题,就是一个模拟,如果是+,就加上,如果是‘-’,就判断一下,如果不够,就记录下来. 代码如下: #include <iostream> #include <cmath> #include <cstdlib> #include <set>…
这道题委实无语了,刚开始以为是很一般的方位模拟题,懒得看样例直接写的代码,然后敲了好几个switch结果样例居然没出来.. 仔细分析了样例之后才发现原来随着宇航员方位的改变他的左手方向以及头顶方向是跟着变的.. 然后就要时刻记录.. 方向感和想象力不行就只能在纸上画好久了 .. 无奈了..   Time Limit:2000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Description 问题描述:  宇航员在太空…
ICE的AMI和AMD: AMI:指的是客户端的调用.一般有一个代理类,一个回调类. 从服务端返回的数据在回调类中的ice_response函数中. AMD:指的是服务端对客户端的调用分派.一般进行业务处理需要自己写一个类继承于_Disp类.重载method_async(AMD_CALLBACK __cb,arg1,arg2,...)函数.在这个函数中调用__cb的ice_response(result)往回调类中写result.这样客户端就能够接收到回写的结果      还有一点很有特色的是,…