A AFei Loves Magic
链接:https://ac.nowcoder.com/acm/contest/338/A
来源:牛客网
题目描述
AFei knew that the magical circle would disappear after t seconds. It meant that after t seconds, he could move and the stones would return to immobility. This also meant that AFei would get those stones. He wondered how many magic stones he could get in the end, including the first one he got when he came to the forest.
输入描述:
The first line contains three integers n, L, t (0≤n≤1000000, 2≤L≤1000000000, 0<=t<=1000000) − the number of stones on the line is n,the length of the line is L meter, and the magic circle will disappear after t seconds.
The following n lines contain description of magic stones on the line. Each i-th of these lines contains two space-separated integers x[i] and d[i] (0<x[i]<L, d[i]∈{1,2} for i<=n), which stand for initial position and direction of motion(1 means from 0 to L, 2 means from L to 0.).
输出描述:
Output a number indicating the amount of the magic stones that AFei will eventually be able to obtain.
说明
The stones are A(1,1), B(5,2), C(6,1), D(9,2). After 1s, they become A(2,1), B(4,2), C(7,1), D(8,2); After 2s, they become A(3,2), B(3,1), C(7,2), D(8,1); After 3s, they become A(2,2), B(4,1), C(6,2), D(9,1); After 4s, they become A(1,2), B(5,2), C(5,1), D reach L and disappears; After 5s, they become A reach 0 and disappears, B(4, 2), C(6,1), D disappeared; After 6s, they become A disappeared, B(3, 2), C(7, 1), D disappeared. AFei finially gets the first one, B and C.
备注:
1,Input guarantees that there will not be two magic stones in one location. 2,If stone A and stone B are located at 4 and 5, respectively, and A's direction is 1, B's direction is 2. Then next second, the position of the two stones have not changed, but they have gone in the opposite direction. 解题思路:不考虑碰撞,直接根据初始坐标,方向,时间和速度求出每一个魔法石的最终坐标,最终坐标在(0,L)的魔法石数量就是剩下的魔法石
#include<stdio.h>
int main()
{
int n ;
int len, t;
int a,b; while(scanf("%d%d%d",&n,&len,&t)!=EOF)
{
int sum = 1;
for(int i = 0; i<n; i++)
{
scanf("%d %d",&a,&b);
if(b == 1)
{
if(a+t<len)
sum++; }
else
{ if(a-t>0)
sum++;
}
}
printf("%d\n",sum);
} }
A AFei Loves Magic的更多相关文章
- 湖南大学第十四届ACM程序设计新生杯(重现赛)
RANK 0 题数 0 期末复习没有参加,补几道喜欢的题. A: AFei Loves Magic 签到 思路 :不需考虑 碰撞 直接计算最终状态即可. #include<bits/stdc ...
- Little Elephant and Magic Square
Little Elephant loves magic squares very much. A magic square is a 3 × 3 table, each cell contains s ...
- F Find the AFei Numbers
链接:https://ac.nowcoder.com/acm/contest/338/F来源:牛客网 题目描述 AFei loves numbers. He defines the natural n ...
- CodeForces-259B]Little Elephant and Magic Square
Little Elephant loves magic squares very much. A magic square is a 3 × 3 table, each cell contains ...
- BestCoder Round #90
有生以来第一场在COGS以外的地方打的比赛.挂成dog了. 主要是没有经验,加之代码能力过弱.还有最后的瞎hack三次,Too Young Too Simple...... 言归正传. (抄一发题解先 ...
- codeforces259B
Little Elephant and Magic Square CodeForces - 259B Little Elephant loves magic squares very much. A ...
- 【Codeforces717F】Heroes of Making Magic III 线段树 + 找规律
F. Heroes of Making Magic III time limit per test:3 seconds memory limit per test:256 megabytes inpu ...
- HDU4876ZCC loves cards(多校题)
ZCC loves cards Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Tot ...
- 多校训练赛2 ZCC loves cards
ZCC loves cards Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) ...
随机推荐
- 体验一把haskell
这几天做到PAT一道比较数据大小的题PAT1065,题目不难,应该说是一道送分题,就是开数组,然后模拟人工计算的过程进行计算,再比较下就行.做完之后,联想到haskell的Integer类型是无限大的 ...
- [SDOI2015]寻宝游戏(LCA,set)
[SDOI2015]寻宝游戏 题目描述 小B最近正在玩一个寻宝游戏,这个游戏的地图中有N个村庄和N-1条道路,并且任何两个村庄之间有且仅有一条路径可达.游戏开始时,玩家可以任意选择一个村庄,瞬间转移到 ...
- 【LeetCode】未分类(tag里面没有)(共题)
[334]Increasing Triplet Subsequence (2019年2月14日,google tag)(greedy) 给了一个数组 nums,判断是否有三个数字组成子序列,使得子序列 ...
- 【CF】38E Let's Go Rolling! (dp)
前言 这题还是有点意思的. 题意: 给你 \(n\) (\(n<=3000\)) 个弹珠,它们位于数轴上.给你弹珠的坐标 \(x_i\) 在弹珠 \(i\) 上面花费 \(C_i\) 的钱 可以 ...
- 前端每日实战:32# 视频演示如何用纯 CSS 创作六边形按钮特效
效果预览 按下右侧的"点击预览"按钮可以在当前页面预览,点击链接可以全屏预览. https://codepen.io/comehope/pen/xjoOeM 可交互视频教程 此视频 ...
- 5.Docker存储卷
一.概述 1.Docker底层存储机制 Docker镜像由多个只读层叠加而成,启动容器时,Docker会加载只读镜像层并在镜像栈顶部添加一个读写层. 如果运行中的容器修改了现有的一个已经存在的文件,那 ...
- springboot cache---本地缓存的使用
使用缓存的几个注解 什么时候需要使用缓存呢?一般是在一个方法的返回值需要被频繁用到.但是返回值很少改变而且执行这个方法会消耗较多的时间,这种情况我们可以考虑将返回值暂时存到内存中,需要时通过对应的唯一 ...
- 每次当浏览到网页,点击tab标签又回到顶部去了!
通常tab的标签使用a链接,而a链接的href值为#,这是一个锚点的属性,因此他会跳转到网页的顶端.如果你的tab包含一个id=tab,也可以设置为href="#tab"这样他就会 ...
- 暴力&打表
_LH巨神好像不太会打表,这里来普及一下 还有暴力这么重要的东西网上讲的人竟然不多…… 一.打表 打表,就是针对一些输入数据比较小的题目的一种骗分技巧,当然有时候也可以在正解或暴力中起一定优化作用. ...
- Airtest断言方法
1,第一种断言方式:验证UI界面 a.存在 b.不存在 2,断言第二种方式:验证数值 assert_equal:断言相等 assert_not_equal:断言不等 3,我发现Airtest一个bug ...