CodeForces - 294A Shaass and Oskols
///////////////////////////////////////////////////////////////////////////////////////////////////////
作者:stxy-ferryman
声明:本文遵循以下协议自由转载-非商用-非衍生-保持署名|Creative Commons BY-NC-ND 3.0
查看本文更新与讨论请点击:http://www.cnblogs.com/stxy-ferryman/
链接被删请百度:stxy-ferryman
///////////////////////////////////////////////////////////////////////////////////////////////////////
Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to n from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are ai oskols sitting on the i-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the i-th wire). Consequently all the birds on the i-th wire to the left of the dead bird get scared and jump up on the wire number i - 1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number i + 1, if there exists no such wire they fly away.
Shaass has shot m birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
The first line of the input contains an integer n, (1 ≤ n ≤ 100). The next line contains a list of space-separated integers a1, a2, ..., an,(0 ≤ ai ≤ 100).
The third line contains an integer m, (0 ≤ m ≤ 100). Each of the next m lines contains two integers xi and yi. The integers mean that for the i-th time Shaass shoot the yi-th (from left) bird on the xi-th wire, (1 ≤ xi ≤ n, 1 ≤ yi). It's guaranteed there will be at least yi birds on the xi-th wire at that moment.
On the i-th line of the output print the number of birds on the i-th wire.
5
10 10 10 10 10
5
2 5
3 13
2 12
1 13
4 6
0
12
5
0
16
3
2 4 1
1
2 2
3
0
3
题意:有个人闲着无聊去打电线上的绿鸭子,而他在打死一只蠢鸭子后,那只鸭子前面的鸭子向左的电线杆上飞,后面的向右飞,如果没了电线杆,它们就飞走了。最后问你每根电线杆上有几只鸭子。
题解:不必多说,模拟即可
代码:
var
a:array[..] of longint;
n,m,x,y,i,j,k:longint;
begin
readln(n);
for i:= to n do
read(a[i]);
readln(m);
for i:= to m do
begin
readln(x,y);
if x> then a[x-]:=a[x-]+y-;
if x<n then a[x+]:=a[x+]+a[x] -y;
a[x]:=;
end;
for i:= to n do
writeln(a[i]);
end.
CodeForces - 294A Shaass and Oskols的更多相关文章
- Codeforce 294A - Shaass and Oskols (模拟)
Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to ...
- Codeforces 294D - Shaass and Painter Robot
294D - Shaass and Painter Robot 思路: 可以用数学归纳法证明一个结论:整个棋盘黑白相间当且仅当边缘黑白相间. 分奇偶讨论又可得出边缘黑色格个数为n+m-2 这样就可以暴 ...
- Codeforces 294B Shaass and Bookshelf:dp
题目链接:http://codeforces.com/problemset/problem/294/B 题意: 有n本书,每本书的厚度为t[i],宽度为w[i] (1<=t[i]<=2, ...
- Codeforces K. Shaass and Bookshelf(动态规划三元组贪心)
题目描述: B. Shaass and Bookshetime limit per test 2 secondsmemory limit per test 256 megabytesinput ...
- Codeforces 294B Shaass and Bookshelf(记忆化搜索)
题目 记忆化搜索(深搜+记录状态) 感谢JLGG //记忆话搜索 //一本书2中状态,竖着放或者横着放 //初始先都竖着放,然后从左边往右边扫 #include<stdio.h> #inc ...
- Codeforces 294E Shaass the Great
树形DP.由于n只有5000,可以直接枚举边. 枚举边,将树分成两个子树,然后从每个子树中选出一个点分别为u,v,那么答案就是: 子树1中任意两点距离总和+子树2中任意两点距离总和+子树1中任意一点到 ...
- CodeForces 294B Shaass and Bookshelf 【规律 & 模拟】或【Dp】
这道题目的意思就是排两排书,下面这排只能竖着放,上面这排可以平着放,使得宽度最小 根据题意可以得出一个结论,放上这排书的Width 肯定会遵照从小到大的顺序放上去的 Because the total ...
- Codeforces Round #178 (Div. 2)
A. Shaass and Oskols 模拟. B. Shaass and Bookshelf 二分厚度. 对于厚度相同的书本,宽度竖着放显然更优. 宽度只有两种,所以枚举其中一种的个数,另一种的个 ...
- OUC_Summer Training_ DIV2_#13 723afternoon
A - Shaass and Oskols Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I ...
随机推荐
- Java http请求和调用
关于http get和post请求调用代码以及示例. 参考:http://www.cnblogs.com/zhuawang/archive/2012/12/08/2809380.html http请求 ...
- PHP中foreach循环传值问题
首先看一段代码: <?php $a=array('ab','cd','ef'); $count=3; foreach($a as $key=>$value){ $value='abcdef ...
- MyBatis框架(三)动态SQL,分页,二进制存入数据库图片
一.动态sql语句,分页 1, <if>条件 <if test="key!=null"> 拼接sql语句 </if> 2, <choose ...
- 极化码之tal-vardy算法(2)
上一节我们了解了tal-vardy算法的大致原理,对所要研究的二元输入无记忆对称信道进行了介绍,并着重介绍了能够避免输出爆炸灾难的合并操作,这一节我们来关注信道弱化与强化操作. [1]<Chan ...
- ioc(Inversion of Control)控制反转和DI
ioc意味着将你设计好的交给容器控制,而不是传统在你的对象中直接控制 谁控制了谁:传统的javaSE程序设计,我们直接在对象内部通过new进行创建对象,是程序主动去创建依赖对象:而ioc是有专门一个容 ...
- Opengl4.5 中文手册—A
因为opengl API 比较庞大,网络上还没有完整的.较新的opengl中文手册 这对很多人很不方便,所以整理了这一系列,用于帮助大家"快速浏览最新的opengl api" 为了 ...
- apache一个ip多个端口虚拟主机
1.打开httpd.conf,查找Listen:80,在下面一行加入Listen:8080:2.查找#Include conf/extra/httpd-vhosts.conf,将此行前面的#去掉:3. ...
- js-注释代码习惯
功能块代码 /** * xxxx */ 定义的函数或方法 /* xxxx */ 调用了某个函数或方法 // <--xxx
- zoj1494 暴力模拟 简单数学问题
Climbing Worm Time Limit: 2 Seconds Memory Limit:65536 KB An inch worm is at the bottom of a we ...
- 扩展js,实现c#中的string.format方便拼接字符串
//"{0}-{1}-{2}".format("xx","yy","zz") //显示xx-yy-zz String.p ...