A - Shaass and Oskols

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to nfrom 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.

Input

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 yibirds on the xi-th wire at that moment.

Output

On the i-th line of the output print the number of birds on the i-th wire.

Sample Input

Input
5
10 10 10 10 10
5
2 5
3 13
2 12
1 13
4 6
Output
0
12
5
0
16
Input
3
2 4 1
1
2 2
Output
3
0
3
这个题就没有什么好说的啦,从头到尾模拟一下就好了。题意也特别好理解哦。
#include<stdio.h>
int main()
{
int a[];
int n,i,j,x,y;
scanf("%d",&n);
for(i = ;i < n;i++)
{
scanf("%d",&a[i]);
}
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&x,&y);
x = x - ;
if(x > && x < n-)
{
a[x - ] += y - ;
a[x + ] += a[x] - y;
a[x] = ;
}
else if(x == )
{
a[] += a[] - y;
a[] = ;
}
else if(x == n - )
{
a[n - ] += y - ;
a[n - ] = ;
}
}
for(i = ;i < n;i++)
{
printf("%d\n",a[i]);
}
return ; }

B - Shaass and Bookshelf

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Shaass has n books. He wants to make a bookshelf for all his books. He wants the bookshelf's dimensions to be as small as possible. The thickness of the i-th book is ti and its pages' width is equal to wi. The thickness of each book is either 1 or 2. All books have the same page heights.

Shaass puts the books on the bookshelf in the following way. First he selects some of the books and put them vertically. Then he puts the rest of the books horizontally above the vertical books. The sum of the widths of the horizontal books must be no more than the total thickness of the vertical books. A sample arrangement of the books is depicted in the figure.

Help Shaass to find the minimum total thickness of the vertical books that we can achieve.

Input

The first line of the input contains an integer n, (1 ≤ n ≤ 100). Each of the next n lines contains two integers ti and wi denoting the thickness and width of the i-th book correspondingly, (1 ≤ ti ≤ 2, 1 ≤ wi ≤ 100).

Output

On the only line of the output print the minimum total thickness of the vertical books that we can achieve.

Sample Input

Input
5
1 12
1 3
2 15
2 5
2 1
Output
5
Input
3
1 10
2 1
2 4
Output
3
这个题是用背包,但是我看不懂也不能完全理解。囧~
#include<stdio.h>
int main()
{
int a[];
int n,i,j,x,y;
scanf("%d",&n);
for(i = ;i < n;i++)
{
scanf("%d",&a[i]);
}
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&x,&y);
x = x - ;
if(x > && x < n-)
{
a[x - ] += y - ;
a[x + ] += a[x] - y;
a[x] = ;
}
else if(x == )
{
a[] += a[] - y;
a[] = ;
}
else if(x == n - )
{
a[n - ] += y - ;
a[n - ] = ;
}
}
for(i = ;i < n;i++)
{
printf("%d\n",a[i]);
}
return ; }
 

OUC_Summer Training_ DIV2_#13 723afternoon的更多相关文章

  1. OUC_Summer Training_ DIV2_#9 719

    其实自己只会做很简单的题,有时都不想写解题报告,觉得不值得一写,但是又想到今后也许就不会做ACM了,能留下来的东西只有解题报告了,所以要好好写,很渣的题也要写,是今后的纪念. B - B Time L ...

  2. OUC_Summer Training_ DIV2_#5

    这是做的最好的一次了一共做了4道题  嘻嘻~ A - Game Outcome Time Limit:2000MS     Memory Limit:262144KB     64bit IO For ...

  3. OUC_Summer Training_ DIV2_#16 725

    今天做了这两道题真的好高兴啊!!我一直知道自己很渣,又贪玩不像别人那样用功,又没有别人有天赋.所以感觉在ACM也没有学到什么东西,没有多少进步.但是今天的B题告诉我,进步虽然不明显,但是只要坚持努力的 ...

  4. OUC_Summer Training_ DIV2_#12(DP1) 723

    这一次是做练习,主要了解了两个算法,最大子矩阵和,最长上升子序列. 先看题好啦. A - To The Max Time Limit:1000MS     Memory Limit:32768KB   ...

  5. OUC_Summer Training_ DIV2_#14 724

    又落下好多题解啊...先把今天的写上好了. A - Snow Footprints Time Limit:1000MS     Memory Limit:262144KB     64bit IO F ...

  6. OUC_Summer Training_ DIV2_#2之解题策略 715

    这是第一天的CF,是的,我拖到了现在.恩忽视掉这个细节,其实这一篇只有一道题,因为这次一共做了3道题,只对了一道就是这一道,还有一道理解了的就是第一篇博客丑数那道,还有一道因为英语实在太拙计理解错了题 ...

  7. OUC_Summer Training_ DIV2_#7 718

    是18号做的题啦,现在才把报告补上是以前不重视报告的原因吧,不过现在真的很喜欢写报告,也希望能写一些有意义的东西出来. A - Dragons Time Limit:2000MS     Memory ...

  8. OUC_Summer Training_ DIV2_#11 722

    企鹅很忙系列~(可惜只会做3道题T_T) A - A Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d &am ...

  9. OUC_Summer Training_ DIV2_#4之数据结构

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=26100#problem/A A - A Time Limit:1000MS     Me ...

随机推荐

  1. php 如何将image图片转化为字符串(GD库操作及imagick两种实现方式)

    前两天研究php中的 imagick 扩展的时候,突发奇想实现的一个小功能感觉挺有意思,在这里记录一下: 将一张image图片转化为字符串的形式,先上一张效果图.(运行笔记中的代码需要先安装 php_ ...

  2. 二、LinkList及其源码解析

    1.链表介绍 链表是一种物理单元上非连续,非顺序的存储结构.链表由一系列的姐点组成,结点可以在运行时动态生成.每个结点包含两个部分,一个是存储数据元素的数据域,一个是存储下一个结点的指针域 双链表是链 ...

  3. CHD-5.3.6集群安装

    我是基于Apache-hadoop2.7.3版本安装成功后,已有的环境进行安装chd-5..6 已用的环境: JDK版本: java version "1.8.0_191" Jav ...

  4. 第四章、前端之BOM和DOM

    目录 第四章.前端之BOM和DOM 一.解释BOM和DOM 二.window对象 三.window子对象 四.弹出框 五.计时相关 六.HTML的DOM树 七.查找元素 八.节点操作 九.JS操作CS ...

  5. Mysql系列-性能优化神器EXPLAIN使用介绍及分析

    MySQL 提供了一个 EXPLAIN 命令, 它可以对 SELECT 语句进行分析, 并输出 SELECT 执行的详细信息, 以供开发人员针对性优化. EXPLAIN 命令用法十分简单, 在 SEL ...

  6. 在线p图网址

    在线P图网址 如果你是简易的P图,不用那么麻烦的去下载安装Photoshop,可以使用以下网址在线编辑 https://www.uupoop.com/ps/?hmsr=ps_menu

  7. 7.caffe:create_lmdb.sh(数据预处理转换成lmdb格式)

    个人实践代码如下: #!/usr/bin/env sh # Create the imagenet lmdb inputs # N.B. set the path to the imagenet tr ...

  8. [USACO14OPEN]GPS的决斗Dueling GPS's

    题目概况 题目描述 给你一个\(N\)个点的有向图,可能有重边. 有两个\(GPS\)定位系统,分别认为经过边\(i\)的时间为\(P_i\),和\(Q_i\). 每走一条边的时候,如果一个系统认为走 ...

  9. 开启aix SFTP日志 是否和链接SFTP有关呢

    1.修改SSH配置 vi /etc/ssh/sshd_config 在sftp配置处添加-l INFO -f AUTH Subsystem sftp /usr/lib64/ssh/sftp-serve ...

  10. 关于IO的同步,异步,阻塞,非阻塞

    上次写了一篇文章:Unix IO 模型学习.恰巧在这次周会的时候,@fp1203 (goldendoc成员之一) 正好在讲解poll和epoll的底层实现.中途正好讨论了网络IO的同步.异步.阻塞.非 ...