Codeforces Round #378 (Div. 2) C. Epidemic in Monstropolis 模拟
1 second
256 megabytes
standard input
standard output
There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.
Soon, monsters became hungry and began to eat each other.
One monster can eat other monster if its weight is strictly greater than the weight of the monster being eaten, and they stand in the queue next to each other. Monsters eat each other instantly. There are no monsters which are being eaten at the same moment. After the monster A eats the monster B, the weight of the monster A increases by the weight of the eaten monster B. In result of such eating the length of the queue decreases by one, all monsters after the eaten one step forward so that there is no empty places in the queue again. A monster can eat several monsters one after another. Initially there were n monsters in the queue, the i-th of which had weight ai.
For example, if weights are [1, 2, 2, 2, 1, 2] (in order of queue, monsters are numbered from 1 to 6 from left to right) then some of the options are:
- the first monster can't eat the second monster because a1 = 1 is not greater than a2 = 2;
- the second monster can't eat the third monster because a2 = 2 is not greater than a3 = 2;
- the second monster can't eat the fifth monster because they are not neighbors;
- the second monster can eat the first monster, the queue will be transformed to [3, 2, 2, 1, 2].
After some time, someone said a good joke and all monsters recovered. At that moment there were k (k ≤ n) monsters in the queue, the j-th of which had weight bj. Both sequences (a and b) contain the weights of the monsters in the order from the first to the last.
You are required to provide one of the possible orders of eating monsters which led to the current queue, or to determine that this could not happen. Assume that the doctor didn't make any appointments while monsters were eating each other.
The first line contains single integer n (1 ≤ n ≤ 500) — the number of monsters in the initial queue.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial weights of the monsters.
The third line contains single integer k (1 ≤ k ≤ n) — the number of monsters in the queue after the joke.
The fourth line contains k integers b1, b2, ..., bk (1 ≤ bj ≤ 5·108) — the weights of the monsters after the joke.
Monsters are listed in the order from the beginning of the queue to the end.
In case if no actions could lead to the final queue, print "NO" (without quotes) in the only line.
Otherwise print "YES" (without quotes) in the first line. In the next n - k lines print actions in the chronological order. In each line print x — the index number of the monster in the current queue which eats and, separated by space, the symbol 'L' if the monster which stays the x-th in the queue eats the monster in front of him, or 'R' if the monster which stays the x-th in the queue eats the monster behind him. After each eating the queue is enumerated again.
When one monster eats another the queue decreases. If there are several answers, print any of them.
6
1 2 2 2 1 2
2
5 5
YES
2 L
1 R
4 L
3 L
5
1 2 3 4 5
1
15
YES
5 L
4 L
3 L
2 L
5
1 1 1 3 3
3
2 1 6
NO
In the first example, initially there were n = 6 monsters, their weights are [1, 2, 2, 2, 1, 2] (in order of queue from the first monster to the last monster). The final queue should be [5, 5]. The following sequence of eatings leads to the final queue:
- the second monster eats the monster to the left (i.e. the first monster), queue becomes [3, 2, 2, 1, 2];
- the first monster (note, it was the second on the previous step) eats the monster to the right (i.e. the second monster), queue becomes [5, 2, 1, 2];
- the fourth monster eats the mosnter to the left (i.e. the third monster), queue becomes [5, 2, 3];
- the finally, the third monster eats the monster to the left (i.e. the second monster), queue becomes [5, 5].
Note that for each step the output contains numbers of the monsters in their current order in the queue.
题意:给你一个序列a,代表有a个怪物,给你每个怪物的重量,要求重量大的可以吃掉相邻重量小的,重量相加,问能否得到b序列;
思路:模拟,找每个区间得到b,对于每个区间找最大值,看等否吃掉旁边的,能则能合并;
搓代码;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int a[N],sum[N];
int ans[N];
struct is
{
int l,r;
}p[N];
int check(int l,int r)
{
if(l==r)return ;
int maxx=;
for(int i=l;i<=r;i++)
maxx=max(maxx,a[i]);
for(int i=l;i<=r;i++)
{
if(a[i]==maxx&&i==l&&a[i]>a[i+])
{
return ;
}
else if(a[i]==maxx&&i==r&&a[i]>a[i-])
{
return ;
}
else if(l!=i&&r!=i&&a[i]==maxx&&(a[i]>a[i+]||a[i]>a[i-]))
{
return ;
}
}
return ;
}
void getans(int l,int len,int n)
{
int maxx=;
for(int i=l;i<=l+len-;i++)
maxx=max(maxx,a[i]);
while(len!=)
{
for(int i=l;i<=l+len-;i++)
{
if(i==l&&a[i]>=maxx&&a[i]>a[i+])
{
printf("%d R\n",i);
a[i]+=a[i+];
maxx=max(maxx,a[i]);
for(int j=i+;j<=n;j++)
a[j]=a[j+];
break;
}
if(i==l+len-&&a[i]>=maxx&&a[i]>a[i-])
{
printf("%d L\n",i);
a[i-]+=a[i];
maxx=max(a[i-],maxx);
for(int j=i;j<=n;j++)
a[j]=a[j+];
break;
}
if(i!=l+len-&&i!=l&&a[i]>=maxx&&a[i]>a[i+])
{
printf("%d R\n",i);
a[i]+=a[i+];
maxx=max(a[i],maxx);
for(int j=i+;j<=n;j++)
a[j]=a[j+];
break;
}
if(i!=l+len-&&i!=l&&a[i]>=maxx&&a[i]>a[i-])
{
printf("%d L\n",i);
a[i-]+=a[i];
maxx=max(a[i-],maxx);
for(int j=i;j<=n;j++)
a[j]=a[j+];
break;
}
}
len--;
}
}
int main()
{
int n,m;
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]),sum[i]=sum[i-]+a[i];
scanf("%d",&m);
for(int i=;i<=m;i++)
scanf("%d",&ans[i]);
int pre=;
int hh=;
for(int i=;i<=m;i++)
{
for(int j=pre+;j<=n;j++)
{
if(sum[j]-sum[pre]==ans[i])
{
p[hh].l=pre+;
p[hh++].r=j;
pre=j;
break;
}
}
}
if(hh!=m||p[hh-].r!=n)
return puts("NO\n");
for(int i=;i<hh;i++)
{
if(check(p[i].l,p[i].r)==)
return puts("NO\n");
}
printf("YES\n");
for(int i=;i<m;i++)
getans(i+,p[i].r-p[i].l+,n);
return ;
}
Codeforces Round #378 (Div. 2) C. Epidemic in Monstropolis 模拟的更多相关文章
- Codeforces Round #378 (Div. 2)-C. Epidemic in Monstropolis
C. Epidemic in Monstropolis time limit per test 1 second memory limit per test 256 megabytes input s ...
- Codeforces Round #378 (Div. 2) D题(data structure)解题报告
题目地址 先简单的总结一下这次CF,前两道题非常的水,可是第一题又是因为自己想的不够周到而被Hack了一次(或许也应该感谢这个hack我的人,使我没有最后在赛后测试中WA).做到C题时看到题目情况非常 ...
- Codeforces Round #367 (Div. 2) B. Interesting drink (模拟)
Interesting drink 题目链接: http://codeforces.com/contest/706/problem/B Description Vasiliy likes to res ...
- Codeforces Round #378 (Div. 2) A B C D 施工中
A. Grasshopper And the String time limit per test 1 second memory limit per test 256 megabytes input ...
- Codeforces Round #378 (Div. 2)F - Drivers Dissatisfaction GNU
http://codeforces.com/contest/733/problem/F 题意:给你一些城市和一些路,每条路有不满意程度和每减少一点不满意程度的花费,给出最大花费,要求找出花费小于s的最 ...
- Codeforces Round #378 (Div. 2) D. Kostya the Sculptor 分组 + 贪心
http://codeforces.com/contest/733/problem/D 给定n个长方体,然后每个长方体都能选择任何一个面,去和其他长方体接在一起,也可以自己一个,要求使得新的长方体的最 ...
- Codeforces Round #378 (Div. 2)
A: 思路: 水题,没啥意思; B: 思路: 暴力,也没啥意思; C: 思路: 思维,可以发现从前往后和为b[i]的分成一块,然后这一块里面如果都相同就没法开始吃,然后再暴力找到那个最大的且能一开始就 ...
- Codeforces Round #378 (Div. 2) C D
在实验室通宵 一边做水题一边准备随时躲起来以免被门卫大爷巡查发现..结果居然没来.. 本来以为可以加几分变个颜色..结果挂了CD...状态有点差...思维不太活跃 沉迷暴力不能自拔 D 给出n个长方体 ...
- Codeforces Round #378 (Div. 2) D - Kostya the Sculptor
Kostya the Sculptor 这次cf打的又是心累啊,果然我太菜,真的该认真学习,不要随便的浪费时间啦 [题目链接]Kostya the Sculptor &题意: 给你n个长方体, ...
随机推荐
- Hibernate,JPA注解@DynamicInsert和@DynamicUpdate,Hibernate如何插入sysdate
@DynamicInsert属性:设置为true,设置为true,表示insert对象的时候,生成动态的insert语句,如果这个字段的值是null就不会加入到insert语句当中.默认false. ...
- 《More Effective C++》
new operator申请内存,并调用构造函数: 而operator new只申请内存: new operator会调用operator new来申请内存:operator new可以重写或重载: ...
- Houdini Krakatoa Render Plugin
HDK真实个混蛋,都懒得写个解释.凭着函数英文意思猜测.. plugin sample video: 在极其残忍的开发环境,"Particle Voxel Render" 产生了( ...
- js 对象toString()方法
({}+{}).length == 30; ({}).toString() '[object Object]' 当对象需要调用toString()方法时会被自动调用.
- Android Studio解决unspecified on project app resolves to an APK archive which is not supported
出现该问题unspecified on project app resolves to an APK archive which is not supported as a compilation d ...
- cocos2dx lua bug之module 'lsqlite3' not found
05-27 15:41:01.360: D/cocos2d-x debug info(7261): [LUA-print] -------------------------------------- ...
- D类 E类地址
D类地址不分网络地址和主机地址,它的第1个字节的前四位固定为1110.⑵ D类地址范围:224.0.0.0到239.255.255.255D类地址用于多点播送.D类IP地址第一个字节以“lll0”开始 ...
- SlickGrid example 2: 按想要的风格展示列
先上效果图. 代码: <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http:// ...
- c# Beginlnvoke 委托
using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; usin ...
- poj3667 Hotel
此题不难却易出错,很能考察思维的严谨性. 指定ll为区间内左端顶格数的连续可利用房间,rr为右端顶格数的数值,mm为区间内最长的连续可利用房间数. 在查询的时候,由于要返回最靠左的区间左端点,使得在该 ...