cf581B Luxurious Houses
The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all the houses with larger numbers. In other words, a house is luxurious if the number of floors in it is strictly greater than in all the houses, which are located to the right from it. In this task it is assumed that the heights of floors in the houses are the same.
The new architect is interested in n questions, i-th of them is about the following: "how many floors should be added to the i-th house to make it luxurious?" (for all i from 1 to n, inclusive). You need to help him cope with this task.
Note that all these questions are independent from each other — the answer to the question for house i does not affect other answers (i.e., the floors to the houses are not actually added).
The first line of the input contains a single number n (1 ≤ n ≤ 105) — the number of houses in the capital of Berland.
The second line contains n space-separated positive integers hi (1 ≤ hi ≤ 109), where hi equals the number of floors in the i-th house.
Print n integers a1, a2, ..., an, where number ai is the number of floors that need to be added to the house number i to make it luxurious. If the house is already luxurious and nothing needs to be added to it, then ai should be equal to zero.
All houses are numbered from left to right, starting from one.
5
1 2 3 1 2
3 2 0 2 0
4
3 2 1 4
2 3 4 0
对于每个a[i]询问要加上多少值才能大于a[i]右边的最大的一个
这显然从后往前搜一遍完了
#include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<deque>
#include<queue>
#include<bitset>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
#define inf 0x7fffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,a[],ans[];
int main()
{
n=read();
for(int i=;i<=n;i++)a[i]=read();
int mx=a[n]+;
for(int i=n-;i>=;i--)
{
ans[i]=max(,mx-a[i]);
mx=max(mx,a[i]+);
}
for(int i=;i<=n;i++)printf("%d ",ans[i]);
}
cf581B
cf581B Luxurious Houses的更多相关文章
- CF581B Luxurious Houses 模拟
The capital of Berland has n multifloor buildings. The architect who built up the capital was very c ...
- CF581B Luxurious Houses 题解
Content 一条大街上有 \(n\) 个房子,第 \(i\) 个房子的楼层数量是 \(h_i\).如果一个房子的楼层数量大于位于其右侧的所有房屋,则房屋是豪华的.对于第 \(i\) 个房子,请求出 ...
- Codeforces Round #322 (Div. 2) B. Luxurious Houses 水题
B. Luxurious Houses Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/pr ...
- Luxurious Houses
The capital of Berland has n multifloor buildings. The architect who built up the capital was very c ...
- 【Henu ACM Round#19 B】 Luxurious Houses
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 从右往左维护最大值. 看到比最大值小(或等于)的话.就递增到比最大值大1就好. [代码] #include <bits/std ...
- Codeforces Round #322 (Div. 2)
水 A - Vasya the Hipster /************************************************ * Author :Running_Time * C ...
- CodeForces - 581B-Luxurious Houses
The capital of Berland has n multifloor buildings. The architect who built up the capital was very c ...
- UESTC 1817 Complete Building the Houses
Complete Building the Houses Time Limit: 2000MS Memory Limit: 65535KB 64bit IO Format: %lld & %l ...
- cdoj 04 Complete Building the Houses 暴力
Complete Building the Houses Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/# ...
随机推荐
- cocos2d-x精灵移动的方法
1.飞机发射子弹型 想要实现子弹的的飞行效果,又不想在每帧Update里计算位置,最好的方法调用CCMoveTo 例在发射子弹的时机,能够ccTouchBegan时: CCMoveTo* move = ...
- Java基础知识强化69:基本类型包装类之Character案例(统计字符串中大写小写以及数字的次数)
我们直接看案例如下: package cn.itcast_03; import java.util.Scanner; /* * 统计一个字符串中大写字母字符,小写字母字符,数字字符出现的次数.(不考虑 ...
- 【网络流#8】POJ 3469 Dual Core CPU 最小割【ISAP模板】 - 《挑战程序设计竞赛》例题
[题意]有n个程序,分别在两个内核中运行,程序i在内核A上运行代价为ai,在内核B上运行的代价为bi,现在有程序间数据交换,如果两个程序在同一核上运行,则不产生额外代价,在不同核上运行则产生Cij的额 ...
- 单线程与多线程的简单示例(以Windows服务发短信为示例)
单线程示例: public delegate void SM(); SM sm = new SM(() => { while (true) ...
- 一位学长的ACM总结(感触颇深)
发信人: fennec (fennec), 信区: Algorithm 标 题: acm 总结 by fennec 发信站: 吉林大学牡丹园站 (Wed Dec 8 16:27:55 2004) AC ...
- spring-data-solr官方学习文档介绍
spring-data-solr文档介绍如下: 通过http://www.springframework.org/schema/data/solr/spring-solr-1.0.xsd(spring ...
- asp.net+Sqlserver 通过存储过程读取数据
Sqlserver代码 创建存储过程如下: /*根据父id获取类别总数*/ IF EXISTS (SELECT name FROM sysobjects WHERE name = N'getsite ...
- asp.net 连接oracle,报错误“System.Data.OracleClient 需要 Oracle 客户端软件 8.1.7 或更高版本
1.http://www.oracle.com/technetwork/database/features/instant-client/index-097480.html 下载对用版本的Instan ...
- Eclipse 导入项目乱码问题(中文乱码)
1.编码不对 a.对某文件或某工程更改编码: 鼠标移到工程名或文件名,右键->Properties->Resource->Text file enCoding ->更改编码 ...
- errno.h 错误码描述.
描述:一般说的Linux源码的目录,默认是基于 /usr/include/ 的. 使用 char *strerror(int errnum); 函数打印错误代码的描述.我简单对比了一下,发现描述大体一 ...