Codeforces Round #322 (Div. 2) B. Luxurious Houses 水题-白红宇

Codeforces Round #322 (Div. 2) B. Luxurious Houses 水题

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发布时间：2019-06-19

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B. Luxurious Houses

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/581/problem/B

Description

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).

Input

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.

Output

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.

Sample Input

1 2 3 1 2

Sample Output

3 2 0 2 0

HINT

题意

在一个一维坐标系上有n个house

每个house显得厉害，就是他的楼层要比比他编号大的楼层都要高

然后问你每栋楼至少得加高多少

题解：

扫一遍就好了，其实就是一个查询最大值的过程

代码：

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                    typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 1205000#define mod 1000000007#define eps 1e-9#define e exp(1.0)#define PI acos(-1)#define lowbit(x) (x)&(-x)const double EP = 1E-10 ;int Num;//const int inf=0x7fffffff;const ll inf=999999999;inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}//*************************************************************************************int Max[maxn];int a[maxn];int main(){ int n=read(); for(int i=1;i<=n;i++) a[i]=read(); int tmp = 0; for(int i=n;i>=1;i--) { Max[i]=tmp+1; tmp = max(tmp,a[i]); } for(int i=1;i<=n;i++) printf("%d ",Max[i]>a[i]?Max[i]-a[i]:0);}

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