CF div2 322 B

B. Luxurious Houses

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 h_i (1 ≤ h_i ≤ 10⁹), where h_i equals the number of floors in the i-th house.

Output

Print n integers a₁, a₂, ..., a_n, where number a_i 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 a_i should be equal to zero.

All houses are numbered from left to right, starting from one.

Sample test(s)

Input

5
1 2 3 1 2

Output

3 2 0 2 0 

Input

4
3 2 1 4

Output

2 3 4 0 

题目意思是对于第i个数,查询区间【i+1，n】的最大值，然后让当前的数跟区间最大的值比较，输出其增量。
裸的线段树。。。。。。。

 1 #include <cstring>
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 const int M = 1000005;
 9 
10 long long  res[M*4];
11 long long  a[M];
12 
13 void PushUp(int rt)
14 {
15     res[rt] = max(res[rt<<1],res[rt<<1|1]);
16 }
17 
18 void build(int l,int r,int rt)
19 {
20     if( l == r) {res[rt] = a[r];return;}
21     int mid = (l+r)>>1;
22     build(l,mid,rt<<1);
23     build(mid+1,r,rt<<1|1);
24     PushUp(rt);
25 }
26 
27 long long query(int L,int R,int l,int r,int rt)
28 {
29     if(L<=l&&r<=R){
30         return res[rt];
31     }
32 
33     int mid =  (l+r)>>1;
34     long long  ret = 0;
35     if(L<=mid) ret = max(ret,query(L,R,l,mid,rt<<1));
36     if(R>mid)  ret = max(ret,query(L,R,mid+1,r,rt<<1|1));
37     return ret;
38 }
39 
40 int main()
41 {
42     int n;
43     scanf("%d",&n);
44     for(int i=1;i<=n;i++){
45         scanf("%lld",&a[i]);
46     }
47 
48     build(1,n,1);
49     for(int i=1;i<n;i++){
50        long long ans = query(i+1,n,1,n,1)-a[i];
51         if(ans >= 0) printf("%lld ",ans+1);
52         //if(ans == 0) printf("%lld ",ans+1);
53         if(ans < 0) printf("0 ");
54     }
55     printf("0\n");
56 
57     return 0;
58 }

View Code

 

转载于:https://www.cnblogs.com/lmlyzxiao/p/4846116.html