【codechef】Towers(比较难想到方法的模拟题)

There are N towers in a city. All N towers are placed in a single row.Score of tower i is the the number of towers visible to its right side.

Tower j will be visible to Tower i if it satisfies the following condition,

  • If height of tower j is less than the height of tower i, then tower j will be visible to tower i unless it is blocked by some tower whose height is greater than or equal to the height of tower i
  • If height of tower j is greater than height of tower i, then Tower j will be visible if it not blocked by any tower whose height is greater than or equal to the height of Tower j

Find the score of each tower. 

Input

First line consists of the number T which denotes the number of test cases.

In each test case, the first line contains N, which denotes the number of towers.

Next line contains N integers which denote the height of towers.

Output

For each test case, print N integers which denote the score of each tower.

Constraints

1 <= T <= 100

1 <= N <= 10^5

1 <= height of each tower <= 10^9

Example

Input:
2
9
4 2 3 1 5 2 7 6 9
3
1 2 2

Output:
6 4 4 3 3 2 2 1 0
1 1 0

 

Explanation

Example case 1.

                                 【codechef】Towers(比较难想到方法的模拟题)_第1张图片

For tower 1, Towers 2,3,4,5 will be visible (as they are not blocked by any tower whose height is greater than tower 1). Tower 6 will be blocked by 6. Tower 7 will be visible as it is not blocked by any bigger tower in front of it. Tower 8 will be blocked by tower 7. Tower 9 will be visible because it's height is greater than the blocking tower (which is tower 5).

Example Case 2. For tower 1,

                                 【codechef】Towers(比较难想到方法的模拟题)_第2张图片

Tower 2 will be visible. Tower 3 will be blocked by tower 2.

http://www.codechef.com/KRKS2015/problems/K15C

当时做的时候超时,方法不太好。大神的代码方法很巧妙~

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<algorithm>  
using namespace std;   
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		long long int n;
		scanf("%lld",&n);
		long long int arr[n],b[n];
		long long int i,k,j;
		for(i=0;i<n;i++){
			scanf("%lld",&arr[i]);
			b[i]=0;
		}
		for (i=n-1;i>=0;i--){
			for(k=i+1;k<n;k++){
				if(arr[k]>arr[i]){
					b[i]=1+b[k]; //求后面比它高的塔的个数 
					break;
				}
			}
		}
		for(int i=0;i<n;++i)
			cout<<b[i]<<" ";
		cout<<endl;
		for (i=0;i<n;i++){
			long long temp=0;
			for (k=i+1;k<n;k++)	{
				if(arr[k]<=arr[i]) //找后面比它矮的 
					temp++;
				else{
					temp+=(b[k]+1);
					break;
				} //找到比它高的,加上,再break 
			}
			printf("%lld ",temp);
		}
	}
	return 0; 
} 


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