【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:294 2 3 1 5 2 7 6 931 2 2Output:6 4 4 3 3 2 2 1 01 1 0

 

Explanation
Example case 1. 
                                 
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, 
                                 
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; }