hdu4455之树状数组+DP

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1571    Accepted Submission(s): 459

Problem Description

XXX has an array of length n. XXX wants to know that, for a given w, what is the sum of the distinct elements’ number in all substrings of length w. For example, the array is { 1 1 2 3 4 4 5 } When w = 3, there are five substrings of length 3. They are (1,1,2),(1,2,3),(2,3,4),(3,4,4),(4,4,5)
The distinct elements’ number of those five substrings are 2,3,3,2,2.
So the sum of the distinct elements’ number should be 2+3+3+2+2 = 12

 

Input

There are several test cases.
Each test case starts with a positive integer n, the array length. The next line consists of n integers a ₁,a ₂…a _n, representing the elements of the array.
Then there is a line with an integer Q, the number of queries. At last Q lines follow, each contains one integer w, the substring length of query. The input data ends with n = 0 For all cases, 06, 0<=Q<=10 ⁴, 0<= a ₁,a ₂…a _n <=10 ⁶

 

Output

For each test case, your program should output exactly Q lines, the sum of the distinct number in all substrings of length w for each query.

 

Sample Input

7 1 1 2 3 4 4 5 3 1 2 3 0

 

Sample Output

7 10 12

/*分析：
dp[i]表示区间长度为i时的个数
则从区间长度i-1到i减少了最后一个i-1长度的区间
记last[i]表示最后一个区间长度为i的不同数的个数
所以dp[i]=dp[i-1]-last[i-1]
同时从区间i-1长度到i增加了n-i+1个数
增加的数为a[i],a[i+1],a[i+2]...a[n]
则增加的这些数表示区间长度为i且以a[i...n]结尾
有多少是不能增加的呢？
比如a[1]~a[i-1]存在a[i]，以a[i]结尾的区间长度为i的区间就不用增加a[i]这个数 
在这里预处理时以a[i]结尾的区间长度为l~r不能增加
则l=i-pos[a[i]]+1;//pos[a[i]]表示i位置前出现的a[i]
r=i-1+1+1
用树状数组能快速的累加
所以最终增加的是n-i+1-Query(i) 
*/
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