pre[i]第i位数往前走多少位碰到和它同样的数
dp[i]表示长度为i的子串,dp[i]能够由dp[i-1]加上从i到n的pre[i]>i-1的数减去最后一段长度为i-1的断中的不同的数得到....
爆int+有点卡内存....
Substrings
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2300 Accepted Submission(s): 716 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, 0<w<=n<=10 6, 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
Source
/* ***********************************************Author :CKbossCreated Time :2015年08月17日 星期一 22时06分06秒File Name :HDOJ4455.cpp************************************************ */#include#include #include #include #include #include #include #include #include #include #include