Substrings
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2340 Accepted Submission(s): 731
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
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 a1,a2…an, 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<=106, 0<=Q<=104, 0<= a1,a2…an <=106
Each test case starts with a positive integer n, the array length. The next line consists of n integers a1,a2…an, 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<=106, 0<=Q<=104, 0<= a1,a2…an <=106
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
解题：其实训练的时候基本都想到了，卡在如何把大的区间个数累加到小区间的个数上了。。。哎。。。
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 1000010; 5 LL dp[maxn],delta; 6 int d[maxn],pre[maxn],len[maxn],n,q; 7 bool vis[maxn]; 8 int main() { 9 while(~scanf("%d",&n),n) { 10 memset(len,0,sizeof len); 11 memset(vis,false,sizeof vis); 12 memset(pre,-1,sizeof pre); 13 for(int i = 0; i < n; ++i) { 14 scanf("%d",d+i); 15 len[i - pre[d[i]]]++; 16 pre[d[i]] = i; 17 } 18 for(int i = n-1; i >= 0; --i) len[i] += len[i+1]; 19 dp[0] = 0; 20 dp[1] = n; 21 delta = 1; 22 vis[d[n-1]] = true; 23 for(int i = 2; i <= n; ++i) { 24 dp[i] = dp[i-1] + len[i] - delta; 25 if(!vis[d[n-i]]) { 26 delta++; 27 vis[d[n-i]] = true; 28 } 29 } 30 scanf("%d ",&q); 31 while(q--) { 32 scanf("%d",&n); 33 printf("%I64d ",dp[n]); 34 } 35 } 36 return 0; 37 }