HDU4455

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2581    Accepted Submission(s): 798

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 a₁,a₂…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<=10⁶, 0<=Q<=10⁴, 0<= a₁,a₂…a_n <=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

2012 Asia Hangzhou Regional Contest

 

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
#define LL long long
#define N 1000000 + 10
int n, a[N], pre[N];
LL dp[N], cnt[N];

int main()
{
    while(~scanf("%d", &n) && n)
    {
        memset(pre, 0, (n + 2) * sizeof(int));
        memset(cnt, 0, (n + 2) * sizeof(int));

        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            cnt[i - pre[a[i]]]++;
            pre[a[i]] = i;
        }

        for(int i = n - 1; i >= 1; i--) cnt[i] += cnt[i + 1];

        memset(pre, 0, (n + 2) * sizeof(int));
        dp[1] = n;
        LL num = 1;
        pre[a[n]] = 1;
        for(int i = 2; i <= n; i++)
            {
                dp[i] = dp[i - 1] + cnt[i] - num;
                if(pre[a[n - i + 1]] == 0)
                {
                    num++; 
                    pre[a[n - i + 1]]++;
                }
            }
        int q, w;
        scanf("%d", &q);
        while(q--)
        {
            scanf("%d", &w);
            printf("%I64d\n", dp[w]);
        }
    }
    return 0;
}