HDU 4455 Substrings

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

 

 

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

 

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。。。

 

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1000010;
LL dp[maxn],delta;
int d[maxn],pre[maxn],len[maxn],n,q;
bool vis[maxn];
int main() {
    while(~scanf("%d",&n),n) {
        memset(len,0,sizeof len);
        memset(vis,false,sizeof vis);
        memset(pre,-1,sizeof pre);
        for(int i = 0; i < n; ++i) {
            scanf("%d",d+i);
            len[i - pre[d[i]]]++;
            pre[d[i]] = i;
        }
        for(int i = n-1; i >= 0; --i) len[i] += len[i+1];
        dp[0] = 0;
        dp[1] = n;
        delta = 1;
        vis[d[n-1]] = true;
        for(int i = 2; i <= n; ++i) {
            dp[i] = dp[i-1] + len[i] - delta;
            if(!vis[d[n-i]]) {
                delta++;
                vis[d[n-i]] = true;
            }
        }
        scanf("%d\n",&q);
        while(q--) {
            scanf("%d",&n);
            printf("%I64d\n",dp[n]);
        }
    }
    return 0;
}