HDOJ 题目4455 Substrings（DP）

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2592    Accepted Submission(s): 801

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

71 1 2 3 4 4 531230

 

Sample Output

71012

 

Source

2012 Asia Hangzhou Regional Contest

 

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题目大意：给个n，下边n个数表示这个序列，然后一个q，q个询问，每个询问一个w，问所有长度的w的连续的子序列中不同数的个数和

思路:http://blog.csdn.net/a601025382s/article/details/12283581

ac代码

153681062015-11-03 19:46:04Accepted44551404MS29128K1221 BC++XY_

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#define LL long longusing namespace std;int a[1000100],c[1000100],sum[1000100],num[1000100];int pre[1000100];LL dp[1000100];int main(){    int n;    while(scanf("%d",&n)!=EOF,n)    {        int i;        memset(pre,0,sizeof(pre));        memset(c,0,sizeof(c));        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        for(i=1;i<=n;i++)        {           // scanf("%d",&a[i]);            c[i-pre[a[i]]]++;            pre[a[i]]=i;        }        sum[n]=c[n];        for(i=n-1;i>=1;i--)            sum[i]=sum[i+1]+c[i];        num[1]=1;        memset(c,0,sizeof(c));        c[a[n]]++;        for(i=2;i<=n;i++)        {            if(c[a[n-i+1]]==0)            {                num[i]=num[i-1]+1;                c[a[n-i+1]]=1;            }            else                num[i]=num[i-1];        }        dp[1]=n;        for(i=2;i<=n;i++)            dp[i]=dp[i-1]-num[i-1]+sum[i];        int q;        scanf("%d",&q);        while(q--)        {            int w;            scanf("%d",&w);            printf("%lld\n",dp[w]);        }    }}