hdu 4455 Substrings(找规律&DP)

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1161    Accepted Submission(s): 351

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

2012 Asia Hangzhou Regional Contest

 

题意：

给你一个数组{a1,a2,a3........an}。然后定义了一个询问。给你一个w。要你求出。所有a[i+1],a[i+2]......a[i+w]。中不同元素个数的和。i+w<=n。

思路：

dp[i]表示w为i时的答案。我们考虑dp[i+1]即长度增加一个后的情况。

对于n=3时。dp[3]已知

[1 1 2 ]3 4 4 5

对于n=4时。

[1 1 2 3] 4 4 5

相当于在以前三元集中加入一个数。所以如果该元素出现过dp[4]的值和dp[3]值一样。

如果不一样就要在dp[3]的基础上减一个。但是对于数组的最后三个数已经不能算作dp[4]的值。但dp[3]把它们

算作在内。所以要把他们减出来。我们对于数组中的每一个数。维护一个数组pre[v]表示数值v上次出现的位置。那么i-pre[v]即两数的不重复区间长度。

哎。。。比赛时居然为一个__int64卡了半天。还是不专业呀。都没有分析数据范围。。。还是最后十分钟做出来的。

不然应该有机会做做其它题目的。

详细见代码：

 

#include <iostream>

#include<string.h>

#include<stdio.h>



using namespace std;

const int maxn=1000010;

int a[maxn],pre[maxn],len[maxn];

__int64 dp[maxn],rest;//注意数据范围呀！！

bool vis[maxn];

int main()

{

    int n,q,w,i;



    while(scanf("%d",&n),n)

    {

        memset(pre,-1,sizeof pre);

        memset(len,0,sizeof len);

        memset(vis,0,sizeof vis);

        for(i=0;i<n;i++)

        {

            scanf("%d",a+i);

            len[i-pre[a[i]]]++;//统计各长度的数目

            pre[a[i]]=i;

        }

        for(i=n-1;i>=0;i--)

            len[i]+=len[i+1];//len[i]代表长度大于等于i的个数

        dp[0]=0;

        dp[1]=n;

        rest=1;

        vis[a[n-1]]=true;

        for(i=2;i<=n;i++)

        {

            dp[i]=dp[i-1]+len[i]-rest;//rest为长度不足i的部分

            if(!vis[a[n-i]])

            {

                rest++;

                vis[a[n-i]]=true;

            }

        }

        scanf("%d",&q);

        while(q--)

        {

            scanf("%d",&w);

            printf("%I64d\n",dp[w]);

        }

    }

    return 0;

}