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
",dp[w]);
        }
    }
    return 0;
}