hdu 4455 Substrings （DP 预处理思路）

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1727    Accepted Submission(s): 518

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

 

1、
非常明显，长度为1的答案为dp[1]=n;

2、
长度为2的为dp[2]=dp[1]+x-y=7+4-1=10；

X为添加的一个数和前边不同的个数，{1,1}，{1,2}，{2,3}，{3,4}，{4,4}，{4,5} 为4；

Y为去掉的不足2的区间有几个不同数字，长度为1的最后一个区间{5},须要舍去。为1；

3、

长度为3的为dp[3]=dp[2]+x-y=10+4-2=12。

X为添加的一个数和前边不同的个数，{1,1,2}。{1,2,3}，{2,3,4}。{3,4,4}，{4,4,5}为4；

Y为去掉的不足3的区间有几个不同数字。长度为2的最后一个区间{4,5},须要舍去，为2；

所以，我们须要得到当前数字和它上次出现的位置差的大小。详细实现看代码。。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 1000005
#define LL __int64
const int inf=0x1f1f1f1f;
int a[N],len[N],pre[N],vis[N],f[N];
LL dp[N];
int main()
{
    int i,n,m;
    while(scanf("%d",&n),n)
    {
        memset(pre,-1,sizeof(pre)); //记录一个值上次出现的位置
        memset(len,0,sizeof(len)); //len[i]：有几个间隔为i的数
        memset(dp,0,sizeof(dp));   //记录终于答案
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            len[i-pre[a[i]]]++;
            pre[a[i]]=i;
        }
        for(i=n-1;i>=0;i--)
            len[i]+=len[i+1];
        memset(f,0,sizeof(f));   //f[i]从后往前记录后i个数有几个不同值
        memset(vis,0,sizeof(vis));
        for(i=1;i<n;i++)
        {
            f[i]=f[i-1];
            if(!vis[a[n-i]])
            {
                vis[a[n-i]]=1;
                f[i]++;
            }
        }
        dp[1]=n;
        for(i=2;i<=n;i++)
            dp[i]=dp[i-1]+len[i]-f[i-1];
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d",&i);
            printf("%I64d\n",dp[i]);
        }

    }
    return 0;
}