HDU-4455 Substrings DP递推

传送门：http://acm.split.hdu.edu.cn/showproblem.php?pid=4455

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3426    Accepted Submission(s): 1050

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

 

思路：

i+1的答案是i的答案减去最后一段，加上前面各段增长1增加的情况数。统计与前面相同数字的距离大于等于i的即为增加的情况数。

#include <cstdio>#include <cstring>using namespace std;const int N=1000005;int a[N];int dist[N];///与前一个相同数字距离为i的个数int distsum[N];///距离大于等于i的个数long long dp[N];///用int会爆int ls[N];int last[N];int main(){    int n;    while (scanf("%d",&n)!=EOF)    {        if (n==0)            break;        memset(dist,0,sizeof(dist));        memset(last,0,sizeof(last));        memset(ls,0,sizeof(ls));        for (int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            dist[i-last[a[i]]]++;            last[a[i]]=i;        }        memset(last,0,sizeof(last));        last[a[n]]=1;        ls[n]=1;        for (int i=n-1;i>=1;i--)        {            if (last[a[i]])                ls[i]=ls[i+1];            else            {                ls[i]=ls[i+1]+1;                last[a[i]]=1;            }        }        memset(distsum,0,sizeof(distsum));        for (int i=n;i>=1;i--)            distsum[i]=distsum[i+1]+dist[i];        dp[1]=n;        for (int i=2;i<=n;i++)        {            dp[i]=dp[i-1]+distsum[i]-ls[n+2-i];        }        int w;        scanf("%d",&w);        while (w--)        {            int x;            scanf("%d",&x);            printf("%lld\n",dp[x]);        }    }    return 0;}