HDU3450-Counting Sequences

Counting Sequences

                                                                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
                                                                                                     Total Submission(s): 2362    Accepted Submission(s): 826

Problem Description

For a set of sequences of integers｛a1，a2，a3，...an｝, we define a sequence｛ai1,ai2,ai3...aik｝in which 1<=i1<i2<i3<...<ik<=n, as the sub-sequence of ｛a1，a2，a3，...an｝. It is quite obvious that a sequence with the length n has 2^n sub-sequences. And for a sub-sequence｛ai1,ai2,ai3...aik｝，if it matches the following qualities: k >= 2, and the neighboring 2 elements have the difference not larger than d, it will be defined as a Perfect Sub-sequence. Now given an integer sequence, calculate the number of its perfect sub-sequence.

 

Input

Multiple test cases The first line will contain 2 integers n, d（2<=n<=100000，1<=d=<=10000000） The second line n integers, representing the suquence

 

Output

The number of Perfect Sub-sequences mod 9901

 

Sample Input

4 21 3 7 5

 

Sample Output

4

 

Source

2010 ACM-ICPC Multi-University Training Contest（2）——Host by BUPT

题意：给出n个数，统计满足相邻两个数之差不超过d的子序列个数。

解题思路：树状数组

    对于每项a[i]，则以[a[i]-d,a[i]+d]范围内的数结尾的子串，加上a[i]，就可以成为符合要求的另一个子串，用树状数组维护。

    树状数组sum(x)记录的是以1~x结尾的符合要求的子串个数，则sum(y)-sum(z-1)+1,就是[z,y]范围子串加上a[i]后新增子串，还要加上a[i]本身成串。 

#include<cmath>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int mod = 9901;const int N = 4e5 + 10;int n,d,a[N],b[N],f[N];int low(int x){    return x&-x;}int sum(int x){    int sum=0;    while(x>0)    {        sum+=f[x];        sum%=mod;        x-=low(x);    }    return sum;}void add(int x,int v){    while(x<=n)    {        f[x]+=v;        f[x]%=mod;        x+=low(x);    }}int main(){    while(~scanf("%d %d",&n,&d))    {        memset(f,0,sizeof f);        for(int i=1; i<=n; i++)            scanf("%d",&a[i]),b[i]=a[i];        sort(b+1,b+1+n);        int ans=0;        for(int i=1; i<=n; i++)        {            int k=lower_bound(b+1,b+1+n,a[i])-b;            int l=lower_bound(b+1,b+1+n,a[i]-d)-b;            int r=upper_bound(b+1,b+1+n,a[i]+d)-b-1;            int x=(sum(r)-sum(l-1));            x=(x+mod)%mod;            ans=(ans+x)%mod;            add(k,(x+1)%mod);        }        printf("%d\n",ans);    }    return 0;}