hdu3450 Counting Sequences（dp+离散化+树状数组优化）

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 2
1 3 7 5

Sample Output
4

大致题意：给你n个数，让你找出找出长度大于2，且相邻两个数的差值不大于d的子序列的个数

思路：和hdu2227那题差不多，也是先离散化，然后用树状数组来优化dp，最后取模的时候要注意下。

代码如下

#include<cstring> #include<cstdio> #include<iostream>   #include <algorithm>  #define ll long long int   using namespace std;const int N=1e5+5; const int mod=9901;int dp[N];int n,d;int a[N],b[N],HASH[N];int tot;int lowbit(int x){    return x&-x;}int sum(int x){    int s=0;    while(x>0)    {        s=(s+dp[x])%mod;        x=x-lowbit(x);    }    return s;}void add(int x,int date){    while(x<=n)    {        dp[x]=(dp[x]+date)%mod;        x=x+lowbit(x);    }}int findR(int v){    int i=1,j=tot+1,mid;    while(i<j)    {        mid=(i+j)/2;        if(HASH[mid]<=v)                i=mid+1;        else             j=mid;    }    return i-1;}int findL(int v){    int i=1,j=tot+1,mid;    while(i<j)    {        mid=(i+j)/2;        if(HASH[mid]<v)              i=mid+1;        else             j=mid;    }    return i-1;}int main()  {    while(~scanf("%d%d",&n,&d))    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b[i]=a[i];        }        sort(b+1,b+n+1);        tot=1;        HASH[1]=b[1];        for(int i=2;i<=n;i++)            if( b[i]!=b[i-1] )                HASH[++tot]=b[i];        for(int i=1;i<=n;i++)        {            int L=findL(a[i]-d),id=findL(a[i])+1,R=findR(a[i]+d);            int tmp=sum(R)-sum(L);            add(id,tmp+1);        }        printf("%d\n",((sum(tot)-n)%mod+mod)%mod);    }       return 0;  }