hdu 3450 Counting Sequences （线段树+二分）

Counting Sequences

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 2849    Accepted Submission(s): 996

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

 

Source

题意：找长度大于1的且相邻两个数字的差值不超过d的子序列。

思路：这些数很大，需要离散化处理，我们可以发现以[a[i]-d,a[i]+d]为结尾的子序列，再加上a[i]也是符合题意的子序列。我没们就用线段树按照输入顺序一个个加到树里面。

代码：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e5+10;
const int mod=9901;
int n,d;
int sum[N<<2];
int a[N],b[N];
void update(int k,int num,int l,int r,int x)
{
    if(l==r&&l==k)
    {
        sum[x]+=num;
        sum[x]%=mod;
        return;
    }
    int mid=(l+r)/2;
    if(k<=mid)update(k,num,l,mid,x*2);
    else update(k,num,mid+1,r,x*2+1);
    sum[x]=(sum[x*2]+sum[x*2+1])%mod;
}
int query(int l,int r,int L,int R,int x)
{
    if(l<=L&&r>=R)return sum[x];
    int mid=(L+R)/2;
    int ans=0;
    if(l<=mid)ans+=query(l,r,L,mid,x*2);
    if(r>mid)ans+=query(l,r,mid+1,R,x*2+1);
    return ans%mod;
}
int main()
{
    while(~scanf("%d %d",&n,&d))
    {
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        sort(b+1,b+n+1);
         int cnt = unique(b+1, b+1+n)-b-1;
         int ans=0;
         for(int i=1;i<=n;i++)
         {
             int l=lower_bound(b+1,b+1+cnt,a[i]-d)-b;
             int r=upper_bound(b+1,b+1+cnt,a[i]+d)-b-1;
             int pos=lower_bound(b+1,b+1+cnt,a[i])-b;
             int ans1=query(l,r,1,cnt,1);
             ans=(ans+ans1)%mod;
             update(pos,ans1+1,1,cnt,1);

         }
         printf("%d\n",ans);
    }
    return 0;
}