Counting Sequences

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 3012 Accepted Submission(s): 1067

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

Source

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

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题意：

给出一段序列，让你找出一段子序列（长度>=2）满足序列中每相邻的两个数之间的差 <=d，求出这样的序列的个数。

分析：

首先我们可以看出是一个DP，我们设dp[i]表示以a[i]为结尾的满足条件序列的个数。

状态转移方程：

dp[i]= $\sum$ dp[j] + 1 |a[i]-a[j]|<=d

接下来就是如何取出【a[i]-d,a[i]+d】区间里的所有dp，暴力肯定超时，树状数组去维护。

当然这题还需要离散化，数据量过大，我们在树状数组存相应的编号就行。

#include<stdio.h>
#include<string>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const int MAXN=200000+5;//最大元素个数
int d,n;//元素个数
ll c[MAXN];
int a[MAXN],b[MAXN];//c[i]==A[i]+A[i-1]+...+A[i-lowbit(i)+1]
//返回i的二进制最右边1的值
int lowbit(int i)
{
    return i&(-i);
}
//返回A[1]+...A[i]的和
ll sum(int x){
    ll sum = 0;
    while(x){
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}
//令A[i] += val
void add(int x, ll val){
	
    while(x <= n){
        c[x] += val;
        x += lowbit(x);
    }
}
const int mod=9901;
int main()
{
	while(scanf("%d%d",&n,&d)!=-1)
	{
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			b[i]=a[i];
		}
		sort(a+1,a+n+1);
		int len=unique(a+1,a+n+1)-a-1;
		memset(c,0,sizeof(c));
		int ans=0;
		for(int i=1;i<=n;i++)
		{
			int pos1=upper_bound(a+1,a+1+len,b[i]+d)-(a+1); ///右界
			
			int pos2=lower_bound(a+1,a+1+len,b[i]-d)-(a+1);///左界
			int pos=lower_bound(a+1,a+1+len,b[i])-a;       ///本身位置
			ll sum1=((sum(pos1)-sum(pos2))%mod+mod)%mod;   ///查询以a[i]为结尾的完美序列
			ans=(ans+sum1)%mod;
			add(pos,sum1+1);
		}
		cout<<ans<<endl;
	}
	return 0;
}

hdu3450Counting Sequences DP+树状数组

Counting Sequences

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