hdu3450——Counting Sequences

Counting Sequences

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1872    Accepted Submission(s): 635

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

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

 

Recommend

线段树+dp，dp方程很简单,dp[i]表示以第i个元素结尾的完美序列的个数
dp[i] += dp[j] + 1(j < i)

先把数据离散化，然后按顺序插入到线段树上，维护区间和

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
const int N = 100010;
__int64 dp[N];
int xis[N];
int arr[N];
int cnt;

struct node
{
	int l, r;
	__int64 sum;
}tree[N << 2];

int BinSearch(int val)
{
	int l = 1, r = cnt, mid, ans;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (xis[mid] == val)
		{
			ans = mid;
			break;
		}
		else if (xis[mid] > val)
		{
			r = mid - 1;
		}
		else
		{
			l = mid + 1;
		}
	}
	return ans;
}

int BinSearch_left(int val)
{
	int l = 1;
	int r = cnt, mid, ans;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (xis[mid] >= val)
		{
			ans = mid;
			r = mid - 1;
		}
		else
		{
			l = mid + 1;
		}
	}
	return ans;
}

int BinSearch_right(int val)
{
	int l = 1;
	int r = cnt, mid, ans;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (xis[mid] <= val)
		{
			ans = mid;
			l = mid + 1;
		}
		else
		{
			r = mid - 1;
		}
	}
	return ans;
}

void build(int p, int l, int r)
{
	tree[p].l = l;
	tree[p].r = r;
	tree[p].sum = 0;
	if (l == r)
	{
		return;
	}
	int mid = (l + r) >> 1;
	build(p << 1, l, mid);
	build(p << 1 | 1, mid + 1, r);
}

void update(int p, int pos, __int64 val)
{
	if (tree[p].l == tree[p].r)
	{
		tree[p].sum += val;
		return;
	}
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (pos <= mid)
	{
		update(p << 1, pos, val);
	}
	else
	{
		update(p << 1 | 1, pos, val);
	}
	tree[p].sum = tree[p << 1].sum + tree[p << 1 | 1].sum;
	tree[p].sum %= 9901;
}

__int64 query(int p, int l, int r)
{
	if (tree[p].l >= l && tree[p].r <= r)
	{
		return tree[p].sum;
	}
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (r <= mid)
	{
		return query(p << 1, l, r);
	}
	else if (l > mid)
	{
		return query(p << 1 | 1, l, r);
	}
	else
	{
		return query(p << 1, l, mid) + query(p << 1 | 1, mid + 1, r);
	}
}

int main()
{
	int n, d;
	while(~scanf("%d%d", &n, &d))
	{
		memset (dp, 0, sizeof(dp));
		__int64 ans = 0;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &arr[i]);
			xis[i] = arr[i];
		}
		sort(xis + 1, xis + n + 1);
		cnt = unique(xis + 1, xis + n + 1) - xis - 1;
		build(1, 1, cnt);
		dp[1] = 0;
		update(1, BinSearch(arr[1]), dp[1] + 1);
		int l, r;
		for (int i = 2; i <= n; i++)
		{
			l = BinSearch_left(arr[i] - d);
			r = BinSearch_right(arr[i] + d);
			dp[i] = query(1, l, r);
			dp[i] %= 9901;
			int x = BinSearch(arr[i]);
			update(1, x, dp[i] + 1);
			ans += dp[i];
			ans %= 9901;
		}
		printf("%I64d\n", ans);
	}
	return 0;
}