HDU3450_Counting Sequences

题意:

让你从所给的序列中找到他的子序列,使他们相邻之间差距不超过d,问有多少个转移的子序列

这题第一眼大概就知道是状态转移,sum[i]表示以前i个中有多少个,那么sum[i+1]比sum[i]

多了一个以第i+1为结尾的子序列,那么只需要知道前面当中以x(x与第i+1距离不超过d)结尾的子序列个数和,那么这个时候在用dp[x]表示当前以x结尾有多少个子序列,但是数字太大不能直接记录,直接求和.

所以需要在状态转移时候运用到一些技巧,树状数组(也可以用线段树)和离散化;

先读入所以数字,然后排序编号,并用树状数组维护

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 

 

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

using namespace std;
const int maxn = 2e5+100;
const int MOD = 9901;
#define pr(x) cout << #x << " = " << x << " ";
#define prln(x) cout << #x << " = " << x <<endl;
#define ll long long

ll cnt;
map<ll,ll> ID;
ll a[maxn],b[maxn],dp[maxn],n;
void getID(ll x) {
    if(!ID.count(x)) {
        ID[x] = ++cnt;
    }
}
ll lowbit( ll x )
{
    return x & (-x);
}

void add(ll x,ll d)
{
    while( x <= n)
    {
        dp[x] = (dp[x] + d)%MOD;;
        x += lowbit(x);
    }
}

ll sum (ll x)
{
    int ans = 0;
    while(x)
    {
        ans = (ans + dp[x])%MOD;
        x -= lowbit(x);
    }
    return ans%MOD;
}

int main(){
#ifdef LOCAL
    freopen("C:\\Users\\User Soft\\Desktop\\in.txt","r",stdin);
  //freopen("C:\\Users\\User Soft\\Desktop\\out.txt","w",stdout);
 #endif
     ll d;
    while( cin >> n >> d) {
        ID.clear();cnt = 0;
        memset(dp,0,sizeof dp);
        for(int i = 0; i  < n; ++i) {
            scanf("%lld", &a[i]);
            b[i] = a[i];
        }
        sort(b,b+n);
        for(int i = 0; i < n; ++i) getID(b[i]);
        for(int i = 0; i < n; i++) {
            int l = lower_bound(b,b+n,a[i] - d) -b;
            int r = upper_bound(b,b+n,a[i] + d) - b-1;
            //if(r == l)
            l = ID[b[l]],r = ID[b[r]];
            ll num = (sum(r) - sum(l-1) +1)%MOD;
            add(ID[a[i]],num );
        }
        cout << (sum(cnt) + 20*MOD- n)%MOD << endl;

    }
    return 0;
}