hdu 3450 Counting Sequences

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3450

题目大意:寻找某个整数列中的完美子序列(相邻之间差不超过d)个数.

题目思路:dp[x]=dp[i]+...+dp[j](a[]为整数列,a[i],a[j]的值和a[x]的差的觉得值不超过d),dp[]中存的为一当前位置结尾完美子序列的个数.),求dp[i]+...+dp[j]同样用线段树或树状数组,查找i,j的值可以把x-d和x+d一起存到进行离散化这样就不用写上下界函数了,当然数组要开3倍.

代码:

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define ull unsigned __int64
#define ll __int64
//#define ull unsigned long long
//#define ll long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define middle (l+r)>>1
#define MOD 9901
#define esp (1e-10)
const int INF=0x3F3F3F3F;
//const double pi=acos(-1.0);
const int N=300010;
int n,m;
ll a[N],sum[N],hash[N];

void mod(ll& x){x= x<MOD? x:x%MOD;}

int lowbit(int x){return x&(-x);}

void Add(int pos,ll c){
	while(pos<=m) mod(sum[pos]+=c),pos+=lowbit(pos);
}

ll Sum(int pos){
	ll r=0;
	while(pos>0) mod(r+=sum[pos]),pos-=lowbit(pos);
	return r;
}

int bs(ll key,int size,ll A[]){
	int l=0,r=size-1,mid;
	while(l<=r){
		mid=middle;
		if(key>A[mid]) l=mid+1;
		else if(key<A[mid]) r=mid-1;
		else return mid;
	}
	return -1;
}

int main(){
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	int i,j,k,pos;
	ll ret,d,tmp;
	char op[2];
	//int T,cas;scanf("%d",&T);for(cas=1;cas<=T;cas++)
	while(~scanf("%d%I64d",&n,&d)){
		for(i=j=0;i<n;i++){
			scanf("%I64d",&a[i]);
			hash[j++]=a[i];
			hash[j++]=a[i]-d;
			hash[j++]=a[i]+d;
		}
		sort(hash,hash+j);
		for(i=m=1;i<j;i++) if(hash[i]!=hash[i-1]) hash[m++]=hash[i];
		memset(sum,0,sizeof(sum));
		for(i=0,ret=0;i<n;i++){
			tmp=Sum(bs(a[i]+d,m,hash)+1)+MOD-Sum(bs(a[i]-d,m,hash));
			mod(ret+=tmp);
			Add(bs(a[i],m,hash)+1,tmp+1);
		}
		printf("%I64d\n",ret);
	}
	return 0;
}