Counting Sequences

Counting Sequences

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 337    Accepted Submission(s): 115

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

主要还是线段树了。。

#include<iostream>

#include<algorithm>

using namespace std;

#define MOD(x) (x)%9901

const int inf=0xfffffff;

const int N=100010;

struct seg

{

int l,r;

int sum;

};

seg s[N*3];

int dia[N],n,dis,old[N],cnt;

int dp[N];

int findpos(int num)

{

int mid,l=1,r=cnt,res;

while(l<=r)

{

mid=(l+r)>>1;

if(dia[mid]>=num)

{

res=mid;

r=mid-1;

}

else 

l=mid+1;

}

return res;

}

void maketree(int l,int r,int c)

{

s[c].l=l;

s[c].r=r;

s[c].sum=0;

if(l==r) return;

int mid=(l+r)>>1;

maketree(l,mid,c<<1);

maketree(mid+1,r,(c<<1)+1);

}

int find(int l,int r,int c)

{

int mid;

if(l<=s[c].l&&r>=s[c].r)

return MOD(s[c].sum);

else

{

mid=(s[c].l+s[c].r)>>1;

if(r<=mid)

return MOD(find(l,r,c<<1));

else if(l>mid)

return MOD(find(l,r,(c<<1)+1));

else

return MOD(find(l,mid,c<<1)+find(mid+1,r,(c<<1)+1));

}

}

void updata(int pos,int num,int c)

{

s[c].sum=MOD(s[c].sum+num);

if(s[c].l==s[c].r)

return ;

int mid = (s[c].l+s[c].r) >> 1;

if(pos<=mid)

updata(pos,num,c<<1);

else

updata(pos,num,(c<<1)+1);

}

int main()

{

int i,j;

while(scanf("%d%d",&n,&dis)!=EOF)

{

memset(dp,0,sizeof(dp));

for(i=1;i<=n;i++)

{

scanf("%d",&dia[i]);

old[i]=dia[i];

}

sort(dia+1,dia+1+n);

cnt=1;

for(i=2;i<=n;i++)

if(dia[i]!=dia[i-1])

dia[++cnt]=dia[i];

dia[++cnt]=inf;

int ll,rr,temp,nowpos,re,ans=0;

maketree(1,cnt,1);

for(i=1;i<=n;i++)

{

ll=findpos(old[i]-dis);

rr=findpos(old[i]+dis);

if(dia[ll]<old[i]-dis)

ll++;

if(dia[rr]>old[i]+dis)

rr--;

temp=find(ll,rr,1);

nowpos=findpos(old[i]);

if(dia[nowpos]!=old[i])

nowpos++;

ans=MOD(ans+temp);

dp[nowpos]=++temp;

updata(nowpos,temp,1);

}

printf("%d/n",ans);

}

return 0;

}