HDU 3450 (tree array + dp)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 3450

Counting Sequences Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/65536 K (Java/Others)
Total submission (s): 1815 accepted submission (s): 618


Problem descriptionfor 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.
 
Inputmultiple 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
Outputthe Number of perfect sub-sequences mod 9901
 
Sample Input

4 21 3 7 5

 
Sample output

4

 
Source2010 ACM-ICPC multi-university training Contest (2) -- host by BUPT
Recommendwe have carefully selected several similar problems for you: 3030 1542 1255 2688 3449 question: a set has n numbers, and then you need to find all the perfect sub-sequences to modulo 9901 ~ Idea: it is easy to think of DP. We can use DP [I] to represent the number of perfect subsequences ending with the number of I, so sum (the total number of perfect subsequences) = DP [2] + dp [3] + dp [4] + ..... DP [N]; then what are the conditions for satisfying the perfect subsequence? | X-A [I] | <= d then there is a [I]-D <= x <= A [I] + D; then, use the tree array for Dynamic Maintenance ~~~

#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <cstdio>#include <algorithm>#include <cmath>const int N=1e5+100;const int mod=9901;using namespace std;struct node{ int v,id;}a[N];bool cmp(node a,node b){ return a.v<b.v;}int n,d,c[N],b[N];int lowbit(int x){  return x&(-x);}void update(int x,int d){  while(x<=n)  {   c[x]+=d;   if(c[x]>mod)c[x]%=mod;   x+=lowbit(x);  }}int getsum(int x){  int ans=0;  while(x>0)  {   ans+=c[x];   if(ans>mod)ans%=mod;   x-=lowbit(x);  }  return ans;}int find(int x){  if(x>=a[n].v)return n;  if(x<a[1].v)return 0;  int l=1,r=n,ret=0;  while(l<=r)  {   int mid=(l+r)/2;   if(x>=a[mid].v)    {      ret=mid;      l=mid+1;    }   else    r=mid-1;  }  return ret;}int main(){        while(scanf("%d%d",&n,&d)!=EOF)        {          memset(c,0,sizeof(c));          memset(b,0,sizeof(b));          for(int i=1;i<=n;i++)          {            scanf("%d",&a[i].v);            a[i].id=i;          }          sort(a+1,a+1+n,cmp);          for(int i=1;i<=n;i++)b[a[i].id]=i;          int sum=0;          for(int i=1;i<=n;i++)          {            int p=find(a[b[i]].v+d);            int q=find(a[b[i]].v-d-1);            int temp=getsum(p)-getsum(q);                temp=(temp+mod)%mod;                sum=(sum+temp)%mod;                update(b[i],temp+1);          }          printf("%d\n",sum);        }        return 0;}

HDU 3450 (tree array + dp)