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hdu2852——KiKi's K-Number

2014-08-26 14:24:00.0 hdu 树状数组  
导读:Problem DescriptionFor the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the contai...。。。
Problem Description
For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations.

Push: Push a given element e to container

Pop: Pop element of a given e from container

Query: Given two elements a and k, query the kth larger number which greater than a in container;

Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?
 

Input
Input some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values:
If p is 0, then there will be an integer e (0
If p is 1, then there will be an integer e (0
If p is 2, then there will be two integers a and k (0

Output
 

Sample Input
50 51 20 62 3 22 8 170 20 20 42 1 12 1 22 1 32 1 4
 

Sample Output
No Elment!6Not Find!224Not Find!
 

Source
 

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PS: 树状数组果然吊,我喜欢这种灵活的不能轻易掌握的东西,正如男人喜欢掌握不了的女人

#include#include#include#includeusing namespace std;int tree[100010];int num[100010];inline int lowbit(int x){ return x&(-x);}void add(int x,int val){ for(int i=x;i<=100010;i+=lowbit(i))  tree[i]+=val;}inline int sum(int x){ int ans=0; for(int i=x;i;i-=lowbit(i))  ans+=tree[i]; return ans;}int main(){ int m; while(~scanf("%d",&m)) {  int q,a,k,mid,l,r,e;  memset(tree,0,sizeof(tree));  memset(num,0,sizeof(num));  while(m--)  {   scanf("%d",&q);   if(q==0)   {    scanf("%d",&e);    add(e,1);    num[e]++;   }   else if(q==1)   {    scanf("%d",&e);    if(num[e]==0)     printf("No Elment!\n");    else    {     num[e]--;     add(e,-1);    }   }   else   {    scanf("%d%d",&a,&k);    l=a;    r=100010;    int ans=100010,low=sum(a),MID;    while(l<=r)    {     mid=(l+r)>>1;     MID=sum(mid);     if(MID-low>=k)     {      r=mid-1;      ans=min(ans,mid);     }     else      l=mid+1;    }    if(ans==100010)     printf("Not Find!\n");    else     printf("%d\n",ans);   }  } } return 0;}


(编辑: Guard_Mine)

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