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hdu 2852 KiKi's K-Number

2024-07-15 14:26:40   227人阅读

L - KiKi‘s K-Number
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit Status

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 <e <100000), means press element e into Container.

If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container  

If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.
 

Output

For each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".
 

Sample Input

5
0 5
1 2
0 6
2 3 2
2 8 1
7
0 2
0 2
0 4
2 1 1
2 1 2
2 1 3
2 1 4
 

Sample Output

No Elment!
6
Not Find!
2
2
4
Not Find!
询问的时候，（a,k）
对a+1,10 0000区间进行二分
getnum(x)得到是大于等于x数的个数
所以我们要找到getnum(a+1)-getnum(mid)==k
并且使得mid最小，最小的mid-1就是比a大的第k个数
好题！！！
 1 #include<iostream> 2 #include<string> 3 #include<cstdio> 4 #include<vector> 5 #include<queue> 6 #include<stack> 7 #include<algorithm> 8 #include<cstring> 9 #include<stdlib.h>10 #include<string>11 #include<cmath>12 using namespace std;13 #define pb push_back14 int p[101010],m,mm;15 void update(int pos,int num){16     while(pos>0){17         p[pos]+=num;18         pos-=pos&(-pos);19     }20 }21 int getnum(int pos){22     int sum=0;23     while(pos<=100000){24         sum+=p[pos];25         pos+=pos&(-pos);26     }27     return sum;28 }29 void Find(int pos,int k){30     int l=pos+1,r=100000,mid,mm=10000000,be=getnum(pos+1);31     while(l<=r){32         mid=(l+r)/2;33         int tmp=be-getnum(mid);34         if(tmp<k) l=mid+1;35         else{36             mm=min(mm,mid-1);37             r=mid-1;38         }39     }40     if(mm!=10000000) printf("%d\n",mm);41     else printf("Not Find!\n");42 43 }44 int main(){45      #ifndef ONLINE_JUDGE46             freopen("input.txt","r" ,stdin);47         #endif // ONLINE_JUDGE48     while(cin>>m){49         memset(p,0,sizeof(p));50         while(m--){51             int a,b,c;52             scanf("%d",&a);53             if(a==0){54                 scanf("%d",&b);55                 update(b,1);56             }57             if(a==1){58                 scanf("%d",&b);59                 if(getnum(b)-getnum(b+1)>0) update(b,-1);60                 else printf("No Elment!\n");61             }62             if(a==2){63                 scanf("%d%d",&b,&c);64                 Find(b,c);65             }66        }67     }68 }

 


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