bestcoder#36 1002 哈希

bestcoder#36 1002 哈希

Gunner

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 726    Accepted Submission(s): 326

Problem Description

Long long ago, there is a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds and n trees. The i−th bird stands on the top of the i−th tree. The trees stand in straight line from left to the right. Every tree has its height. Jack stands on the left side of the left most tree. When Jack shots a bullet in height H to the right, the bird which stands in the tree with height H will falls.
Jack will shot many times, he wants to know how many birds fall during each shot.

a bullet can hit many birds, as long as they stand on the top of the tree with height of H .

 

 

Input

There are multiple test cases (about 5), every case gives n,m in the first line, n indicates there are n trees and n birds, m means Jack will shot m times.

In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which describes the height of the trees.

In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which describes the height of the Jack’s shots.

Please process to the end of file.

[Technical Specification]

1≤n,m≤1000000(106)

1≤h[i],q[i]≤1000000000(109)

All inputs are integers.

 

 

Output

For each q[i] , output an integer in a single line indicates the number of birds Jack shot down.

 

 

Sample Input

4 3

1 2 3 4

1

1

4

 

 

Sample Output

1

0

1

Hint

Huge input, fast IO is recommended.

 

#include<iostream>

#include<string>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<ctype.h>



using namespace std;



const int maxn=1000100;

const int HM=1000100;



int n,m;

struct Hash

{

    int date;

    int cnt;

    Hash *next;

};

Hash H[maxn];



inline int read()

{

    char c=getchar();

    while(!isdigit(c)) c=getchar();

    int f=c-'0';

    while(isdigit(c=getchar())) f=f*10+c-'0';

    return f;

}



int h(int x)

{

    return x%HM;

}



void insert(Hash*H,int date)

{

    int key=h(date);

    Hash *pre=&H[key],*p=pre->next;

    while(p!=NULL){

        if(p->date==date){

            p->cnt++;

            return;

        }

        pre=p;

        p=p->next;

    }

    p=(Hash*)malloc(sizeof(Hash));

    pre->next=p;

    p->next=NULL;

    p->date=date;

    p->cnt=1;

}



void find_del_output(Hash*H,int date)

{

    int key=h(date);

    Hash *pre=&H[key],*p=pre->next;

    while(p!=NULL){

        if(p->date==date){

            printf("%d\n",p->cnt);

            pre->next=p->next;///删掉结点

            free(p);

            return;

        }

        pre=p;

        p=p->next;

    }

    puts("0");

}



int main()

{

    while(cin>>n>>m){

        memset(H,0,sizeof(H));

        while(n--){

            int h=read();

            insert(H,h);

        }

        while(m--){

            int q=read();

            find_del_output(H,q);

        }

    }

    return 0;

}

View Code